Word Break

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class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        n = len(s)
        dp = [False]*(n+1) # Initialization
        dp[0] = True # As an empty string can always be segmented

        for i in range(1, n + 1):
            for j in range(i):
                if dp[j] and s[j:i] in wordDict:
                    dp[i] = True
                    break # Once we found a match, no need to check further

        return dp[-1] # Return the final result

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class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
		    n = len(s)
		    dp = [False] * (n + 1)
		    dp[0] = True
		    wordSet = set(wordDict)

		    for i in range(1, n + 1):
		        for j in range(i):
		            if dp[j] and s[j:i] in wordSet:
		                dp[i] = True
		                break

		    return dp[n]

Identifying Problem Isomorphism

“Word Break” has an approximate isomorph: “Longest Substring Without Repeating Characters”.

“Word Break” requires you to determine whether a given string can be segmented into a sequence of one or more words in a dictionary of known words.

An easier problem with a similar concept is “Longest Substring Without Repeating Characters”. In this problem, you must identify the longest substring from a given string that has no repeating characters. This requires dividing the string into substrings, a simpler form of the segmentation required in “Word Break”.

A more complex problem that builds upon the principles of “Word Break” is “Word Break II”, which extends the problem by requiring you to return all possible sentences that can be formed by breaking the input string into valid words. This requires not just identifying whether a break is possible, but also how many different ways it can be done.

The reasoning is that all three problems involve dividing or segmenting a string, but they increase in complexity:

“Longest Substring Without Repeating Characters” requires you to identify substrings without repeating characters, a more basic form of segmentation.
“Word Break” builds on this concept by requiring segmentation into valid words from a dictionary.
“Word Break II” adds another layer of complexity by requiring all possible segmentations to be found.

They share core principles of string segmentation and manipulation.

10 Prerequisite LeetCode Problems

For this, the following is a good preparation:

Subsets - LeetCode 78: This problem helps you to understand how to generate all possible combinations of elements, which is a fundamental concept in this type of problem.
Palindrome Partitioning - LeetCode 131: In this problem, you will need to break down a string into segments, like in the “Word Break” problem, but with the additional condition that each segment must be a palindrome.
Partition Equal Subset Sum - LeetCode 416: It’s another problem where you have to determine if a given set can be partitioned in a certain way. It will be helpful in getting familiar with the concept of partitioning in the context of dynamic programming problems.
Implement Trie (Prefix Tree) - LeetCode 208: Understanding prefix trees (tries) can be useful for this problem, as you could use a trie to store the dictionary words for efficient lookup.
Longest Substring Without Repeating Characters - LeetCode 3: This problem deals with substrings and can help with understanding how to manipulate and work with substrings.
Concatenated Words - LeetCode 472: This problem requires you to determine if one word can be formed by concatenating other words in the list, which is similar to the “Word Break” problem.
Valid Anagram - LeetCode 242: This problem also involves comparing and matching characters in strings.
Uncommon Words from Two Sentences - LeetCode 884: In this problem, you’ll practice handling and comparing words from different collections, a skill that can also be useful for the “Word Break” problem.
Unique Substrings in Wraparound String - LeetCode 467: This problem involves the creation of substrings in a different context. It will improve your understanding of substring operations in different scenarios.
Find All Anagrams in a String - LeetCode 438: Understanding how to find all anagrams of a word in a string could help you to better understand how to approach the “Word Break” problem.

These cover substrings, partitioning a set into subsets, handling anagrams, and understanding prefix trees (tries).

Problem Classification

This problem is a dynamic programming problem that involves strings and dictionaries. It is mainly concerned with string segmentation and the utilization of a dictionary data structure. Here’s how the problem can be analyzed:

The problem belongs to the domain of string manipulation as we need to determine whether the input string ’s’ can be segmented into space-separated words, which exist in the dictionary provided. The main action here is segmenting and comparing substrings of ’s’ to elements in ‘wordDict’. Also, this problem involves finding an optimal solution and has overlapping subproblems (determining if a certain segment is a word in the dictionary can be reused in other segmentations), which makes it a Dynamic Programming problem.

Components:

String ’s’: This is the string that we want to check if it can be segmented into a sequence of one or more dictionary words.
List ‘wordDict’: This is the dictionary of strings that we will use to determine if the segmentation is possible.

This problem is a decision problem as the goal is to determine if ’s’ can be segmented into a sequence of words based on ‘wordDict’. It’s also an optimization problem because we aim to maximize the use of all characters in the string ’s’ by segmenting it into dictionary words. The presence of overlapping subproblems and the need for an optimal solution makes this problem fall under the Dynamic Programming paradigm.

The problem involves both string manipulation (segmenting and matching substrings) and a specific data structure (list), making it a good problem for learning how to solve problems that involve multiple domains and require the application of dynamic programming concepts.

Clarification Questions

When seeking to clarify a problem like this, consider asking the following questions:

Is the dictionary list (“wordDict”) always provided with unique words or should I handle duplicate words?
Can words in the dictionary be reused multiple times when segmenting the string?
If the string can be segmented in multiple ways, is any segmentation acceptable as long as all parts are valid words according to the dictionary?
What should be the output if the input string ’s’ is an empty string?
If the string cannot be segmented into dictionary words completely but partially, should we still return ‘false’?
Are all characters in the string and the dictionary words lowercase?
Is it safe to assume that both the input string and the dictionary words will only contain English alphabets and no other characters like digits, special characters, or whitespaces?
Is there a maximum or minimum length that the input string ’s’ or the dictionary words can be?
What should be returned if the ‘wordDict’ is empty?

Problem Analysis and Key Insights

From analyzing the problem statement, here are the key insights:

Segmentation: The problem is about dividing the given string ’s’ into segments where each segment is a valid word present in the given dictionary. This is clear from the problem statement “return true if s can be segmented into a space-separated sequence of one or more dictionary words.”
Reuse of Words: We are allowed to use the same dictionary word multiple times. This insight is gained from the note in the problem statement.
Existence of Solution: The problem is about finding whether a solution exists or not, rather than finding all possible solutions or the best solution. If the string ’s’ can be segmented into words that exist in the dictionary, we return true; otherwise, we return false.
Case sensitivity: Since the problem constraints mention that s and wordDict[i] consist of only lowercase English letters, we don’t need to worry about case sensitivity.
Uniqueness of Dictionary Words: All words in the dictionary are unique. This insight can simplify the way we approach the problem.
Constraints: The length of the string and the dictionary have upper bounds. Understanding this will help us determine if a solution with a certain time complexity is feasible or not.

These insights will be helpful in formulating a strategy to solve this problem.

Problem Boundary

The scope of the “Word Break” problem is to develop an algorithm that can determine whether a given string can be segmented into a sequence of one or more words from a given dictionary. The following factors outline the scope:

Input: The input will consist of a string ’s’ and a list of words ‘wordDict’. The length of ’s’ is between 1 and 300 characters and the dictionary contains between 1 and 1000 words, where each word has a length of 1 to 20 characters. The string ’s’ and the words in ‘wordDict’ only consist of lowercase English letters.
Output: The output is a boolean value. If the string ’s’ can be segmented into a sequence of words present in ‘wordDict’, the algorithm should return true, otherwise, it should return false.
Processing: The algorithm must be able to handle repeated usage of words from the dictionary in forming the string ’s’. Furthermore, the algorithm needs to perform under the constraints of time and space complexity as described in the problem statement.
Constraints: The problem must be solved while respecting the constraints given in the problem statement, including the lengths of the string ’s’ and the word dictionary ‘wordDict’, as well as the uniqueness of words in the dictionary.

The scope does not include finding all possible ways to segment the string or the most optimal way. It is simply about determining if it is possible to do so with the given dictionary.

Establishing the boundary of a problem involves defining its constraints and limitations. Here are the boundaries for the “Word Break” problem:

Input Constraints: The input string ’s’ has a length between 1 and 300. The word dictionary ‘wordDict’ has between 1 and 1000 words, each word being between 1 and 20 characters long. The string ’s’ and the words in ‘wordDict’ are composed of only lowercase English letters.
Output Constraints: The output is a boolean value, indicating whether or not the string ’s’ can be broken into a sequence of words from the dictionary ‘wordDict’.
Algorithmic Constraints: The algorithm needs to perform under the constraints of time and space complexity as described in the problem statement.
Problem Statement Constraints: Words from the dictionary may be reused multiple times in the segmentation. All the strings in ‘wordDict’ are unique.

The problem does not involve finding all possible segmentations or the most optimal segmentation. The boundary is to determine whether it is possible to segment the string ’s’ into a sequence of dictionary words. It also does not account for upper-case letters or special characters, nor does it involve changing the order of characters in the input string.

Distilling the Problem to Its Core Elements

Fundamental Concept: This problem is based on the concept of Dynamic Programming, specifically involving the subproblem optimization and overlapping subproblems. The main idea is to reuse the solutions of the smaller problems to solve larger ones.
Simple Description: Imagine you have a long string of letters and a dictionary of words. Your task is to find out if you can split this long string into valid words from the dictionary. You can use a word from the dictionary as many times as you want.
Core Problem: The core problem we are trying to solve is determining whether or not a given string can be segmented into a sequence of one or more dictionary words. We can simplify the problem statement as “Can we break up a given string into a sequence of valid words present in a provided dictionary?”
Key Components:
- Input: A string ’s’ and a dictionary of strings ‘wordDict’.
- Processing: Determining if ’s’ can be split into a sequence of one or more words present in ‘wordDict’.
- Output: A boolean value indicating whether the string can be segmented or not.
Minimal Set of Operations:
- Initialization: Begin with an array (dp) with length of string plus one, where the first value is true (as an empty string can always be segmented).
- Iteration: For every index in the array, if the current index is marked true (meaning the string until this index can be segmented), check for all words in the dictionary and mark the length of the word forward from this index as true if it matches with the dictionary word.
- Return: The last index in the array will indicate if the string can be segmented or not.

Visual Model of the Problem

Visualizing this problem can be done by imagining the input string and the dictionary words, and then trying to divide the string into segments where each segment is a word in the dictionary.

Let’s take an example:

String: “applepenapple”
Dictionary: [“apple”, “pen”]

We start from the leftmost part of the string, and try to match segments of the string with words in the dictionary:

applepenapple
^
|
Segment: "a" -> Not in the dictionary.

applepenapple
 ^^
 |
Segment: "ap" -> Not in the dictionary.

applepenapple
  ^^^
  |
Segment: "app" -> Not in the dictionary.

applepenapple
   ^^^^
   |
Segment: "appl" -> Not in the dictionary.

applepenapple
    ^^^^^
    |
Segment: "apple" -> In the dictionary!

Now we move to the next segment:

applepenapple
     ^^^^^^
     |
Segment: "p" -> Not in the dictionary.

applepenapple
      ^^^^^^
      |
Segment: "pe" -> Not in the dictionary.

applepenapple
       ^^^^^^
       |
Segment: "pen" -> In the dictionary!

Continue this process until the end of the string.

If we reach the end of the string and every segment has matched a word in the dictionary, then the string can be segmented into dictionary words. If there are parts of the string that don’t match any dictionary words, then the string cannot be segmented into dictionary words.

This can be considered a visualization of the problem statement. However, remember that an actual solution to this problem would probably not proceed exactly like this, and would likely use a more efficient approach, such as dynamic programming.

Problem Restatement

Given a string (let’s call it ’s’) and a list of words (we’ll call this ‘wordDict’), you are required to determine if the string can be broken down into individual segments where each segment is a word found in the provided list. It’s important to note that the same word from the list can be used multiple times while segmenting the string.

For example, if the string is “applepenapple”, and the list of words includes “apple” and “pen”, then the string can indeed be segmented into “apple pen apple”, which are all words found in the list.

However, if the string is “catsandog” and the word list includes “cats”, “dog”, “sand”, “and”, “cat”, then we cannot segment the entire string using only these words, so the output in this case would be false.

In terms of constraints, the length of the string will be between 1 and 300 characters. The word list will have between 1 and 1000 words, each word will be between 1 and 20 characters in length. All characters will be lowercase English letters, and all the words in the word list will be distinct.

Abstract Representation of the Problem

Consider the given string as a sequence of characters and the dictionary of words as a set of sequences. The task is to verify if the initial sequence (the string) can be segmented into one or more subsequences such that all these subsequences are present in the set of sequences (the dictionary of words).

This task can be represented in an even more abstract manner: Given a sequence and a set of possible sub-sequences, can the original sequence be completely and exactly covered by an arrangement of the possible sub-sequences? It’s worth noting that each sub-sequence can be used multiple times for this arrangement.

This abstract representation strips away all the specific real-world details, such as words and strings, and focuses solely on the structural aspects of the problem. This can often make it easier to see the underlying logic and algorithmic challenges of the problem.

Terminology

Here are a few terms that are important in understanding this problem:

String: A string is a sequence of characters. In programming, strings are a common data type that is used to represent text.
Segmentation: In this context, segmentation means splitting a string into multiple substrings or segments. Each segment must be a valid word that exists in the provided dictionary.
Dictionary (of words): A dictionary in this context refers to a set of valid words. In the problem, the string is to be segmented into words that exist within this dictionary.
Dynamic Programming: Dynamic Programming (often referred to as DP) is a problem-solving method used in computer programming for optimizing problems by breaking them down into simpler subproblems and storing the results of these subproblems to avoid redundant calculations.
Subsequence: A subsequence is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements.

In the context of this problem, a string s needs to be segmented into valid words (subsequences) that exist in the dictionary (set of valid subsequences). Dynamic Programming can be used to solve this problem efficiently by breaking down the problem into subproblems that check if each segment of s is a valid word and can be connected to the other valid words to form the original string.

Problem Simplification and Explanation

Let’s break down the problem into simpler terms:

The problem can be seen as a task where you are given a long, unbroken chain of letters and a dictionary of words. Your task is to break this chain into pieces such that each piece is a word from the dictionary.

Think of this task as similar to a jigsaw puzzle. The string “s” represents the complete puzzle, and the words in the dictionary “wordDict” are the individual puzzle pieces. Just like a jigsaw puzzle, the task is to determine whether it’s possible to assemble the complete puzzle (string “s”) using only the individual pieces (words from “wordDict”).

Moreover, just as some puzzle pieces might be very similar, or even identical (you can use the same word multiple times according to the problem statement), figuring out where they fit in the bigger picture can be tricky, which is why you need a methodical approach - such as Dynamic Programming - to solve it.

You start by checking each section of the string (“s”) to see if it matches a word from the dictionary (“wordDict”). You then build upon that, just like you would put together sections of a jigsaw puzzle, until you have checked the entire string.

The interaction between the key concepts - strings, segmentation, dictionary of words, and dynamic programming - work together to allow us to solve this problem in a step-by-step and efficient manner. Just like in a jigsaw puzzle, the image (or in this case, the string) can only be completed if the pieces (words) fit together in the correct way.

Constraints

Here are a few specific characteristics or conditions that can be exploited:

The dictionary words are unique: We don’t have to worry about duplicate words in the dictionary, which means we can use a data structure like a set to store the dictionary words for quick look-up.
The length of dictionary words are not very large: The maximum length of a word in the dictionary is 20. This makes it feasible to check all substrings of length up to 20 in the given string “s”.
The length of the string s is not very large: The maximum length is 300, which means we can use dynamic programming to solve the problem efficiently, as we can afford the memory usage of O(n), where n is the length of the string.
Strings consist only of lowercase English letters: We don’t need to handle cases of uppercase letters, digits, or special characters, which simplifies the problem.
Reuse of words: The same word in the dictionary can be reused multiple times in the segmentation, which means that we cannot simply count the frequency of words in the dictionary and in the string “s”, we have to check every possible segmentation.
Dynamic Programming Applicability: This problem exhibits both optimal substructure and overlapping subproblems properties - hallmarks of problems that can be solved using dynamic programming. The problem of whether we can segment the entire string “s” can be broken down into subproblems of whether we can segment the string “s” up to a certain index, which allows us to use dynamic programming to solve this problem efficiently.

Analyzing the constraints of a problem can give us crucial insights into the possible approaches for the problem and which data structures or algorithms might be most suitable. Here’s what we can glean from the constraints of this problem:

String and Dictionary Size: The string s and the word dictionary wordDict are relatively small (s length <= 300, wordDict length <= 1000, wordDict[i] length <= 20). This tells us that solutions involving time complexities of O(n^2), or even some with O(n^3), might be acceptable, as n is relatively small.
Uniqueness of Words in Dictionary: All the strings of wordDict are unique. This allows us to store these words in a HashSet for constant time lookups, avoiding duplicate handling.
Lowercase English Letters: s and wordDict[i] consist of only lowercase English letters, which simplifies the problem as we don’t need to handle case sensitivity or other types of characters.
Reuse of Words: The same word in the dictionary may be reused multiple times in the segmentation. This indicates that a simple one-to-one matching might not work and we’ll have to consider all possible combinations and orders of words.
Problem Nature: The constraints also hint at the nature of this problem. It’s a decision problem (returning true/false) and involves finding if a certain condition (segmentation of the string into dictionary words) can be fulfilled. This suggests that it could potentially be solved using dynamic programming, which can be effective for decision problems where the solution involves making a series of choices, and we want to find if a successful choice sequence exists.
Absence of Output Constraints: There are no constraints related to the number of possible segmentations or the order of the words in the segmentation. We are only required to find whether a valid segmentation exists. This means we don’t need to generate all possible segmentations or find a particular type of segmentation, which simplifies the problem.

Case Analysis

Let’s provide some additional examples that cover various edge and boundary conditions for the problem. This will help us in generating a deeper understanding of the problem.

Minimum Input Case: (Let’s call this “The Atom Case”)
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s = "a" wordDict = ["a"] # Output: True
This case tests the smallest possible input size. Here, the string s is equal to a word in wordDict, so it can obviously be segmented into a sequence of dictionary words.
Case with Multiple Word Choices: (Let’s call this “The Multiple Choice Case”)
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s = "applepen" wordDict = ["apple", "pen", "applepen"] # Output: True
This case demonstrates that there can be multiple ways to segment the string, but as long as there exists at least one valid segmentation, the output should be True.
Case with Repeatable Words: (Let’s call this “The Repeat Case”)
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s = "appleapple" wordDict = ["apple"] # Output: True
This case tests the condition that words in wordDict can be used multiple times for segmentation. Here, “apple” is used twice to segment the string s.
Case with Overlapping Words: (Let’s call this “The Overlap Case”)
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s = "racecar" wordDict = ["race", "ace", "car"] # Output: False
This case demonstrates that even if all the words in s exist in wordDict, a valid segmentation might not exist if the words overlap in s. Here, “race” and “ace” overlap in “racecar”, and so there’s no valid way to segment the string.
Case with No Words in the Dictionary: (Let’s call this “The Empty Case”)
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s = "apple" wordDict = [] # Output: False
This case tests the scenario where wordDict is empty. Since there are no words to segment the string, the output should be False.
Case with a Non-Segmentable String: (Let’s call this “The Non-Segmentable Case”)
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s = "applepenpineapple" wordDict = ["apple", "pen"] # Output: False
This case is interesting because even though each individual word in the string is present in the dictionary, the entire string cannot be segmented into a sequence of dictionary words due to the substring “pineapple” which is not in wordDict.

Remember, it’s not just about whether the words in the string exist in the dictionary, but also about how they are sequenced in the string. The string should be able to be segmented into a sequence of one or more dictionary words.

Visualizing the problem can be done by thinking of it as a sequence of decisions, where at each character we decide whether or not we “break” the string to start a new word. Here’s how you might visualize the provided cases:

The Atom Case:
For string “a” and dictionary [“a”], we have:
a -> ‘a’ is in the dictionary -> Return True
Here, the entire string is a word from the dictionary.
The Multiple Choice Case:
For string “applepen” and dictionary [“apple”, “pen”, “applepen”], we have:
applepen -> Break at ‘apple’ (from the dictionary) -> pen -> ‘pen’ is in the dictionary -> Return True
or
applepen -> ‘applepen’ is in the dictionary -> Return True
Here, the string can be segmented in multiple ways but we stop at the first valid segmentation we find.
The Repeat Case:
For string “appleapple” and dictionary [“apple”], we have:
appleapple -> Break at ‘apple’ -> apple -> ‘apple’ is in the dictionary -> Return True
Here, the same word is used twice for the segmentation.
The Overlap Case:
For string “racecar” and dictionary [“race”, “ace”, “car”], we have:
racecar -> Cannot break at ‘race’ or ‘ace’ or ‘car’ to get valid words for the entire string -> Return False
Even though the individual words are present in the string, the words overlap in a way that prevents the string from being fully segmented.
The Empty Case:
For string “apple” and dictionary [], we have:
apple -> Cannot find ‘apple’ or any part of it in the dictionary -> Return False
Here, since the dictionary is empty, we cannot segment the string.
The Non-Segmentable Case:
For string “applepenpineapple” and dictionary [“apple”, “pen”], we have:
applepenpineapple -> Cannot break at ‘apple’ or ‘pen’ to get valid words for the entire string -> Return False
Here, although ‘apple’ and ‘pen’ are in the dictionary, the entire string cannot be segmented due to the substring ‘pineapple’.

Analyzing these different cases, we can glean several key insights:

Atom Case Insight: Even single characters can form valid segments if they are present in the dictionary. This indicates that our solution should be flexible enough to handle both single character and multi-character words in the dictionary.
Multiple Choice Case Insight: There can be more than one valid way to segment a string given a dictionary. This suggests that our solution needs to be able to explore different possibilities rather than stopping at the first valid segmentation.
Repeat Case Insight: A word from the dictionary can be repeated multiple times in the string. Our solution should accommodate repeated words in its segmentation process.
Overlap Case Insight: The presence of dictionary words within the string does not always guarantee a valid segmentation. The order and overlap of words matter, which implies our solution should not only check for the presence of words but also consider their arrangement.
Empty Case Insight: An empty dictionary implies that no segmentation is possible. This is an edge case that our solution should handle explicitly.
Non-Segmentable Case Insight: Despite having dictionary words within the string, it might not be possible to segment the entire string. This suggests that our solution should be able to recognize when complete segmentation is impossible.

Overall, these insights hint at the need for a solution that can handle a variety of scenarios, from single characters to overlapping words, and that can explore different segmentation possibilities. This points towards a dynamic programming approach where we can build up solutions for smaller sub-problems to solve larger ones.

The term “Atom Case” is used to refer to the simplest, most basic case in a problem space. It is a term used to describe a case that cannot be broken down further or simplified any more.

In the context of the problem we’re discussing about segmenting a string into dictionary words, the atom case refers to the scenario where a single character string is also a valid word in the dictionary.

It’s like considering the “atoms” or most fundamental units in the problem space. By understanding how to solve these basic cases, we can often understand how to build up solutions for more complex scenarios.

The term “Atom Case” is not a standard computer science terminology. However, the concept behind it is commonly applied in many computer science problem-solving scenarios. For instance, when solving problems using divide-and-conquer strategies or recursive algorithms, we often identify and solve the simplest sub-problem (similar to an “Atom Case”) and then build up to more complex situations. However, the term “base case” is more typically used in this context. It’s also possible that “Atom Case” is used in certain circles or it could be a term introduced by a particular person or group for teaching or discussion purposes. It’s always a good idea to clarify such terminology if you come across it and it’s not familiar or widely recognized.

You can interpret “Atom Case” as referring to the smallest or simplest instance of a problem. It is similar to the “base case” in recursion, which represents the simplest form of the problem that can be solved directly without further recursion. Understanding the “Atom Case” or smallest instance can be very important in problem-solving, particularly for developing and understanding recursive algorithms or algorithms based on the principle of dynamic programming. It helps you understand how to build the solution for more complex instances based on simpler ones.

Identification of Applicable Theoretical Concepts

Here are a few concepts that could be relevant in the context of the “Word Break” problem:

Dynamic Programming: This is an algorithmic paradigm that solves a complex problem by breaking it down into simpler sub-problems in a recursive manner. In this case, we could break down the problem of finding whether a string can be segmented into a sequence of dictionary words by considering smaller and smaller substrings of the original string.
String Matching: The problem inherently involves the task of checking if certain substrings exist in a given dictionary. This calls for concepts related to string manipulation and matching.
Recursion: As we breakdown our string into smaller substrings, we can see that the problem possesses the optimal substructure property. This means an optimal solution to the problem can be constructed efficiently from optimal solutions of its sub-problems. This property indicates that we could potentially use a recursive approach.
Boolean Algebra: We are required to return a boolean value (true/false) as the result. The solution would involve logical operations which map well onto boolean algebra.
Graph Theory: If we visualize our string as a directed graph where each node represents a prefix of the string and there is an edge between two nodes if the substring between the indices of two nodes is a word in the dictionary, this problem can be solved using a graph traversal algorithm. This might not be the most straightforward approach, but it could offer a different perspective on the problem.

Remember, identifying these concepts doesn’t mean they all should be used. It simply provides a set of tools you might consider when creating your approach to the problem.

Simple Explanation

Imagine you’re playing with a set of toy blocks, each with a different word on it. Now, you’ve been given a bigger word, and your task is to see if you can build that bigger word using the blocks you have.

The only rule is you can’t break any of the word blocks apart. You have to use the whole word on the block or not at all. And if you have more than one of the same block, you can use them as many times as you like.

So, let’s say you’ve been given the word “applepenapple”, and your blocks have the words “apple” and “pen” on them. Can you build the word “applepenapple” with your blocks? Yes, you can! You use the “apple” block, then the “pen” block, then the “apple” block again.

But, what if you’ve been given the word “catsandog”, and your blocks have the words “cats”, “dog”, “sand”, “and”, “cat” on them? Can you build the word “catsandog”? In this case, no matter how you arrange your blocks, you won’t be able to build the word “catsandog”.

This is essentially what this coding problem is asking. You’re given a string (the “big word”) and a list of words (the “blocks”). The task is to find out if you can build the string using the words in the list.

Problem Breakdown and Solution Methodology

Identify Problem Components: The problem is essentially asking us to divide the input string into parts such that each part is a word present in our dictionary (wordDict). So we need to identify a way to keep track of where a division is possible.
Define Subproblems: We can make use of dynamic programming to solve this problem, as the solution to the larger problem depends on solutions to smaller subproblems. Specifically, we’ll define our subproblem as: Is it possible to divide the string from the beginning to index i into valid words in the dictionary?
Build Dynamic Programming Table: We initialize an array (let’s call it dp) of size equal to the length of the string plus one. dp[i] will represent whether the string till index i can be segmented into words from the dictionary. We start with dp[0] = True because an empty string can always be segmented.
Iterate and Populate the DP Table: Next, we iterate over the string, and for each index i, we iterate backward to all previous indexes j. If dp[j] is True (meaning that the string till index j can be segmented) and the substring from index j to i is in the dictionary, we set dp[i] to True.
Analyze Result: After populating the dp array, dp[n] (where n is the length of the string) will tell us if the whole string can be segmented into words from the dictionary.

Imagine it as trying to cross a river by hopping onto stones (representing words from the dictionary). You start on the shore (empty string, dp[0] = True) and each hop you take is equivalent to segmenting part of the string with a word from the dictionary. If you can hop your way to the other side (end of the string), then it’s possible to segment the whole string with the words from the dictionary.

For example, if the input string is “applepenapple” and the dictionary contains [“apple”, “pen”], the process would look something like this:

Initialize dp = [True, False, False, False, …, False] (size = length of string + 1)
At i = 5 (the end of the first “apple”), we find that dp[0] is True and “apple” is in the dictionary, so we set dp[5] = True
At i = 8 (the end of “pen”), we find that dp[5] is True and “pen” is in the dictionary, so we set dp[8] = True
At i = 13 (the end of the second “apple”), we find that dp[8] is True and “apple” is in the dictionary, so we set dp[13] = True
Finally, we return dp[n] (dp[13] in this case), which is True, indicating that “applepenapple” can be segmented as “apple pen apple”.

If the dictionary or the string changes, the content of the dp array will adjust accordingly, making this a dynamic solution. For instance, if “pen” is removed from the dictionary, dp[8] would remain False, and so would dp[13], and the function would correctly return False.

Inference of Problem-Solving Approach from the Problem Statement

Let’s go through the key terms or concepts of this problem:

String: In this problem, the input is a string that we need to segment. Strings are a fundamental concept in computer science and programming, and many problems involve some form of string manipulation or analysis.
Dictionary of words: This is a list of acceptable segments that the string can be broken down into. This is essentially a lookup table, which can be used to check the presence of a segment quickly.
Segmentation: This is the process of breaking the string into smaller pieces or segments. In this context, each segment must be a word present in the given dictionary.
Dynamic Programming (DP): This is a method for solving complex problems by breaking them down into simpler subproblems and solving these subproblems in a bottom-up manner (i.e., solve smaller subproblems first, then use these solutions to construct solutions to larger subproblems). In this problem, the DP approach is essential as the solution to the larger problem (can the entire string be segmented?) depends on solutions to smaller subproblems (can the string up to a certain point be segmented?).

As for visualizing these concepts, a DP table would be quite useful. For a given string s and dictionary wordDict, you can draw a table with rows representing each character in the string (including an empty string at the start), and columns representing the state of whether the string can be segmented up to that point. Initially, all entries in the DP table are False (except the first one, as an empty string can always be segmented), and you update the table entries based on the DP formulation we discussed earlier.

It helps to understand the state transitions in the problem and how solutions to smaller subproblems can be used to build up the solution to the larger problem. It also allows you to visualize the iterative process of populating the DP table.

The problem of segmenting a string into words from a dictionary can be solved using dynamic programming due to a couple of key reasons:

Subproblem Overlap: This problem has the property of overlapping subproblems, which is one of the main indicators of a problem that can be solved using dynamic programming. The subproblem here is checking if a prefix (substring from start to a given point) of the string can be segmented into words from the dictionary. This subproblem repeats multiple times for different prefixes of the original string.
Optimal Substructure: This problem also exhibits optimal substructure, another key property of dynamic programming problems. This means that the optimal solution for the entire problem can be composed from optimal solutions to its subproblems. If we know that a substring of the string can be segmented into words from the dictionary, then the answer for the whole string depends on whether the remainder of the string can be segmented.

Hence, a dynamic programming approach can be used to break down the original problem into smaller subproblems (checking if substrings of the original string can be segmented), solve each subproblem only once, store the results of each subproblem, and use these stored results to construct the solution to the larger original problem. This approach leads to an efficient and clean solution to the problem.

Simple Explanation of the Proof

Let’s break down the Word Break problem and the solution proof in a more digestible way.

Problem Statement: We’re given a string and a list of words (dictionary). Our goal is to determine whether or not the string can be segmented into a sequence of one or more words from the dictionary. If it can, we return true; if not, we return false. The same word from the dictionary can be reused multiple times in the segmentation.

Proof of the Algorithm: We’ll use Dynamic Programming (DP) to solve this problem. The DP approach is generally useful when a problem can be broken down into overlapping subproblems, and the solution to the main problem depends on the solutions to these subproblems.

Initialization: We’ll start by creating a boolean DP array of size n+1 (where n is the length of the string), initializing all elements to false. The i-th element in the DP array represents whether the substring from the start to the i-th character (inclusive) can be segmented into dictionary words. We set DP[0] to true, representing an empty string, which can always be segmented.
Subproblems & Recurrence Relation: Now, for each substring of the given string, we’ll iterate over it and check if it can be broken down into words from the dictionary. We’ll start from i = 1 and go up to n. For each i, we check every substring ending at i and starting at some j (where 0 <= j < i). If the substring from j to i is a word in the dictionary, and if the substring from the start to j can be segmented (i.e., DP[j] is true), then we can say that the substring from the start to i can be segmented (so we set DP[i] to true).
Final Solution: Once we’ve filled out the DP array, the last element (DP[n]) will tell us if the whole string can be segmented into dictionary words. If DP[n] is true, we return true; otherwise, we return false.

In essence, the DP solution to this problem uses the concept of breaking down the problem into subproblems (i.e., checking if each substring can be segmented) and uses the solutions of these subproblems to solve the original problem.

Let’s refine our approach to the Word Break problem.

Understand the problem: First, ensure you understand the problem statement thoroughly. We are given a non-empty string and a list of non-empty words (our dictionary). Our task is to determine if the string can be broken down, or segmented, into a sequence of one or more words from this dictionary.
Identify subproblems: Notice that the task of breaking down the string into words can be broken into smaller tasks - breaking down substrings of the original string into words. This hints towards a dynamic programming approach, as we are identifying overlapping subproblems.
Design a dynamic programming solution: In this case, we can use a boolean array dp where dp[i] represents whether the substring up to the i-th index can be broken down into words from the dictionary.
- Start with the base case: An empty string can always be segmented, so set dp[0] to true.
- Next, fill up the dp array iteratively. For each index i from 1 to n (the length of the input string):
  - Look at each substring ending at i and starting at some j where 0 <= j < i.
  - If the substring from j to i is in the dictionary and the substring from 0 to j can be segmented (dp[j] is true), then we can say that the substring from 0 to i can be segmented. So, we set dp[i] to true.
- This gives us a stepwise process to fill out the dp array.
Implement the solution: Once we’ve filled out the dp array, we return the last element of the array, dp[n], as our final result.
Identify independent parts and repeatable patterns:
- Each entry in the dp array is calculated independently based on previously calculated entries. This independence allows us to compute each entry in a simple loop structure.
- We observe a repeatable pattern in the way the dp array is filled - we always look at a substring ending at i and check if it can be segmented into dictionary words based on the dp value of the starting index of the substring.

This refined approach gives a clear, actionable plan for implementing the solution.

Solution Approach and Analysis

We can visualize our problem as trying to traverse a path from the beginning of the string to the end, stepping only on “stones” (substrings) that exist in our dictionary.

Prepare the Dynamic Programming Table: We’ll prepare a boolean array dp with size n + 1, where n is the length of the string s. The i-th entry in dp will tell us whether the substring s[0:i] can be segmented into dictionary words.
Initialize: We set dp[0] as true because an empty string can always be segmented.
Fill Up the Table: We then iterate from 1 to n for each character in the string, for each i:
- We take another loop iterating j backwards from i to 0, trying to find a point j such that the dp[j] is true and the substring s[j:i] is in the dictionary. This is like trying to find a stone to step on that’s within our reach (the substring s[j:i] is in our dictionary) and safe (the preceding substring can be segmented, i.e., dp[j] is true).
- If we find such a j, we set dp[i] to true and break the inner loop. This is like confirming that we’ve found a safe stone to step on, and there’s no need to check for other stones in this step.
Solution: The value of dp[n] will give us the answer. If dp[n] is true, that means we can segment the entire string into dictionary words. If dp[n] is false, that means we can’t.

For example, let’s take s = "leetcode" and wordDict = ["leet", "code"]. Our dp array will start like this: [true, false, false, false, false, false, false, false, false].

For i = 1, 2, 3, there are no words in the dictionary that match the substrings, so dp[1], dp[2], dp[3] remain false.
For i = 4, s[0:4] (“leet”) is in the dictionary and dp[0] is true, so we set dp[4] to true.
For i = 5, 6, 7, there are no words in the dictionary that match the substrings “leetc”, “leetco”, “leetcod”, so dp[5], dp[6], dp[7] remain false.
For i = 8, s[4:8] (“code”) is in the dictionary and dp[4] is true, so we set dp[8] to true.

So the final dp array is [true, false, false, false, true, false, false, false, true]. As dp[8] (which corresponds to the length of the string) is true, we can break “leetcode” into a sequence of dictionary words.

As we can see, the solution largely depends on the composition of the word dictionary. Different dictionaries would yield different results, and may require more or less computational effort depending on their size and the lengths of the words they contain. Additionally, longer strings would naturally require more computational time, as there would be more substrings to check against the dictionary.

Identify Invariant

In the context of this problem, an invariant would be a property that remains unchanged during the execution of the algorithm.

The invariant for this problem can be considered as the following:

“The dp[i] value (for all 0 <= i <= n) is always correct. That is, dp[i] correctly represents whether the substring s[0:i] can be segmented into words from the dictionary.”

This invariant is maintained throughout the execution of our dynamic programming solution. When we iterate over the string and fill up our dp table, we ensure that each dp[i] accurately represents whether the substring up to the i-th index can be segmented into dictionary words.

The maintenance of this invariant is crucial for the correctness of the algorithm, as it allows us to build up the solution iteratively. In each step, we rely on the truth of this invariant for the previously computed values in the dp table to correctly compute the current dp value.

The base case, dp[0] = true, signifies that an empty string can always be segmented, establishing the invariant at the start. For each subsequent i, we only update dp[i] to true if we can find a j (0 <= j < i) such that dp[j] is true and the substring from j to i is in the dictionary. This ensures that the invariant is maintained after each update, leading to the correct final answer in dp[n].

Identify Loop Invariant

In the context of the Word Break problem, a loop invariant is a condition that is initially true and remains true after each iteration of a loop. For the dynamic programming solution to this problem, the loop invariant can be formulated as follows:

“For any point i in the loop iterating over the string s, dp[j] is true for any j (0 <= j < i) if and only if the substring s[0:j] can be segmented into one or more dictionary words.”

This means that for any point in our loop, the previous elements in our dp array correctly represent whether those substrings can be broken down into words from the dictionary.

At the beginning, before the loop starts, this condition is trivially true as dp[0] is set to true, meaning an empty string can be segmented.

During each iteration, we only update dp[i] to true if there exists a j < i such that dp[j] is true and the substring s[j:i] is in the dictionary. This ensures the loop invariant is maintained at each iteration.

By the end of the loop, because of this invariant, we know that dp[n] (where n is the length of the string s) correctly represents whether the entire string can be segmented into dictionary words.

In this problem context, the invariant and loop invariant refer to the same condition. Specifically, they refer to a condition that remains true before, during, and after each iteration of the loop.

A loop invariant is a type of invariant used specifically in the context of loops. However, in general computer science, an invariant can also be a condition that’s true at various points in a program, not just in loops. For example, in a class in object-oriented programming, an invariant might be a condition that’s always true after the constructor runs or after any public method is called.

But in this problem, since we are using a loop in the dynamic programming approach to traverse the string, the terms ‘invariant’ and ’loop invariant’ would essentially be referring to the same thing: the property of the dp array that we maintain throughout the execution of the loop.

Identify Recursion Invariant

In the context of recursive solutions, an invariant is a condition that remains true before and after each recursive call. These invariants can guide the design of the recursive function and help verify its correctness.

In the problem at hand, a recursive solution can be designed using dynamic programming, where the invariant is the fundamental definition of our problem. The invariant during recursion in this problem is:

“The return value of our recursive function, let’s call it canSegment(i), is true if and only if the substring s[i:] can be segmented into a space-separated sequence of one or more dictionary words.”

This invariant holds at all levels of recursion. This is how we structure the base case (for an empty string, the function should return true as an empty sequence is a valid segmentation) and how we ensure the correctness of the recursive case (by iterating over all possible segmentations and checking if any of them is valid).

This invariant gives us confidence that the recursion is correctly structured and will yield the right result, as it is a property that holds before and after every recursive call.

While both invariants and invariants during recursion refer to conditions or properties that remain unchanged, their context of use might differ.

Invariant: This is a broad term that could refer to any condition or property that remains true throughout the execution of an algorithm or a program. It’s used to describe something that holds under all circumstances within a particular context. In this problem, one invariant might be the definition of the problem itself – that is, our task is to segment the given string into space-separated sequence of one or more dictionary words.
Invariant During Recursion: This specifically refers to a condition or property that remains true before and after each recursive call. It’s an instance of the broader concept of invariants, but specifically in the context of recursive function calls. In this problem, the invariant during recursion is the property that “The return value of our recursive function, canSegment(i), is true if and only if the substring s[i:] can be segmented into a space-separated sequence of one or more dictionary words.”

So while the terms “invariant” and “invariant during recursion” can be related and might even refer to similar conditions, they’re used in different contexts – one for the algorithm as a whole and the other specifically within recursive calls. They might be the same or different depending on the specific problem at hand.

Thought Process

Let’s breakdown the problem and build a Python solution for it.

Problem: Given a string s and a dictionary of strings wordDict, we need to determine if s can be segmented into a space-separated sequence of one or more dictionary words. It’s also mentioned that the same word in the dictionary may be reused multiple times in the segmentation.

Step 1: Understanding the problem The problem is basically asking to break the given string into words that exist in the provided dictionary. For instance, if the string is “leetcode” and the dictionary contains [“leet”, “code”], then we can break the string as “leet code”. So we need to find such breaks, if exist.

Step 2: Insights from problem statement The problem statement implies that we can use a dynamic programming approach, where we solve the problem by breaking it down into smaller subproblems. The subproblem here is to check whether each substring of the given string can be formed using the words in the dictionary.

Step 3: Approach We can use a boolean DP array, dp[], where dp[i] is True if the first i characters of the string can be segmented into one or more dictionary words. Therefore, dp[0] = True (since an empty string can be segmented). For all characters from 1 to n, we check each substring ending at position i, and if the substring exists in the dictionary and the remaining string (0, j - 1) can be segmented, then dp[i] = True. The final result would be dp[n].

Step 4: Coding the solution Here’s the Python code:

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class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        n = len(s)
        dp = [False]*(n+1) # Initialization
        dp[0] = True # As an empty string can always be segmented

        for i in range(1, n + 1):
            for j in range(i):
                if dp[j] and s[j:i] in wordDict:
                    dp[i] = True
                    break # Once we found a match, no need to check further

        return dp[-1] # Return the final result

Remember, the keys to identifying this as a dynamic programming problem were the words “segmented” and “one or more dictionary words”. It’s a sign that we need to break the problem down into smaller subproblems (i.e., checking whether each substring of the given string can be formed using the words in the dictionary).

Establishing Preconditions and Postconditions

Parameters:
- The method has two inputs: s which is a string, and wordDict which is a list of strings.
- s is the string that we need to check whether it can be segmented into a sequence of one or more dictionary words. wordDict is the list of words that we can use for segmentation.
Preconditions:
- There is no specific state that the program or some part of it must be in before this method is called.
- s must be a non-empty string, and wordDict must be a non-empty list of strings. These are the constraints given in the problem statement.
- Each string in wordDict must be unique, as stated in the problem.
Method Functionality:
- The method is expected to return a boolean value indicating whether the string s can be segmented into a sequence of one or more dictionary words.
- It examines the input string s and the list of words wordDict and uses dynamic programming to determine if s can be segmented into a sequence of words from wordDict.
Postconditions:
- After the method is called and returns, the method does not modify the state of the program or the values of the parameters.
- The return value indicates whether the string s can be segmented into a sequence of words from wordDict.
- The method does not have any side effects.
Error Handling:
- If the preconditions are not met, for example, if s or wordDict is empty, the method does not throw an exception. It simply returns False because an empty string cannot be segmented into a sequence of dictionary words, and if the dictionary is empty, there are no words to use for segmentation.
- If s is not a string or if wordDict is not a list of strings, Python will raise a TypeError during execution.

Problem Decomposition

Problem Understanding:
- The problem requires us to determine if a given string s can be broken down into a sequence of words, where these words are part of a given dictionary wordDict. Each word in the dictionary can be reused multiple times in the segmentation of s.
Initial Breakdown:
- The problem can be broken down into two broad subproblems:
  1. Parsing the string s into possible word sequences.
  2. Checking if each word in a sequence is present in the dictionary wordDict.
Subproblem Refinement:
- The subproblems can be further broken down into smaller tasks.
  - Parsing the string s: This can involve starting from the first character and incrementally considering longer substrings, checking each time if they form a word in the dictionary.
  - Checking if each word is in the dictionary: This is a simple task of checking if a given string is present in a list of strings.
Task Identification:
- The task of checking whether a word is in the dictionary is repeated for each possible word sequence.
Task Abstraction:
- The task of checking whether a word is in the dictionary is sufficiently abstracted. It makes sense in the context of the problem and can be used each time a new possible word is formed.
Method Naming:
- The method can be named canSegment, as it checks whether the string s can be segmented into a sequence of dictionary words.
Subproblem Interactions:
- The parsing of the string s into possible word sequences needs to be done before checking if each word is in the dictionary. There is a dependency between the two subproblems in that the second subproblem cannot be performed without the completion of the first. The order of these subproblems is therefore fixed.

From Brute Force to Optimal Solution

Let’s start with a brute force approach to this problem:

Brute Force Approach:

A naive solution would be to generate all possible partitions of the given string s and for each partition, check if all the substrings are in the dictionary. If we find any partition where all substrings are in the dictionary, we return True, else we return False.

However, the issue with this approach is its inefficiency. There can be an exponential number of partitions of a string of length n. For example, a string of length n has 2^(n-1) possible partitions. So, this approach would take O(2^n) time complexity, which is not feasible for large inputs.

Optimized Approach:

Instead of generating all partitions and then checking them, we can solve this problem more efficiently using Dynamic Programming (DP).

Create a boolean DP table dp of size n+1, where n is the length of the string s. Initialize dp[0] as True and all other elements as False. The i-th index in dp will represent whether the substring s[0...i-1] can be segmented into words in the dictionary.
The outer loop runs from 1 to n. For every index i, we check if dp[i] is False. If False, this means that the substring s[0...i-1] cannot be segmented into dictionary words, so we move on to the next index.
If dp[i] is True, then we need to check the substring s[i...j] (for every i < j <= n) and update dp[j] to True if the substring s[i...j] is present in the dictionary.
Finally, return dp[n] as the result, which represents whether the entire string can be segmented into dictionary words.

Python Code:

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def canSegment(s, wordDict):
    n = len(s)
    dp = [False] * (n+1)
    dp[0] = True

    for i in range(1, n+1):
        for j in range(i):
            if dp[j] and s[j:i] in wordDict:
                dp[i] = True
                break
    return dp[n]

This solution works by effectively breaking the problem into smaller sub-problems, where each sub-problem represents a prefix of the input string s and whether it can be segmented into dictionary words. We use the results of these smaller sub-problems to incrementally build up to the solution for the full problem.

This approach has a time complexity of O(n^2) and a space complexity of O(n), where n is the length of the input string s. This is a significant improvement over the exponential time complexity of the brute force approach.

The space complexity is due to the DP array, which has a size equal to the length of the input string. The time complexity is due to the nested loop structure, which in the worst case scenario can result in n*(n-1)/2 comparisons, which simplifies to O(n^2).

By avoiding unnecessary calculations and using the results of previous computations, the dynamic programming approach provides a much more efficient solution to this problem.

Code Explanation and Design Decisions

Initial Parameters: The initial parameters are the string s and the dictionary wordDict. The string s represents the input string we want to segment, while wordDict contains the set of words against which we will compare segments of s. The significance of these parameters is that they together provide the data we need to solve the problem, i.e., the string to segment and the words to segment it into.
Primary Loop: The outer loop iterates over each character in the string s from left to right (1 to n). Each iteration of this loop represents an attempt to partition the string into two parts: a prefix ending at the current character and a suffix starting from the next character. If the prefix can be segmented into words from the dictionary (which we check using our DP array), we then look at all possible suffixes to see if they are in the dictionary.
Conditions or Branches within the Loop: There is a condition within the loop, if dp[j] and s[j:i] in wordDict. This condition checks if the current prefix s[0...j-1] can be segmented into dictionary words (dp[j] is True) and if the current suffix s[j...i] is a word present in the dictionary. The significance of this condition is that it ensures we only mark a segment as possible (dp[i] = True) if it results in a valid partition of the string into dictionary words.
Updates or Modifications to Parameters within the Loop: The primary modification within the loop is the update to dp[i]. If we find a valid partition that ends at the i-th character (with the prefix s[0...j-1] being segmentable into dictionary words and the suffix s[j...i] being a dictionary word), we mark dp[i] as True. This update is crucial because it allows us to record valid partitions as we progress through the string, effectively building up a solution incrementally.
Invariant: The invariant here is the dp array. At any given point in the loop, dp[i] is True if and only if the substring s[0...i-1] can be segmented into dictionary words. This invariant is critical because it allows us to make local decisions (about segmenting the string at each character) based on global information (about the segmentability of the prefixes of the string).
Final Output: The final output of the algorithm is dp[n], which represents whether the entire string s can be segmented into words from the dictionary. This output is significant because it directly answers the problem’s question: Can the input string be segmented into a sequence of one or more dictionary words?

Coding Constructs

High-Level Problem-Solving Strategies: The code uses dynamic programming as its primary problem-solving strategy. Dynamic programming is a method for solving complex problems by breaking them down into simpler, overlapping subproblems. The code also uses a technique called string slicing to create substrings of the given string.
Purpose Explanation for Non-Programmer: Imagine that you have a long string of letters and a dictionary of words. This code is like a smart pair of scissors that tries to cut the long string into pieces such that each piece is a word found in the dictionary. If it can do this for the entire string, it tells us “Yes, it’s possible”; if not, it says “No, it’s not possible.”
Logical Constructs: The code uses iteration (a loop that runs from the first character to the last in the input string), condition checking (to see if a particular segment of the string is a word in the dictionary and whether the rest of the string can be broken down into words), and an array to store results of subproblems (so it can use those results later rather than recalculating them).
Algorithmic Approach in Plain English: The code starts by preparing a list (or a tally) that keeps track of whether or not we can break down the first ’n’ characters of the string into words from the dictionary. Then, it walks through our string letter by letter. For each position, it looks backward to see if it can “cut” the string there, such that the first part of the string can be broken down into words and the second part is a word itself. If it can do so, it notes down in the tally that up to this position, the string can be broken down into words. It continues this until it has walked through the whole string. At the end, it checks its tally to see if the entire string can be broken down into words.
Key Steps or Operations:
- The code initializes an array (dp) of boolean values to keep track of whether substrings of the input string can be segmented into dictionary words.
- It then iterates over the string, at each character checking every possible prefix of the string ending at that character.
- For each prefix, it checks if the prefix itself can be segmented into words and if the rest of the string up to the current character is a word.
- If both these conditions are met, it updates the dp array to indicate that the substring up to the current character can be segmented into dictionary words.
- Finally, it returns whether the entire string can be segmented into dictionary words.
Algorithmic Patterns or Strategies:
- The code uses the “tabulation” method of dynamic programming, where it builds up a table (dp array) iteratively.
- It uses a nested loop to check all possible prefixes for each character in the string.
- It uses the “optimal substructure” property of dynamic programming problems, where the solution to a problem relies on solutions to smaller instances of the same problem. In this case, the ability to segment the string up to a given character relies on the ability to segment the prefixes of the string.
- It uses the “memoization” strategy, storing the results of smaller subproblems to avoid duplicate work, which is done via the dp array in this context.

Language Agnostic Coding Drills

Concept Identification:

Here are the distinct concepts used in the code, presented in a language-agnostic manner:

Variable Declaration and Initialization: This involves setting up variables with initial values. In the case of this problem, we need an array (or a similar data structure) to keep track of whether substrings of the input string can be segmented into dictionary words.
Iteration: This concept involves looping over a set of data, performing some operation for each element. In this problem, we iterate over the characters of the input string.
Sub-string Extraction or Slicing: This is used to extract parts of the string. We use this concept to form substrings, and check if they form valid words according to the dictionary.
Condition Checking: This involves using conditional statements (like if-else) to perform different actions based on different conditions. Here, we use conditions to check if a substring is present in the dictionary, and if we can segment the prefix string into words.
Array Manipulation: This involves changing the values stored in an array. In our problem, we manipulate an array to keep track of the possibility of segmenting substrings into dictionary words.
Dynamic Programming: This advanced concept involves solving complex problems by breaking them down into simpler, overlapping subproblems, and storing the results of these subproblems to avoid redundant computations.

Concepts Listed in Order of Increasing Difficulty:

Variable Declaration and Initialization: This is a fundamental concept and is typically one of the first things learned when starting to code.
Iteration: This concept is slightly more advanced, but is still fundamental to most programming tasks.
Condition Checking: This concept, while not difficult, requires understanding of logical operators and boolean values.
Sub-string Extraction or Slicing: This requires a good understanding of string manipulation and indexing, which can be a bit more complex than the previous concepts.
Array Manipulation: This concept requires a good understanding of data structures (specifically arrays) and can involve more complex operations.
Dynamic Programming: This is a complex concept, requiring a deep understanding of problem-solving techniques, and an ability to identify overlapping subproblems and optimal substructures. It’s typically encountered later in a programming curriculum.

Problem-Solving Approach:

Here is a high-level strategy for how these drills contribute to solving the problem:

Start by declaring and initializing an array that will keep track of whether substrings of the input string can be segmented into dictionary words.
Next, iterate over the input string from the beginning to the end. For each character in the string, perform another iteration that goes backwards to the start of the string.
In the inner iteration, use slicing to get a substring of the string that ends at the current character and starts at each possible position before it.
Use condition checking to see if the substring can be segmented into dictionary words (according to the data in the array) and if the remaining part of the string (up to the current character) forms a word in the dictionary.
If the condition is satisfied, manipulate the array to indicate that the substring ending at the current character can be segmented into dictionary words.
This approach of storing results of subproblems and using them to solve larger problems is a key feature of dynamic programming.
Continue this process until the entire string has been iterated over.
Finally, the array will indicate whether the entire string can be segmented into dictionary words.

Targeted Drills in Python

Variable Declaration and Initialization

This is a simple concept, here’s how you can create and initialize a list in Python:

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# Create a list with 10 elements, all set to False
dp = [False] * 10

Iteration

Python provides the ‘for’ loop to iterate over elements. Here’s a basic drill:

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# Print numbers from 0 to 9
for i in range(10):
    print(i)

Condition Checking

Python uses ‘if-elif-else’ constructs for condition checking. Here’s an example:

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# Check if a number is positive, negative, or zero
num = 5
if num > 0:
    print("Positive")
elif num < 0:
    print("Negative")
else:
    print("Zero")

Sub-string Extraction or Slicing

Python’s string slicing is very useful. Here’s a simple drill:

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# Get a substring from a string
s = "Hello, World!"
print(s[7:12])  # Prints "World"

Array Manipulation

Python’s lists are dynamic and easy to manipulate:

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# Update elements in a list
dp = [False] * 10
dp[5] = True  # Set the 6th element to True
print(dp)

Dynamic Programming

A basic example of dynamic programming would be computing Fibonacci numbers:

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def fib(n):
    dp = [0, 1] + [0] * (n-1)
    for i in range(2, n + 1):
        dp[i] = dp[i - 1] + dp[i - 2]
    return dp[n]

print(fib(10))  # prints 55

For the problem-specific concept, it is breaking a word into possible segments and checking each one for being a valid word. We can simulate this with a simplified example:

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# A simplified word segmentation drill
def can_segment(word, dictionary):
    n = len(word)
    dp = [False] * (n + 1)
    dp[0] = True

    for i in range(1, n + 1):
        for j in range(i):
            if dp[j] and word[j:i] in dictionary:
                dp[i] = True
                break
    return dp[n]

dictionary = {"apple", "pie"}
print(can_segment("applepie", dictionary))  # prints True

To assemble these drills into the final solution, we would first declare our dp array. Then we would iterate over the string, and for each character, we would use another iteration that goes backwards to the start of the string. In this inner iteration, we would slice out a substring and use condition checking to see if we can segment the substring into dictionary words and if the rest of the string forms a dictionary word. If it does, we update the dp array. We continue this process until we have iterated over the entire string, at which point our dp array can tell us if we can segment the entire string into dictionary words.

Q&A

I don’t understand why this should output false given that the description says at least one dictionary word. It has at least one dictionary word in catsandog, so why output false? What am I missing?

The problem is not asking if the string contains at least one dictionary word. Rather, the problem is asking if the entire string can be segmented into dictionary words, with no extra characters left over.

In other words, you must be able to divide the entire string such that each segment (which can be of varying length) is a word in the dictionary. If this is possible, the function should return True. If it’s not possible to divide the string such that each segment is in the dictionary, then the function should return False.

In the case of “catsandog”, even though it contains dictionary words like “cat” and “dog”, it can’t be completely segmented into dictionary words because “sand” is not in the dictionary provided. Thus, the function returns False.

Explanation: I wish the question was explained in bit more detail.

The question should be reframed as; Is there a combination of words in wordDict that can be used to recreate the original string s. Notice that I mentioned the word combination, this does not mean that you will need all the words in wordDict to recreate s (Using some words from wordDict as long as they perfectly recreate the string s; is a valid solution or word break). Also note that you can use the words in wordDict more than once.

Take this example for reference:

Input: “bb”, [“a”,“b”,“bbb”,“bbbb”] Expected: true

Here, word “b” from wordDict can be used to perfectly to recreate the string s = “b” + “b” , which is why the expected output is True. Note that we did not use all the words in the dictionary and still found a valid word break.

Now, let’s take another example: Input: s = “catsandog”, wordDict = [“cats”,“dog”,“sand”,“and”,“cat”] Output: false

Here, no combination of words exist in wordDict that can perfectly recreate the input string s. “cats” + “dog” will require “san” in wordDict in order to recreate s “cats” + “and” will require “og” in wordDict in order to recreate s “cat” + “sand” will require “og” in wordDict in order to recreate s … and so on.

The intuition is that you will have consider every combination of words in the wordDict, moreover these words can be used more than once when computing the combinations.

That’s exactly right! The word break problem asks us to determine whether the entire input string s can be broken down into a sequence of one or more words from the dictionary wordDict. It’s not enough to just find some words from the dictionary within the string. The key requirement is that we use the words from the dictionary to build up the entire string s, potentially using some words multiple times.

As you mentioned, the core of this problem lies in identifying all possible combinations of words from the dictionary that can be put together to recreate the string s. This involves iterating through the string, breaking it down into substrings, and checking if these substrings exist in the word dictionary.

A brute force approach could involve generating all possible substrings and checking if each one is in the dictionary. However, this would be quite inefficient. A more efficient approach is to use dynamic programming, where we keep track of which parts of the string can be broken down into dictionary words, and use this information to guide how we break down the rest of the string.

I hope this explanation helps clarify the problem further. Feel free to ask if you have any more questions!

“cats” + “dog” will require “an” in wordDict in order to recreate s

Yes, you’re correct. In order to recreate the string “catsandog” using “cats” and “dog” from the word dictionary, you would indeed need “an”, not “san”.

Here’s the correct breakdown:

“cats” + “an” + “dog” = “catsandog”

But in the given word dictionary [“cats”,“dog”,“sand”,“and”,“cat”], “an” is not present. So, we can’t break down “catsandog” into a sequence of dictionary words, and the result is False.

For the test case

s=“a” wordDict = [“b”]

when clicking submit, leetcode claims my code outputs “True” when the output should be “False”. When I run the code locally or when I change the custom test case to match, they both return “False”. What is wrong with leetcodes compiler??

I can propose a couple of potential issues:

Different initial conditions: Make sure that your code doesn’t rely on some global state that could be different between runs. For instance, if you have a global variable that isn’t reset between different tests, that could be causing inconsistencies.
Edge case handling: If your code doesn’t correctly handle edge cases, you may see inconsistent results. For example, if your code assumes the input string or wordDict is always non-empty, it could fail in certain cases.

Why this input is false?

“catsandog” [“cats”,“dog”,“sand”,“and”,“cat”]

and this one is true

“cars” [“car”,“ca”,“rs”]

The problem is asking if the entire string s can be segmented into one or more dictionary words.

Let’s look at the two examples:

For s="catsandog" and wordDict=["cats","dog","sand","and","cat"]:
We can see that the words “cats”, “and”, and “dog” are in wordDict, but the leftover part “og” from “sandog” is not in the wordDict, so we cannot break up the entire string s using words from wordDict. Therefore, the answer is False.
For s="cars" and wordDict=["car","ca","rs"]:
The string s can be broken up into “car”+“s” or “ca”+“rs”. Since both “car”, “ca”, “rs” and “s” are in wordDict, we can segment the entire string s into dictionary words. Therefore, the answer is True.

Remember, the goal of the problem is to check if we can break up the entire string s into words present in the wordDict, not just a part of it.

Has anybody thought about the solution when the problem is changed such that each word in the dictionary can be only used once?

My initial thoughts: • DP: At each point i where the string is breakable, we need to use a set to remember which words we have used to break the string up to this point. And when we are considering a later point j, based on which previous breakable point we are using as a pivot, we will need to consider the “breakability” of the string at index j with (input set - set stored at index j)

• DFS + Memo: The memo is not only keyed by the starting index, but should also encode the information of what words are still available in the set, because both of those together represent the state of the DFS (whether the substring starting at index i, with the remaining usable words, is breakable). But I am not exactly sure how to encode a set of strings as a key. (sort the strings and concatenate?)

Great thinking! You are right on track.

If each word in the dictionary can only be used once, it does become a more complex problem.

For a DP solution, you would indeed need to track the used words at each possible break point. For each new word you encounter, you’d need to check if it can be formed using the remaining words in the dictionary. This could potentially be quite inefficient as the size of the word dictionary increases, since you’d be creating and comparing sets of words at each possible break point.

For a DFS + Memo approach, it’s a bit trickier. You can’t just key the memoization structure by the start index anymore, as the remaining words in the dictionary become part of the state.

A potential way to handle this could be to convert the remaining words into a bitmask. This will work well if the number of words in the dictionary is relatively small (up to about 20 words due to the limitation of integer size). Each bit in the bitmask would correspond to whether a particular word is still available. This would allow for efficient comparison and combination of states. Here’s a high-level idea of how the solution might look like:

Convert the word dictionary into an indexed list of words, so each word corresponds to an index.
Initialize a DP table or memoization structure that is keyed by the start index in the string and the bitmask representing the remaining words.
Perform the DFS or DP process as normal, but when marking a word as used, also update the bitmask to indicate that the corresponding word is no longer available.

Remember that this approach would only be feasible for a small dictionary size. For larger dictionaries, a different approach would be required, potentially involving complex data structures to efficiently represent the set of remaining words.

The question asks to check " if s can be segmented into a space-separated sequence of** one or more** dictionary words."

Since it says one or more, shouldn’t we be returning true for below testcase as it has one dictionary word which is cats in sequence?

s = “catsandog”, wordDict = [“cats”,“dog”,“sand”,“and”,“cat”] output - false

The key phrase in the problem statement is that s can be segmented into a “sequence of one or more dictionary words”.

This doesn’t mean that the presence of one or more dictionary words within s would return True. Instead, it means that the entirety of s needs to be able to be segmented into a sequence of one or more dictionary words.

In your example:

s = “catsandog”, wordDict = [“cats”,“dog”,“sand”,“and”,“cat”]

Though “cats”, “dog”, “sand”, “and”, “cat” are present within s, there’s no way to segment the entire string s into a sequence of one or more dictionary words without any leftover characters. The leftover characters “an” and “og” are not part of the word dictionary.

Hence, the output is False.

If your string was “catsand” and your wordDict = [“cats”,“and”], the output would be True, because you can segment the entire string into dictionary words: “cats” and “and”.

How is the time complexity of the solution O(n^2) for Recursion with memoization?

The time complexity of the recursive solution with memoization for this problem is O(n^2), where n is the size of the input string. Here’s why:

Recursion Depth / State Space: We’re making a recursive call for each prefix of the string that matches a word in the dictionary, up to n. Therefore, we have up to n different states.
Work Done Per State: For each state, i.e., for each index in the string, we’re doing a constant amount of work: we’re checking if the remaining suffix of the string is a word we’ve seen before (which is an O(1) operation because we’ve assumed that word lookup in the dictionary is constant time), and if it is, we’re making a recursive call.
Memoization: Because of memoization, each state (each possible start of a suffix) is only computed once. Subsequent calls to the same state are returned in constant time from the memoization table.
Number of States: We have n states and we’re doing O(n) work per state, leading to O(n^2) time complexity overall.

However, note that the actual time complexity could be worse than O(n^2) if the average length of the words in the dictionary is large because string slicing in Python isn’t a constant-time operation. If the word lengths are small and/or roughly equal, though, it can be considered as approximately O(n^2).

I am wondering whether using Trie to hold the words in the dictionary will be faster than just using Set. It seems that using Trie will reduce the number of repetitive operations. Do anyone have tried out this data structure and got a better result?

Yes, a Trie (also known as a prefix tree) could indeed be a beneficial data structure for this problem. It allows us to efficiently find all words in the dictionary that can serve as a prefix for a given string, which aligns well with our problem requirements.

A Trie provides O(L) lookup where L is the length of the word we are looking for. Inserting words from the dictionary into the Trie also takes O(L) time. These are roughly equivalent to the performance characteristics of a Set, but there are some key differences when considering our specific problem.

Finding All Prefixes: Given a string s, we can use the Trie to find all the dictionary words that are prefixes of s in a single pass through s. This can be more efficient than the approach of checking each possible prefix against a Set, which would require separate checks for each prefix.
Reducing Repetition: We can terminate early if we know a prefix of s cannot lead to a valid segmentation. With a Set, we might perform unnecessary computations for longer prefixes that share the same starting part.

However, there are trade-offs:

Space Complexity: Tries can be more space-intensive than Sets. They store a node for each character of each word, which can add up, particularly if the words in the dictionary are long.
Implementation Complexity: Tries are more complex to implement and use than Sets. This might make your solution harder to write and debug.

So, if your dictionary has many long words with common prefixes, a Trie might speed up your solution. However, remember that Tries use more memory and are harder to implement than Sets. Always consider the trade-offs based on your specific use case.

Which range should this problem be classified into? DP? Backtracking? By definition from web, a problem that has optimal substructure and overlapping subproblems and searches for a optimal solution should be defined as a DP problem. However, this problem seems to judge if a breaking rule exists. And backtracking seems to be an inituitive solution. How can we associate this problem with DP if we don’t meet it in LeetCode? Can you help me for giving the DP transfer equation?

This problem can indeed be solved using both dynamic programming (DP) and backtracking. However, the dynamic programming approach is generally more efficient due to its ability to avoid re-computation of overlapping subproblems. Here’s how we can formulate this problem in terms of dynamic programming:

Step 1: Understand the problem statement

The problem is asking if we can break up the input string into words that exist in the given dictionary. This suggests that we can solve smaller instances of the same problem and use them to solve the larger problem, hinting at the optimal substructure property required for DP.

Step 2: Define the state

Let’s define dp[i] to be True if the string up to the i-th index can be segmented into one or more dictionary words, and False otherwise. The state dp[i] depends on the states of all the indices before i.

Step 3: Formulate the DP transfer equation

For dp[i] to be True, there needs to be an index j before i such that dp[j] is True and the string from the (j+1)-th index to i-th index is a dictionary word. Formally,

dp[i] = True if there exists a j such that dp[j] = True and s[j+1:i+1] is a dictionary word
dp[i] = False otherwise

We initialize dp[0] to True since the empty string can always be segmented.

The final answer will be dp[n], where n is the size of the input string.

This way, we formulate the problem as a DP problem and solve it efficiently by reusing solutions to smaller subproblems. The time complexity of this approach is O(n^2), and the space complexity is O(n) for the dp array.

What is a DP transfer equation?

A Dynamic Programming (DP) transfer equation, also known as a state transition equation, describes how the solution to the problem can be composed from solutions to smaller subproblems. It establishes the relationship between a state and its preceding states.

In the context of DP, a “state” typically represents the solution to a subproblem. The transfer equation explains how the solution to a larger subproblem can be derived from its smaller subproblems.

Here is a simple example using the Fibonacci sequence. The Fibonacci sequence is a series of numbers where a number is the sum of the two preceding ones, usually starting with 0 and 1.

In this case, the state Fib(n) represents the nth Fibonacci number. The transfer equation for the Fibonacci sequence is:

Fib(n) = Fib(n-1) + Fib(n-2)

This equation states that the nth Fibonacci number is the sum of the (n-1)th and (n-2)th Fibonacci numbers, which are smaller subproblems of the original problem of finding the nth Fibonacci number. This is a DP transfer equation because it describes how to transition from one state to another.

Is it the same as recurrence relation?

Yes, a dynamic programming (DP) transfer equation and a recurrence relation share similar ideas. Both involve defining a value or a state based on smaller input values or preceding states. However, they’re used in slightly different contexts.

A recurrence relation is a mathematical construct that expresses the elements in a sequence in terms of previous elements. It’s commonly used in mathematics to define sequences or to solve certain types of problems, such as divide-and-conquer problems.

For example, the Fibonacci sequence can be defined by the recurrence relation F(n) = F(n-1) + F(n-2), with base cases F(0) = 0 and F(1) = 1.

A DP transfer equation, on the other hand, is a concept specific to the field of computer science and dynamic programming. It describes how to build up the solution to a larger problem from the solutions to its subproblems. The subproblems in dynamic programming often overlap, which distinguishes DP from other problem-solving strategies like divide-and-conquer.

For the Fibonacci sequence, the DP transfer equation is the same as the recurrence relation: Fib(n) = Fib(n-1) + Fib(n-2). But in the context of DP, we would usually store the results of each subproblem in a data structure (like an array or a dictionary) to avoid redundant computation of the same subproblem, a technique known as memoization.

So in essence, while the concepts of DP transfer equations and recurrence relations are similar and sometimes used interchangeably, they belong to somewhat different problem-solving paradigms.

I’m wondering what is the reason behind to switch the input parameter from set to list? Isn’t the set could check contains in O(1)? But it’s interesting that it takes more time if I create a hashset. I think the dictionary is probably not big enough though.

Indeed, set operations in Python typically have an average time complexity of O(1), while for lists it’s O(n). That said, the actual performance can depend on various factors, and it’s not always guaranteed that a set will be faster than a list.

One key factor is the size of the data. Sets in Python are implemented as hash tables, and their speed advantage becomes noticeable only when dealing with larger datasets. If the word dictionary is small, the overhead of hashing might outweigh the benefits of quick lookup, making a list operation faster.

In addition, the specific nature of the operations being performed also matters. If you’re doing a lot of adding and removing operations, a set will generally outperform a list, since these operations are O(1) for a set and O(n) for a list. However, if the primary operation is iterating over the elements, a list might be faster, as sets in Python are unordered and iterating over a set could potentially be slower than over a list.

Lastly, it’s worth noting that the time complexity measures are theoretical abstractions that describe the growth of time with respect to the input size. They don’t account for lower-level details like memory access patterns, CPU caching, interpreter overhead, etc., which can all influence the actual run time. Therefore, while complexity analysis provides valuable insight, it doesn’t always perfectly correlate with performance in practice.

Remember, it’s always good practice to benchmark different approaches with realistic datasets to choose the most suitable one for a particular use case.

wordDict always is a null object, waste me lots of time

This soln pass 33/45 test cases can you clear me that where i can make mistake in the given code in java using trie.

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class Solution {
    static class Node{
        Node [] children = new Node[26];
        boolean eow = false;
        Node(){
            for(int i=0;i<26;i++){
                children[i] = null;
            }
        }
    }
    public static Node root = new Node();
    public static void insert(String word){
           Node curr = root;
           for(int level =0;level < word.length();level++){
            int ind = word.charAt(level) -'a';
            if(curr.children[ind] == null){
                curr.children[ind] = new Node(); 
            }
            curr = curr.children[ind];

           }
           curr.eow = true;
    }
   public static boolean search(String key){
    Node curr = root;
     for(int level =0;level<key.length();level++){
        int ind = key.charAt(level)-'a';
        if(curr.children[ind] == null){
            return false;
        }
        curr = curr.children[ind];
     }
    return curr.eow == true;
   }
  public static boolean wordBreak(String key){
    if(key.length() == 0){
        return true;
    }
    for(int i=1;i<=key.length();i++){
      if(search(key.substring(0, i)) && wordBreak(key.substring(i))){
          return true;
      }

    }
    return false;
  }
    public boolean wordBreak(String s, List<String> wordDict) {
        for(String str : wordDict){
            insert(str);
        }
        return wordBreak(s);
    }
} }

The problem seems to be that your recursive function wordBreak is creating an exponential number of recursive calls because it doesn’t remember which substrings it has already checked. You are creating a substring and recursively searching for it in the trie, but you don’t have a way to remember if you have already checked a certain substring. Therefore, it may be rechecking the same substring multiple times leading to inefficiency.

This can be solved by using a technique called memoization. Memoization is an optimization technique used primarily to speed up computer programs by storing the results of expensive function calls and reusing them when the same inputs occur again.

Below is your corrected code:

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class Solution {
    static class Node{
        Node [] children = new Node[26];
        boolean eow = false;
        Node(){
            for(int i=0;i<26;i++){
                children[i] = null;
            }
        }
    }

    public static Node root = new Node();
    
    public static void insert(String word){
        Node curr = root;
        for(int level =0;level<word.length();level++){
            int ind = word.charAt(level) -'a';
            if(curr.children[ind] == null){
                curr.children[ind] = new Node(); 
            }
            curr = curr.children[ind];
        }
        curr.eow = true;
    }

    public static boolean search(String key){
        Node curr = root;
        for(int level =0;level<key.length();level++){
            int ind = key.charAt(level)-'a';
            if(curr.children[ind] == null){
                return false;
            }
            curr = curr.children[ind];
        }
        return curr.eow == true;
    }

    public static boolean wordBreak(String key, boolean[] dp, int start){
        if(start == key.length()){
            return true;
        }
        if(dp[start] != false) return dp[start]; //returning the memoized result
        for(int i=start+1;i<=key.length();i++){
            if(search(key.substring(start, i)) && wordBreak(key, dp, i)){
                return dp[start] = true;  //storing the result in dp for future use
            }
        }
        return dp[start] = false;
    }

    public boolean wordBreak(String s, List<String> wordDict) {
        for(String str : wordDict){
            insert(str);
        }
        boolean[] dp = new boolean[s.length()];  //initialize dp array
        return wordBreak(s, dp, 0);
    }
}

In this modified solution, I’ve added a boolean array dp[] to keep track of the results of subproblems. When you start a new subproblem, you first check if the result for that subproblem has already been calculated and stored in dp[]. If it has, you can simply return the stored result instead of calculating it again. This significantly reduces the number of recursive calls and makes the program more efficient.

Keep in mind that the dp[] array should be initialized to false because by default in Java, boolean arrays are initialized to false and we are using false to indicate that a subproblem has not yet been solved.

Can someone please help me understand what part am missing.

“bb” [“a”,“b”,“bbb”,“bbbb”]

As per my understanding the output should be false since a is not part of the input string but in actual it is expected true.

“a” [“b”]

If I go by above behaviour then this should also be true but it is expected to be false.

I am sure there is something which I am missing from the problem statement. Please help.

You might be misunderstanding the problem statement a bit. In the word break problem, you are given a non-empty string s and a dictionary wordDict containing a list of non-empty words. Your task is to determine if s can be segmented into a space-separated sequence of one or more dictionary words.

So, for your examples:

The input string is “bb” and the word dictionary is [“a”,“b”,“bbb”,“bbbb”]. The string “bb” can be segmented into “b” and “b”, both of which are in the dictionary. Therefore, the expected output is true.
The input string is “a” and the word dictionary is [“b”]. The string “a” cannot be segmented into any word in the dictionary, so the expected output is false.

The dictionary doesn’t need to include all the characters from the input string, rather it should have words which can be used to segment the input string. The words in the dictionary can be reused any number of times to form the string. The segments do not need to be unique, i.e., different segmentations can produce the same string.

For instance, given the string “catsanddog” and dictionary [“cat”, “cats”, “and”, “sand”, “dog”], both segmentations [“cat”, “sand”, “dog”] and [“cats”, “and”, “dog”] are valid.

I have gone through the solutions and I could see that best complexicty in n^3 with n space. n^3: n^2 for all the combinations of substring (i,j) and n for slicing or substring and 1 of finding string in a set = n^2 * n * 1

If I have all the substring precomuted with n^2 complecity using dictionary. (cache[i][j] = substring[i:j]) and we can use one more n^2 complexicty for checking all the combinations of substring.

Your approach to use pre-computation to store substrings can certainly be done but it is generally not used in practice. The reasons are as follows:

Memory Usage: The use of pre-computed substrings in a 2D array would take O(n^2) space. This is because you’re storing all possible substrings of the original string in the 2D array. If n is large, it may lead to a memory overflow error.
Access Time: Accessing elements from a large 2D array can be slow and might not lead to significant speedup in practice.
Overall Time Complexity: The overall time complexity is still O(n^3), because you’re still checking all possible substrings which takes O(n^2), and checking whether a substring is in the dictionary (which takes O(n), even if you use a hash set). Hence, the total time complexity will still be O(n^3).

The more common approach is to use dynamic programming. We maintain a boolean dp array, where dp[i] is true if substring from index 0 to i (inclusive) can be segmented into dictionary words, and false otherwise. The space complexity is O(n) and time complexity is O(n^2), assuming the cost to check if a word exists in the dictionary is O(1).

Here is the approach:

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public class Solution {
    public boolean wordBreak(String s, List<String> wordDict) {
        Set<String> wordDictSet = new HashSet(wordDict);
        boolean[] dp = new boolean[s.length() + 1];
        dp[0] = true;
        for (int i = 1; i <= s.length(); i++) {
            for (int j = 0; j < i; j++) {
                if (dp[j] && wordDictSet.contains(s.substring(j, i))) {
                    dp[i] = true;
                    break;
                }
            }
        }
        return dp[s.length()];
    }
}

This algorithm maintains a DP table of size n+1 (for string of length n) and it fills the table in left to right manner. For filling the DP table, it breaks the given string in all possible ways and checks whether the left substring can be segmented into dictionary words using DP and right substring is present in dictionary or not.

The catsanddog example doesn’t work because you can’t use all the words in the worddict?

But why does this test case equate to true? “leetcode” [“leet”,“can”,“code”]

The word break problem asks if the input string can be segmented into a sequence of one or more words from the word dictionary. The words in the word dictionary can be used as many times as required and it’s not necessary to use all the words from the dictionary. The goal is not to use all the words in the dictionary, but to break down the input string using the words from the dictionary.

So, taking your example:

The input string is “leetcode”, and the word dictionary is [“leet”,“can”,“code”].

You can break “leetcode” into “leet” and “code”, both of which are in the dictionary. Therefore, the expected output is true.

On the other hand, if you tried to break down “catsanddog” with the dictionary [“cat”,“cats”,“and”,“sand”,“dog”], it would work as you can break “catsanddog” into “cats”, “and”, “dog” or “cat”, “sand”, “dog” which are all in the dictionary.

Remember, the words in your dictionary can be reused any number of times to form the string. The segments do not need to be unique. So different segmentations can produce the same string. The ultimate goal is to break the given string using the words present in the dictionary.

Construct an epsilon-NFA of all the words in the dictionary. Starting state “start” is the only accepting state. Transitions will take an input character of a word and transition to the next state.

For example: a word “word” will have the NFA as

(Start) -> W -> WO -> WOR -> WORD

D will perform an epsilon transition back to the start state.

There exists algorithms in automata theory to convert any epsilon-NFA to an NFA, and NFA can be converted to a DFA.

We can run the target string on that DFA. If the program accepts all input characters from the target string and is on the accepting state, return true. All other cases will return false.

The runtime will only be based on the length of the target string which is O(n).

The approach you have outlined is indeed correct and offers an interesting way to tackle the word break problem, particularly in the context of Automata Theory. This approach leverages Finite Automaton (FA), a well-studied concept in computer science.

The idea to build an ε-NFA (Nondeterministic Finite Automaton where ε represents empty string transitions) from the word dictionary, convert it to NFA (without ε-transitions), then to DFA (Deterministic Finite Automaton) is a sound one. By transforming the ε-NFA into a DFA, you can deterministically decide whether a given string can be segmented into words from the dictionary.

The procedure will be as follows:

Construct the ε-NFA where each word in the dictionary corresponds to a path in the automaton from the start state to the accepting state.
Remove ε-transitions to convert the ε-NFA to NFA.
Convert the NFA to a DFA using the powerset construction.
Given the target string, start from the initial state of the DFA and transition based on each character of the string.
- If we end up in an accepting state after consuming all characters, return true.
- Otherwise, if we end up in a non-accepting state or cannot transition based on the current character, return false.

The runtime complexity of the last step (running the target string on the DFA) is indeed O(n), where n is the length of the target string.

However, keep in mind that constructing the automaton (step 1) could be time-consuming, depending on the size and complexity of your word dictionary. The conversion steps (steps 2 and 3) are also not trivial and require time. These steps may not be the most efficient approach for larger dictionaries or when the dictionary changes frequently, as you would need to reconstruct and convert the automaton each time.

In practice, for this specific problem, a simpler and more commonly used approach is the dynamic programming one, as it provides good time complexity without the need to construct and convert automata. But the approach you proposed is a perfect demonstration of the interplay between different areas in computer science and how problems can indeed have multiple solutions.

The problem’s structure is by itself DFA. So no need for NFA

Yes, you’re correct. The structure of this problem can directly map to a DFA. The dictionary words essentially define paths in the DFA from the start state to the accepting state. A word from the dictionary creates a sequence of transitions from the start state to the accepting state, with the last character of the word looping back to the start state via an ε-transition (although, in a strict DFA, ε-transitions are not allowed, so you might just loop back on the next word’s starting character).

Then, for any given string, you would start at the initial state and move according to the characters in the string. If after consuming all characters in the string, you end up in the accepting state, then you return true, otherwise false.

This approach indeed has a linear running time (O(n)) with respect to the length of the target string, because you’re simply traversing the string character by character and following the corresponding transitions in the DFA. But still, you’ll need to consider the time to construct the DFA from the dictionary, especially if the dictionary is large.

In conclusion, while a direct DFA mapping is indeed possible and may offer a performance boost in certain cases, it’s worth mentioning that more straightforward and commonly used approaches, like dynamic programming, can be more accessible and easier to implement.

DFA stands for Deterministic Finite Automaton, a concept in the field of computer science and more specifically in Automata Theory, which is a branch of theoretical computer science.

A DFA is an abstract machine used to model computation or to represent and control execution flow. It’s called deterministic because it has a unique state transition for every state in the machine based on the input.

A DFA has the following components:

A set of states: A finite set of states that the automaton can be in.
An alphabet: A finite set of symbols that can be read as input.
A transition function: A function that takes a state and an input symbol, and returns the next state.
An initial or start state: The state of the automaton before any input has been processed.
A set of accept states: States that are designated as “accepting” or “final”.

The automaton reads a string of symbols from its input, and changes its state based on the current state and the current input symbol, following the rules defined by the transition function. If, after reading the entire input string, the automaton is in an accept state, then it “accepts” the string. If it is not in an accept state, then it “rejects” the string. This behavior is used to classify strings and check if they belong to a certain set or follow certain rules, making DFAs a fundamental concept in the design of lexical analyzers, parsers, and other language processing tools.

I have tried to setting empty string as always a valid break for convenience of initializing the dynamic programming array, and the OJ passed all test cases. But I am wondering that is just because empty string is always added to the wordDict in all the test cases. So, my question is should we check if the empty sting is in the wordDict?

In most contexts of the word break problem, the empty string is considered a valid break. This is because an empty string can be thought of as a string broken up into zero words, which is a valid scenario. This makes it convenient to initialize the dynamic programming array, as you’ve noted.

However, whether or not you should check if the empty string is in the word dictionary depends on the specific rules of the problem you’re solving. If the problem specifies that a valid break must consist of one or more words from the dictionary, then an empty string should not be considered a valid break unless it is explicitly included in the dictionary. But in most versions of the word break problem, an empty string is implicitly considered a valid break even if it’s not in the dictionary, because it can be seen as a string segmented into zero words.

In the context of the dynamic programming solution to the word break problem, treating the empty string as a valid break (i.e., initializing the DP array such that dp[0] is true) is an important base case that allows the algorithm to correctly determine whether or not other, non-empty substrings of the input string are valid breaks. This doesn’t necessarily mean that the empty string itself is considered to be in the word dictionary; it’s just a useful way to structure the problem in order to make the dynamic programming approach work.

see constraints, it can never be empty

In that case, if your problem constraints explicitly state that the input string will never be empty, then you should not need to check for an empty string in the dictionary. The handling of empty strings would depend on the exact problem specifications and constraints.

However, from a dynamic programming perspective for solving the word break problem, it’s common to initialize your DP array or list with dp[0] = true which essentially considers the case of an empty prefix of the input string. This is done not because the empty string is necessarily a valid word according to the dictionary, but because it serves as a useful base case that simplifies the implementation of the DP algorithm.

So, even if the constraints specify that the input string can never be empty, you might still initialize dp[0] = true as a kind of “bootstrap” for your dynamic programming approach. It’s a technical detail of the solution, not really a statement about whether the empty string is a valid word.

For the test cast, “aaaaaaa”, [“aaaa”,“aa”], the compiler deems it false, but I don’t understand why. Shouldn’t it be true because aaaa and aa can be made from aaaaaa?

The word “aaaaaaa” should indeed be breakable into “aaaa” and “aaa” or alternatively “aa”, “aaaa”, both of which are present in the dictionary [“aaaa”, “aa”]. Therefore, the output of the function for this input should be true.

If your program is returning false for this test case, then there might be a bug in your code. The issue might be related to how your code is iterating through the substrings or how it’s managing its dynamic programming state.

Please note that order matters in many word break problem solutions. Your code should be able to accommodate the use of words from the dictionary in any order, and of varying lengths.

If you’re using a dynamic programming approach, a common pattern is to use a boolean array dp of size n + 1 (where n is the size of the input string), and set dp[0] to true. Then, for each i from 1 to n, check for each j from 0 to i whether dp[j] is true and whether the substring from j to i is in the dictionary. If so, set dp[i] to true.

If you’re still having trouble with this test case, I would recommend debugging your code with this input to see exactly what’s going wrong.

can anyone explain that why after using recursion memorization we have O(n^2) complexity?

In the Word Break problem, using recursion with memorization (also known as top-down dynamic programming or DP) allows us to store and reuse results of overlapping subproblems to avoid redundant computation. This significantly reduces the time complexity of the algorithm.

Here’s why it becomes O(n^2):

State: In the context of this problem, a “state” can be considered a suffix of the input string starting at index i. There are ’n’ such states corresponding to ’n’ indices of the input string.
State transition: For each state, we check the possible breaks from index ‘i’ to ’n’, that is, we are examining each suffix of the string starting from ‘i’. So for each state, we could make ’n’ different decisions.

Putting the above two points together, there are ’n’ states and for each state, we could perform ’n’ operations at most (for each of the ’n’ decisions, in worst case). Therefore, the time complexity is O(n^2).

The key is that memoization ensures each state is solved only once and then the result is stored. The next time we need to solve the same state, instead of performing the computation again, we simply return the stored result. This way, we ensure that the time complexity for the computation of each state remains constant. That’s why the overall time complexity is proportional to the number of states times the time complexity for each state, which is O(n) * O(n) = O(n^2).

Keep in mind that this analysis assumes that the dictionary look-up operations (i.e., checking whether a string is in the dictionary) can be done in constant time, such as by using a hash set to store the dictionary words. If dictionary look-ups take longer, that would need to be factored into the time complexity.

The question is, the word should be a space separated sequence of one or more words, the answer is true only if you can take one or more words from wordDict and repeat them(can use one word multiple times) and form s

if s = “aaa” and wordDict = [“a”, “b”] you can take first word from wordDict and repeat it three times to make “aaa” (repetition is allowed)

if s = “abc” and wordDict = [“a”, “b”] no combination of words in wordDict can make “abc”

thr third example: s = “catsandog” and wordDict = [“cats”,“dog”,“sand”,“and”,“cat”] “cats” + “sand” + “dog” doesn’t make “catsanddog” and the result is false

Yes, your understanding of the problem is correct. In the word break problem, you’re trying to determine whether or not the input string ’s’ can be segmented into a space-separated sequence of one or more dictionary words. This means that the input string ’s’ must be able to be completely broken down into words that are in the wordDict, with no additional characters left over.

As per your examples:

For s = “aaa” and wordDict = [“a”, “b”], we can indeed repeat “a” three times to form “aaa”. So, the output would be true.
For s = “abc” and wordDict = [“a”, “b”], there’s no way to form “abc” using only the words in the dictionary, because “c” is not present in the dictionary. So, the output would be false.
For s = “catsandog” and wordDict = [“cats”,“dog”,“sand”,“and”,“cat”], although we can form words like “cats”, “sand”, and “dog”, there’s no way to arrange these words to form “catsandog”. “catsandog” is not just a concatenation of the words, but needs to be a space-separated sequence where the order of the characters in ’s’ is preserved. So, the output would be false.

A common way to solve this problem is by using dynamic programming. We create a boolean DP table, where dp[i] represents whether the first ‘i’ characters of ’s’ can be segmented into one or more dictionary words. By iteratively filling in this table based on the entries of the dictionary, we can eventually determine whether the entire string ’s’ can be segmented into dictionary words.

Can someone tell me why I’m getting a TLE in this?

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class Solution {
public:
unordered_set words;

bool sumi(int ind, string s, string str, vector<vector> &dp){
if(ind==s.length()){
if(str.length()==0) return true;
else return false;
}
if(dp[ind][str.length()]) return dp[ind][str.length()];
str+=s[ind];
bool one=false,two=false,three=false;
if(words.find(str)==words.end()){
one=sumi(ind+1,s,str,dp);
}
else{
three=sumi(ind+1,s,str,dp);
str="";
two=sumi(ind+1,s,str,dp);
}
return dp[ind][str.length()]=(one or two or three);
}

bool wordBreak(string s, vector<string>& wordDict) {
    for(auto x:wordDict) words.insert(x);
    int n=s.length();
    vector<vector<bool>> dp(n,vector<bool> (n+1,false));
    return sumi(0,s,"",dp);
}
};

The problem seems to lie in your memoization approach, and the way you’re using the dp vector to store the results of the subproblems.

Firstly, you’re using dp[ind][str.length()] to represent whether the first ‘ind’ characters of ’s’ can be broken down into words in the dictionary. However, str.length() doesn’t always correctly capture the state of the problem at position ‘ind’. The string str that you’re building up and breaking down during the recursion isn’t directly tied to the state at ‘ind’. This can lead to incorrect results being pulled from the dp table.

A more common and reliable approach is to use a one-dimensional DP table where dp[i] represents whether the first ‘i’ characters of ’s’ can be segmented into one or more dictionary words. Here’s how you could modify your code to implement this approach:

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class Solution {
public:
    unordered_set<string> words;
    vector<int> dp;

    bool wordBreak(string s, vector<string>& wordDict) {
        for(auto x: wordDict) words.insert(x);
        int n = s.length();
        dp.resize(n+1, -1); // -1 indicates that the state has not been solved yet
        return helper(0, s);
    }

    bool helper(int start, string s){
        if(start == s.length()) return true;
        if(dp[start] != -1) return dp[start];
        for(int end = start+1; end <= s.length(); end++){
            if(words.find(s.substr(start, end-start)) != words.end() && helper(end, s)){
                dp[start] = 1;
                return true;
            }
        }
        dp[start] = 0;
        return false;
    }
};

In this code, we use a helper function to do the main work of the dynamic programming. The helper function checks if the substring from ‘start’ to ’end’ is in the dictionary, and if the remaining part of the string can be broken down into words in the dictionary. The result for each ‘start’ position is stored in dp[start].

For the test case s =“cbca” wordDict = [“bc”,“ca”]

Output = True Expected = False

So above is right from my side?

No, the output for the test case with s = "cbca" and wordDict = ["bc","ca"] should be False, not True.

Here’s why:

In order to return True, every character in string s must be part of a word in wordDict, with the words being contiguous (i.e., not scrambled or rearranged). The words from wordDict can be reused any number of times.

In this case, there’s no way to break “cbca” entirely into words from wordDict. The word “bc” does appear in “cbca”, but there’s no way to account for the “c” at the beginning and the “a” at the end with the given wordDict.

You can try it yourself by attempting to combine the words in wordDict in different orders and repetitions, and you’ll see that it’s impossible to create “cbca” with only these words.

So, the output for this test case should indeed be False. If your solution is giving True, there may be a bug in your implementation.

why my memoziation code give tle

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class Solution {
  bool solve(string s, vector < string > & wordDict, string st, unordered_map < string, bool > & mp) {
    if (st == s) {
      return true;
    }
    if (st.length() > s.length()) {
      return false;
    }
    if (mp.find(st) != mp.end()) {
      return mp[st];
    }
    for (int i = 0; i < wordDict.size(); i++) {
      if (solve(s, wordDict, st + wordDict[i], mp)) {
        return mp[st] = true;
      }
    }
    return mp[st] = false;
  }
  public: bool wordBreak(string s, vector < string > & wordDict) {
    unordered_map < string, bool > mp;
    return solve(s, wordDict, "", mp);
  }
};

The reason your code is getting a Time Limit Exceeded (TLE) error is because the approach you’ve taken has a high time complexity.

The problem lies in the fact that you are trying all combinations of words in the wordDict to form the string ’s’. This results in a lot of repeated computation as you may be constructing the same string multiple times.

In your recursive function ‘solve’, you are generating all possible combinations of the words from the wordDict by appending them to ‘st’, and for each combination, you’re checking if it matches ’s’. Since you’re constructing all combinations of words, the complexity of this operation is exponential, which results in the TLE.

A more efficient approach would be to use dynamic programming (DP) where you’d keep track of whether each substring of ’s’ can be broken down into words from wordDict. You could use a boolean DP array where dp[i] represents whether the substring s[0...i] can be broken down into words from the dictionary. This way, you avoid repeated computation and reduce the time complexity significantly.

Here’s a sample implementation of this approach:

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class Solution {
public:
    bool wordBreak(string s, vector<string>& wordDict) {
        unordered_set<string> dict(wordDict.begin(), wordDict.end());
        vector<bool> dp(s.size() + 1);
        dp[0] = true;
        for (int i = 1; i <= s.size(); ++i) {
            for (int j = i - 1; j >= 0; --j) {
                if (dp[j] && dict.find(s.substr(j, i - j)) != dict.end()) {
                    dp[i] = true;
                    break;
                }
            }
        }
        return dp[s.size()];
    }
};

In this code, dp[i] is set to true if there exists a j < i such that dp[j] is true and the substring s[j…i] is in the word dictionary. We start with dp[0] as true because an empty string can always be segmented.

This implementation has a time complexity of O(n^2) and space complexity of O(n), where n is the length of the string ’s’. This is much more efficient than the exponential time complexity of the recursive approach without DP.

Input s =“bb” wordDict =[“a”,“b”,“bbb”,“bbbb”] Expected: true why??? s does’nt contain “a” anywhere… so how it can aspect true ??? am i missing somthing?

The Word Break problem asks if the given string s can be segmented into a space-separated sequence of one or more dictionary words, where each word is from the given wordDict. A word can be reused any number of times.

For your given input s = "bb" and wordDict = ["a", "b", "bbb", "bbbb"], the result is True because the string s can be broken down into words from wordDict like so: s = "b" + "b".

The presence of “a” in wordDict doesn’t mean it must be used to form the string s. It’s just one of the available words that you can use. In this case, it’s not used at all, but that doesn’t affect whether s can be segmented into words from wordDict.

So you are correct, ’s’ doesn’t contain ‘a’, but ’s’ can still be broken down entirely using other words from the wordDict. That’s why the expected output is True.

Does anyone know why using backtracking will TLE in this problem?

The “Word Break” problem can be solved using a backtracking approach, but it often results in a time limit exceeded (TLE) error for larger inputs. Here’s why:

Backtracking is a strategy for finding all (or some) solutions to computational problems, notably constraint satisfaction problems, that incrementally builds candidates for the solutions, and abandons a candidate (“backtracks”) as soon as it determines that the candidate cannot possibly be extended to a valid solution.

The classic backtracking approach for this problem involves trying all possible break points and recursively checking whether each prefix can be broken using the word dictionary. If you try to solve it this way, you’ll end up with a solution that has a time complexity of O(2^n), where n is the length of the string. This is because in the worst-case scenario, you’re trying all possible partitions of the string.

For example, consider a string of length n containing only ‘a’, and a dictionary containing a single word ‘a’. There are 2^(n-1) ways to partition this string. Given that the string length can be up to 300, it’s not feasible to check all these partitions in a reasonable amount of time.

To avoid the TLE, you should use dynamic programming (DP) which reduces the time complexity to O(n^2). The DP solution involves precomputing whether all substrings of the given string can be broken using the word dictionary, and using these precomputed results to calculate the result for the larger string.

can you explain why this happens wordDict = [“a”,“b”,“bbb”,“bbbb”] s = “bb”

In the Word Break problem, you are given a non-empty string s and a dictionary wordDict containing a list of non-empty words. The problem asks whether s can be segmented into a space-separated sequence of one or more dictionary words.

For your given input s = "bb" and wordDict = ["a", "b", "bbb", "bbbb"], we are asked if we can break s into a sequence of one or more words from wordDict.

Here’s how we can do that:

Break s into two words: s = "b" + "b"

Both “b” and “b” are available in wordDict, so the output is True.

Notice that even though “a”, “bbb”, and “bbbb” are in wordDict, we don’t have to use them. The problem only asks if it’s possible to break s into words from wordDict, not whether we can use all the words in wordDict.

So even though s doesn’t contain “a”, “bbb”, or “bbbb”, we can still break it into words from wordDict, and therefore the output is True.

Interestingly you could also start from index 0 and then increase by each word in wordDict

mem[i] = any(s[i, i+len(word)] == word and dp[i + len(word)])

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def is_word_possible(start,mem):
            if start == N:
                return True
            if mem[start] != -1:
                return mem[start]
            matches = []
            for word in wordDict:
                if s[start : start + len(word)] == word: # match
                    matches.append(word)

            if not matches:
                mem[start] = False
                return False # exhuasted all words 
            mem[start] = any(is_word_possible(start + len(word), mem) for word in matches)
            return mem[start]


        N = len(s)
        mem = [-1] * ( N + 1)
        mem[N] = True

        return is_word_possible(0, mem)

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, this code determines if s can be segmented into a space-separated sequence of one or more dictionary words.

This function works by using dynamic programming and depth-first search:

The is_word_possible(start, mem) function checks if it is possible to segment the string s starting from index start. It uses memoization to avoid redundant computations: if mem[start] is not -1, it means this subproblem has already been solved, so it returns the stored result.
For each word in wordDict, if it matches the substring of s starting from index start, it is added to the matches list.
If no word matches, it means it is not possible to segment the string s starting from index start, so it sets mem[start] to False and returns False.
If there are matches, it checks if it is possible to segment the rest of the string s starting from index start + len(word) for each matched word. If it is possible for any word, it sets mem[start] to True and returns True.
It initializes mem to be a list of -1 with length N + 1, where N is the length of s. mem[N] is set to True because it is always possible to segment an empty string.
Finally, it calls is_word_possible(0, mem) to check if it is possible to segment the whole string s.

This function returns True if s can be segmented into a space-separated sequence of one or more dictionary words, and False otherwise. It has a time complexity of O(N^2 * M) and a space complexity of O(N), where N is the length of s and M is the number of words in wordDict.

These approaches treat each letter (index) as a graph node. You can think it as you can move to next node (index) only if you have a word that allows it and that only happens if at index i there is a word that matches the whole word so you would move i + word.size(). If you reach the end, then you are done. If you reach an index you’ve already visited then no need to explore further as you know that it doesn’t reach the end. The solution doesn’t use substring() and instead uses s.startsWith(prefix, beginIdx) making it more optimal since it doesn’t have to iterate the whole word if there is no a match it’d break earlier.

Your understanding is correct. This problem can indeed be viewed as a graph problem, where each index in the string represents a node, and there is a directed edge from node i to node j if and only if the substring from i to j is a word in the dictionary.

This approach is very efficient because it treats the problem as a graph traversal problem and uses depth-first search (DFS) to explore the graph. By using memoization, it avoids redundant computations and unnecessary explorations: if an index has already been visited and does not lead to a solution, there’s no need to visit it again.

The use of s.startsWith(prefix, beginIdx) instead of substring() is also a great optimization. The startsWith() function checks if the string starts with the specified prefix at the specified index, and it can terminate early as soon as a mismatch is found, without needing to check the whole word. This can significantly reduce the time complexity when the word to check is long and does not match the string.

So, overall, this approach is a very efficient and elegant way to solve the Word Break problem, by combining the ideas of graph theory, dynamic programming, and string matching.

In the bfs approach, rather than iterating till end we can just iterate till 20, since length of a word in wordDict is at max 20.

Yes, that’s correct. If you know that the maximum length of a word in your dictionary (wordDict) is 20, you can optimize your breadth-first search (BFS) approach by only checking substrings of up to length 20 at each step. This can greatly reduce the number of iterations in your loop and potentially improve performance.

Here’s an example of what that might look like in Python:

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def wordBreak(s, wordDict):
    wordDict = set(wordDict)
    queue = deque([0])
    visited = [0]*len(s)

    while queue:
        start = queue.popleft()
        if visited[start] == 0:
            for end in range(start+1, min(start+21, len(s)+1)):
                if s[start:end] in wordDict:
                    queue.append(end)
                    if end == len(s):
                        return True
            visited[start] = 1
    return False

In the above code, min(start+21, len(s)+1) is used to ensure that we don’t go beyond the end of the string s. This function will return True if the string s can be segmented into a space-separated sequence of one or more dictionary words, and False otherwise.

Note that the space complexity of this BFS solution is O(n), where n is the length of the string s, because in the worst case scenario, we might need to store all the indices in the queue. The time complexity is O(n*m), where m is the maximum length of a word in the dictionary, because in the worst case scenario, we might need to check all substrings of length up to m for all indices in the string.

I think it’s worth nothing that using a HashSet then iterating through pairs is O(n^2 * k +m) runtime whereas the other DP methods of going through the list wordDict (method 2 and 3 in the article) is O(n * m * k). HashSet approach could be better for if there was a lot of wordDict items (large m).

Yes, you’re absolutely right. The choice of the right data structure and the right algorithm can greatly affect the time complexity and thus the performance of the solution.

In the Word Break problem, using a HashSet for the dictionary and then iterating through pairs in the string has a time complexity of O(n^2 * k + m), where:

n is the length of the string s,
k is the maximum length of a word in the dictionary,
m is the number of words in the dictionary.

This is because, in the worst case, we have to check all pairs of indices in the string, and for each pair, we need to check if the corresponding substring is in the dictionary. The additional m term is for converting the word list into a HashSet.

On the other hand, the dynamic programming (DP) methods that iterate through the word list have a time complexity of O(n * m * k). This is because, for each index in the string, we need to check all words in the dictionary, and for each word, we need to check if it matches the substring of s ending at the current index.

So, if the number of words in the dictionary m is very large, the HashSet approach could indeed be more efficient, because the HashSet allows checking the existence of a word in constant time on average, whereas iterating through the word list takes linear time in the number of words. However, if the length of the string n is much larger than the number of words m, the DP methods could be more efficient. As always, the choice of the right approach depends on the specific characteristics of the problem and the input data.

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public class Solution {
    public boolean wordBreak(String s, Set<String> dict) {
        
        boolean[] f = new boolean[s.length() + 1];
        
        f[0] = true;
        
        
        /* First DP
        for(int i = 1; i <= s.length(); i++){
            for(String str: dict){
                if(str.length() <= i){
                    if(f[i - str.length()]){
                        if(s.substring(i-str.length(), i).equals(str)){
                            f[i] = true;
                            break;
                        }
                    }
                }
            }
        }*/
        
        //Second DP
        for(int i=1; i <= s.length(); i++){
            for(int j=0; j < i; j++){
                if(f[j] && dict.contains(s.substring(j, i))){
                    f[i] = true;
                    break;
                }
            }
        }
        
        return f[s.length()];
    }
}

The provided Java code is a solution to the Word Break problem, which is to determine if a string s can be segmented into a space-separated sequence of one or more dictionary words. The dictionary is given as a Set of strings.

The solution uses dynamic programming to solve the problem, and there are two versions of the dynamic programming approach in the code, but the second one (Second DP) is active while the first one (First DP) is commented out.

Here’s how the active code (Second DP) works:

An array f of boolean values is initialized, where f[i] is true if the substring of s up to index i can be segmented into dictionary words, and false otherwise. The length of f is s.length() + 1 to account for the empty string, which can always be segmented (no words). Thus, f[0] is initialized to true.
The code then iterates over the string s. For each index i, it checks all substrings of s ending at i.
If the substring from j to i is a word in the dictionary, and the substring of s up to j can be segmented into dictionary words (i.e., f[j] is true), then f[i] is set to true.
Finally, the code returns f[s.length()], which indicates if the whole string s can be segmented into dictionary words.

The time complexity of this solution is O(n^2 * m), where n is the length of the string s and m is the average length of a word in the dictionary. This is because, in the worst case, the code needs to check all substrings of s and for each substring, it needs to create a new string and check if it is in the dictionary. The space complexity is O(n), which is the space used by the array f.

The commented-out code (First DP) is a slightly different dynamic programming approach that iterates over the words in the dictionary instead of the substrings of s. Its time complexity is also O(n^2 * m), but it might be slower in practice if the number of words in the dictionary is large, because it checks all words for each index in the string.

First DP: [length of s][size of dict][avg length of words in dict] Second DP: [length of s]^3

BTW, for this kind of problem, which time complexity is [length of s][size of dict][avg length of words in dict]. We can usually remove [size of dict] by using Tire, remove [avg length of words in dict] by KMP, and what’s more, remove both [size of dict] and [avg length of words in dict] by AC-automata. And of course these are all doable for this problem.

Yes, you’re correct about the time complexity analysis:

First DP: The time complexity is O(n * m * k), where n is the length of the string s, m is the number of words in the dictionary, and k is the average length of a word in the dictionary. This is because for each index in the string, the code checks all words in the dictionary, and for each word, it needs to create a new string and check if it equals the word.
Second DP: The time complexity is O(n^3), where n is the length of the string s. This is because the code needs to check all substrings of s, and for each substring, it needs to create a new string and check if it is in the dictionary. The creation of a new string takes linear time in the length of the string, hence the cubic time complexity.

You’re also correct that we can use advanced data structures and algorithms to reduce the time complexity:

Trie: A trie can be used to store the dictionary words and check if a substring is a word in the dictionary in linear time in the length of the substring, which removes the m term from the time complexity.
KMP (Knuth-Morris-Pratt): The KMP algorithm can be used to search for a word in a string in linear time in the length of the string, which removes the k term from the time complexity.
AC-automata (Aho-Corasick): The Aho-Corasick algorithm is a string matching algorithm that can search for all words in a dictionary in a string in linear time in the length of the string and the number of found matches, which removes both the m and k terms from the time complexity.

However, these methods require more complex code and may not be necessary if the size of the dictionary and the length of the words are not very large. It’s always a trade-off between time complexity and code complexity.

I think the second dp could be optimized to O(s.length) * (O(wordDict[i].length) ^ 2) because we can skip if substring is larger than 20 or the max length of element in wordDict (mentioned in constraints that 1 <= wordDict[i].length <= 20 while 1 <= s.length <= 300). Please correct me if I’m wrong.

You’re correct. If we know that the maximum length of a word in the dictionary is bounded (20 in this case), we can indeed optimize the second DP approach by only checking substrings of up to length 20 at each step. This can greatly reduce the number of iterations in your loop.

Here’s the revised time complexity analysis with this optimization:

The outer loop runs in O(n), where n is the length of the string s.
For each index in the string, we check all substrings ending at this index with length up to 20. This takes O(k^2), where k is the maximum length of a word in the dictionary (20 in this case).
For each substring, we need to create a new string and check if it is in the dictionary. The creation of a new string takes O(k) time and the dictionary check takes O(1) time if the dictionary is a HashSet.

So the total time complexity is O(n * k^2). This is a significant improvement over the original O(n^3) time complexity. Note that the space complexity remains O(n), which is the space used by the DP array.

So you’re absolutely correct, and this is a great observation! By understanding the problem constraints and applying them in your algorithm, you can achieve significant optimizations.

Is string.substring() method O(n) ?

Yes, the substring() method in Java is O(n) in terms of time complexity, where n is the length of the substring. This is because substring() returns a new String object that contains a copy of the specified substring. Therefore, it takes time proportional to the length of the substring to create the new string.

Here’s how it works:

substring(int beginIndex): This method returns a new string that is a substring of the original string, starting from beginIndex and extending to the end of the string. The time complexity is O(n - beginIndex), where n is the length of the original string.
substring(int beginIndex, int endIndex): This method returns a new string that is a substring of the original string, starting from beginIndex and extending to endIndex - 1. The time complexity is O(endIndex - beginIndex).

In both cases, the time complexity is proportional to the length of the substring. Therefore, in algorithms that call substring() in a loop, it’s important to consider the cost of this method in the time complexity analysis.

f[i] stands for whether subarray(0, i) can be segmented into words from the dictionary. So f[0] means whether subarray(0, 0) (which is an empty string) can be segmented, and of course the answer is yes.

The default value for boolean array is false. Therefore we need to set f[0] to be true.

Yes, that’s correct. The f[i] in this context is used to represent whether the substring of the input string up to index i can be segmented into words from the dictionary.

As you correctly mentioned, f[0] represents the case where the substring is an empty string. Since an empty string doesn’t need any segmentation to be valid (it’s already “segmented” into zero words), we initialize f[0] to true.

The rest of the f array is initialized to false as the default, representing that we haven’t yet determined if the corresponding substrings can be segmented into dictionary words.

As the algorithm progresses, each f[i] is updated based on the check if there’s some j < i such that f[j] is true and the substring from j to i is a word from the dictionary. The final result, f[s.length()], tells us if the entire string can be segmented into words from the dictionary.

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public boolean wordBreak(String s, List<String> wordDict) {
    boolean[] dp = new boolean[s.length() + 1];
    Set<String> set = new HashSet<>();
    set.addAll(wordDict);

    dp[0] = true;
    for (int i = 1; i <= s.length(); i++) {
        for (int j = i - 1; j >= 0; j--) {
            dp[i] = dp[j] && set.contains(s.substring(j, i));
            if(dp[i]) break;
        }
    }
    return dp[s.length()];
}

The Java code you’ve posted is another solution to the Word Break problem. It uses dynamic programming (DP) and a HashSet to store the words in the dictionary for efficient lookups. Here’s a step-by-step breakdown of how the code works:

The code first initializes an array dp of boolean values, where dp[i] is true if the substring of s up to index i can be segmented into dictionary words, and false otherwise. The length of dp is s.length() + 1 to account for the empty string, which can always be segmented (no words). Thus, dp[0] is initialized to true.
The dictionary words are then added to a HashSet named set for efficient lookups.
The code then iterates over the string s. For each index i, it checks all substrings of s ending at i.
If the substring from j to i is a word in the dictionary, and the substring of s up to j can be segmented into dictionary words (i.e., dp[j] is true), then dp[i] is set to true. It then breaks out of the inner loop, as it has found a valid segmentation up to index i.
Finally, the code returns dp[s.length()], which indicates if the whole string s can be segmented into dictionary words.

The time complexity of this solution is O(n^2 * m), where n is the length of the string s and m is the maximum length of a word in the dictionary. This is because, in the worst case, the code needs to check all substrings of s, and for each substring, it needs to create a new string and check if it is in the dictionary. The space complexity is O(n), which is the space used by the DP array.

This is a correct and efficient solution to the Word Break problem, as it combines dynamic programming with a HashSet for fast word lookups.

so does this check if each letter exists in set? then I don’t understand how example 3 returns false

The Word Break problem is not about checking if each letter exists in the set (dictionary), but rather whether the string s can be segmented into a sequence of one or more words that exist in the dictionary.

The algorithm in the provided code checks substrings of s, not individual letters. For each position i in the string s, it checks all substrings that end at position i and start at some position j (where j ranges from i-1 down to 0). If the substring from j to i is in the dictionary and the substring up to j can be segmented into words from the dictionary, then it sets dp[i] to true.

In the third example from the LeetCode problem (assuming it’s the one with s = "catsandog" and wordDict = ["cats","dog","sand","and","cat"]), the function returns false because there is no way to segment the entire string s into a sequence of words that exist in the dictionary. The substring “andog” cannot be segmented into dictionary words, so dp[s.length()] is false.

So the function doesn’t just check if each letter exists in the set, but rather if each substring of s can be segmented into a sequence of dictionary words.

what is the logic behind assigning dp[i] first and then checking and breaking if dp[i] is true. I tried something like

if(dp[j] && set.contains(s.substring(j, i))) { dp[i] = true; break; } and it did not work. can someone please help me understand why?

The difference between the two versions of code is the order of evaluation in the condition.

In your code, the && operator is short-circuiting. This means that if the first operand is false, the second operand is not evaluated at all because the result of the && operation can be determined to be false without evaluating the second operand. In this case, if dp[j] is false, set.contains(s.substring(j, i)) is not evaluated, and dp[i] is not set to true.

In the original code, dp[i] is assigned the result of dp[j] && set.contains(s.substring(j, i)) regardless of the value of dp[j]. If dp[j] is false, dp[i] is set to false. If dp[j] is true but set.contains(s.substring(j, i)) is false, dp[i] is again set to false. Only if both dp[j] and set.contains(s.substring(j, i)) are true is dp[i] set to true.

The break statement is executed if dp[i] is true after the assignment. This is to avoid unnecessary computations: once we have found a segmentation ending at i that is valid, we don’t need to check other segmentations ending at i.

So the logic behind the original code is to assign dp[i] first and then check if dp[i] is true to possibly break the loop early. Your code only assigns dp[i] = true; when both conditions are met and dp[j] is true, which might not always be the case in the early stages of the DP calculation, leading to different results.

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/**
 * Initial recursion + memoization solution in Java.
 *
 */
class Solution {
    public boolean wordBreak(String s, List<String> wordDict) {
        // create the memoization array to save results and avoid repeat computations
        Boolean[] canBreak = new Boolean[s.length()];

        // convert the list into set for faster lookup
        Set<String> wordSet = new HashSet<>(wordDict);

        // recursion with memoization
        return helper(s, 0, wordSet, canBreak);
    }

    private boolean helper(String s, int startIdx, Set<String> wordSet, Boolean[] canBreak) {
        // in case we've reached the end of string, return true
        if (startIdx == s.length()) return true;
        // else if we've already computed on current substring before
        if (canBreak[startIdx] != null) return canBreak[startIdx]; // auto-unboxing

        boolean res = false;
        // iterate through all indices after startIdx, explore every possible word
        for (int i = startIdx + 1; i <= s.length(); i++) {
            String currWord = s.substring(startIdx, i);
            // skip if this is not a word in the input dictionary
            if (!wordSet.contains(currWord)) continue;
            // recursively call upon the rest of string
            if (helper(s, i, wordSet, canBreak)) {
                res = true;
                break;
            }
        }
        // add result to memo and return the result
        canBreak[startIdx] = res;
        return res;
    }
}

The Java code you’ve posted is a recursive solution with memoization to the Word Break problem. Here’s how it works:

It first initializes an array canBreak of Boolean objects, where canBreak[i] is true if the substring of s from index i to the end can be segmented into dictionary words, and null if this has not been computed yet. It also converts the word list into a HashSet for efficient lookups.
It then calls the recursive helper function on the whole string s with start index 0.
The helper function checks if it has reached the end of the string. If it has, it returns true because an empty string can always be segmented (no words).
If the helper function has not reached the end of the string, it checks if it has already computed the result for the current start index. If it has, it returns the stored result.
If the result for the current start index has not been computed yet, the helper function iterates over all possible words starting from the current start index. If it finds a word from the dictionary, it recursively calls itself on the rest of the string.
If the recursive call returns true, it means the rest of the string can be segmented into dictionary words, so the helper function sets res to true and breaks the loop.
Finally, the helper function stores the result in the canBreak array and returns the result.

The time complexity of this solution is O(n^2), where n is the length of the string s. This is because, in the worst case, the helper function needs to check all substrings of s and for each substring, it needs to create a new string and check if it is in the dictionary. However, because of memoization, each substring is only checked once, so the time complexity is effectively linear in the number of substrings, which is quadratic in the length of s.

The space complexity is O(n), which is the space used by the call stack (in the worst case, the depth of the recursion can be as large as the length of s) and the memoization array.

can you provide an example when below condition returns true or false .

if (canBreak[startIdx] != null) return canBreak[startIdx];

The condition if (canBreak[startIdx] != null) return canBreak[startIdx]; is checking if the result for the substring of s starting from startIdx has been computed before. If it has been computed before, the function immediately returns the stored result, avoiding redundant computation. This is the essence of memoization.

Let’s take a concrete example to illustrate this. Suppose we have the following input:

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String s = "leetcode";
List<String> wordDict = Arrays.asList("leet", "code");

The function wordBreak(s, wordDict) would call helper(s, 0, wordSet, canBreak).

At first, all elements in the canBreak array are null, indicating that the result for all substrings are unknown.

The helper function first checks the substring “leet” (from index 0 to 3), finds it in the dictionary, and recursively calls helper(s, 4, wordSet, canBreak).

This recursive call checks the substring “code” (from index 4 to 7), finds it in the dictionary, and recursively calls helper(s, 8, wordSet, canBreak).

Since startIdx is now equal to s.length(), the function returns true.

The true result is then propagated back up the recursion, and canBreak[4] and canBreak[0] are set to true.

Now, suppose the function is called again with the same string s but a different dictionary:

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wordDict = Arrays.asList("leet", "co", "de");

The function wordBreak(s, wordDict) would again call helper(s, 0, wordSet, canBreak).

But this time, when it checks if (canBreak[startIdx] != null), the condition is true because canBreak[0] is true from the previous computation. So the function immediately returns true, without any further computation.

If the condition if (canBreak[startIdx] != null) is false, it means the result for the current startIdx is not known yet, and the function needs to compute it by checking all possible words starting from startIdx.

Still confused about initial state, empty string can’t be further segmented anyways, so why by f[0]=true?

The f[0] = true is an initial condition of the dynamic programming solution to the Word Break problem. Here f[i] represents whether the substring from index 0 to i (exclusive) can be segmented into words from the dictionary.

When i = 0, the substring from index 0 to i is an empty string. The empty string is considered as being able to be segmented into words from the dictionary, although the segmentation contains no words. This is because there’s no word in the dictionary that needs to be removed from the empty string to make it possible to be segmented into words from the dictionary.

Setting f[0] = true is crucial for the dynamic programming solution. It serves as the base case for the dynamic programming. It allows us to handle the case where a word in the dictionary is a prefix of the string s. If the prefix word ends at index i, we can say that f[i] is true because we can segment the prefix word (from index 0 to i) and the rest of the string (from index i to s.length()) can be considered separately due to the principle of optimality.

So the f[0] = true is not about whether the empty string can be further segmented, but rather about whether it can be segmented into words from the dictionary, which is trivially true because there’s no need to remove any word from the empty string.

Are you assuming that array indexing starts from 1? If it starts from 0 and f is inclusive on the ranges, then how does f[0] represent empty string, shouldn’t it represent the first character? Or do you mean that f[i] represents segmentation of subarray(0,i-1) instead of subarray(0,i)?

Apologies for any confusion. You’re correct. In the context of the problem and the code we’ve been discussing, f[i] represents whether the substring from index 0 to i (exclusive) can be segmented into words from the dictionary. So f[i] actually represents the subarray from 0 to i-1 (inclusive), or equivalently, the substring up to i but not including i.

So when i = 0, f[0] indeed represents an empty string, because there are no characters from index 0 up to but not including 0. Hence, we initialize f[0] = true to denote that the empty string can be segmented into words from the dictionary (trivially, because there are no words).

Then, for i > 0, f[i] represents whether the substring from index 0 to i-1 can be segmented into words from the dictionary. In this context, the indices for f are one step ahead of the indices for s. This is a common pattern in dynamic programming problems involving strings or arrays.

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public class Solution {
    public class TrieNode {
        boolean isWord;
        TrieNode[] c;
        
        public TrieNode(){
            isWord = false;
            c = new TrieNode[128];
        }
    }
    
    public void addWord(TrieNode t, String w) {
        for (int i = 0; i < w.length(); i++) {
            int j = (int)w.charAt(i); 
            if (t.c[j] == null) t.c[j] = new TrieNode();
            t = t.c[j];
        }
        t.isWord = true;
    }
    
    public boolean wordBreak(String s, Set<String> wordDict) {
        TrieNode t = new TrieNode(), cur;
        for (String i : wordDict) addWord(t, i);
        char[] str = s.toCharArray();
        int len = str.length;
        boolean[] f = new boolean[len + 1];
        f[len] = true;
        
        for (int i = len - 1; i >= 0; i--) {
            //System.out.println(str[i]);
            cur = t;
            for (int j = i; cur != null && j < len ; j++) {
                cur = cur.c[(int)str[j]];
                if (cur != null && cur.isWord && f[j + 1]) {
                    f[i] = true;
                    break;
                }
            }
        }
        return f[0];
    }
}

The Java code is a solution to the Word Break problem using a Trie data structure. A Trie is a tree-like data structure that is used to store a collection of strings. It has a number of uses, one of which is efficient prefix matching, which is what’s being utilized in this solution. Let’s go through the code:

A TrieNode class is defined with a boolean field isWord to denote if the node represents the end of a word, and an array c to store references to child nodes.
The addWord(TrieNode t, String w) method adds a word w to the Trie rooted at t. It iterates over the characters in w, for each character creating a new node if necessary, and moving down to the child node corresponding to the current character. When all characters are processed, it sets isWord to true for the last node.
The wordBreak(String s, Set<String> wordDict) method first builds the Trie for the word dictionary. It initializes a boolean array f where f[i] represents whether the substring from index i to the end of s can be segmented into words from the dictionary.
It then iterates over the string s in reverse. For each index i, it tries to match a word in the Trie starting from i. If it finds a word and the substring from the end of the word to the end of s can be segmented (i.e., f[j + 1] is true), it sets f[i] to true.
Finally, it returns f[0], which indicates whether the whole string s can be segmented into words from the dictionary.

This solution has a time complexity of O(n^2) in the worst case where n is the length of the string s. This is because, in the worst case, it needs to check all substrings of s. The space complexity is O(n) for the DP array, plus the space for the Trie, which in the worst case can be O(n*m) where m is the length of the longest word in the dictionary. This Trie-based approach can be more efficient than the previous solutions if there are many words in the dictionary or the words are long, because it can check if a substring is a word in the dictionary in time proportional to the length of the substring.

Could you please help me understand how does searched array helped optimize the solution?

The searched array (or similar structure in other implementations) is used as a form of memoization to avoid unnecessary repeated work. The idea is to remember (memoize) whether we’ve already attempted to solve the subproblem for a given substring. If we have, we don’t need to do it again.

In the context of the Word Break problem, if we’ve already tried to segment a given substring and determined that it can’t be segmented into dictionary words, then if we encounter the same substring again (as a part of a larger string), we can immediately know that it can’t be segmented, without having to go through the same computations again.

Without this optimization, in cases where the string is long and the dictionary doesn’t contain words that can segment the string, the algorithm can end up attempting to segment the same substrings over and over again, leading to unnecessary repeated work.

By keeping track of which substrings we’ve already tried to segment, we can avoid this repeated work and make the algorithm more efficient. This is particularly important in cases where the string is long and/or the dictionary is large. The time saved can be significant because the Word Break problem has an exponential number of possible segmentations in the worst case, but with memoization, each substring is considered at most once.

We can replace the HashSet with a Trie and then the time complexity goes from O(n^3) to O(n^2). Space complexity stays the same (although real world space consumption might go down if there are lots of common prefixes between the words in the dictionary). Using a Trie seems to be the best possible solution.

Yes, you’re correct. In the Word Break problem, if you replace the HashSet with a Trie, you can potentially reduce the time complexity from O(n^3) to O(n^2). This is because with a Trie, you can check if a substring is a word in the dictionary in time proportional to the length of the substring, rather than having to create a new substring and check if it’s in the HashSet.

Trie is a prefix tree, where each node represents a character and each path from the root to a node represents a word. This makes Trie extremely efficient for prefix-based searches, which is exactly what we need for the Word Break problem.

You’re also correct that the space complexity remains the same, O(n), for the dynamic programming array, plus the space for the Trie, which in the worst case can be O(m*l), where m is the number of words in the dictionary and l is the maximum length of a word. However, in practice, if there are many common prefixes between the words in the dictionary, the space consumption might indeed be less.

So yes, using a Trie is a great optimization for the Word Break problem, especially when the dictionary is large or the words are long. It’s a great example of how a suitable data structure can significantly improve the performance of an algorithm.

if you do : for(int j = i-1 ; j >= 0; j–) in the inner loop, will improve performance a little bit

Yes, starting the inner loop from i-1 and decrementing j until it reaches 0 can indeed improve performance in some cases for the Word Break problem. This approach tries the longest possible word first, then the shorter ones.

In cases where there are longer words in the dictionary that match the segments of the string, this approach could lead to an early match, allowing the algorithm to break out of the loop sooner, potentially skipping a lot of unnecessary checks.

However, the overall time complexity remains the same (O(n^2 * m) where n is the length of the string and m is the maximum length of a word in the dictionary) as we’re still performing the same number of operations in the worst-case scenario.

But yes, in practice, this could lead to performance improvements depending on the specific inputs.

the problem doesnt specify but looks like the test cases only have lowercase letters. In that case, c = new TrieNode[26] suffices, as when you get the children node, you can simply (str.charAt(i) - ‘a’) to get accurate letter

Yes, you’re correct. If the problem only involves lowercase letters, then you can indeed optimize the TrieNode by using an array of size 26 instead of 128, as there are 26 lowercase letters in the English alphabet. This would save a significant amount of space when building the Trie.

The line where you get the children node would then look something like this:

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cur = cur.c[(int)str[j] - 'a'];

This is because the character ‘a’ corresponds to ASCII value 97, ‘b’ corresponds to 98, and so on. So str.charAt(i) - 'a' would convert the character at position i in the string into a number between 0 and 25 (inclusive), which can be used as an index in the c array.

This is a good example of how understanding the constraints and characteristics of the input can help optimize an algorithm. However, always be sure to check the problem constraints and test cases to confirm such assumptions.

Can somebody explain why each TrieNode has 128 children? Ascii value of lower case letters are [97-122] and uppercase are [65-90]. So, 122 is enough right? Or is it something else?? Somebody explain please..!!

You’re correct that in the ASCII table, lowercase letters correspond to values between 97 and 122, and uppercase letters correspond to values between 65 and 90. However, the size of the array in the TrieNode doesn’t necessarily have to match these ranges exactly.

Choosing an array size of 128 is a design decision that provides enough space to cover all ASCII characters up to the ‘DEL’ character (ASCII value 127). This covers all standard ASCII characters, including lowercase and uppercase letters, digits, punctuation marks, control characters, and a few others. This allows the Trie to handle any ASCII string, not just lowercase or uppercase letters.

If the problem specifies that only lowercase or uppercase letters will be used, then you could indeed reduce the size of the array to exactly cover the necessary range. For example, if only lowercase letters are used, an array of size 26 would suffice, and the ASCII value could be normalized by subtracting the value of ‘a’ or ‘A’ to get an index in the range 0-25.

However, using an array of size 128 by default allows the Trie to handle any ASCII string without having to know in advance which characters will be used, making it more versatile. It’s a trade-off between memory usage and versatility.

I just read about substring API change, it was O(1) before, do you know why they changed to O(N)?

In earlier versions of Java (prior to Java 7), the substring() method had a time complexity of O(1). It achieved this by reusing the character array of the original string and creating a new String object with a different start index and end index. This approach was memory-efficient when the substring was small compared to the original string.

However, it had a potential memory leak issue. Even if you had a small substring of a very large string, as long as the small substring was in scope, the large character array it referenced couldn’t be garbage collected. This could lead to unexpectedly high memory usage.

For example, consider the case where you have a large text document stored in a string, and you use substring() to extract a small part of it. The small substring would hold a reference to the entire character array of the original large string, even though it only represents a small part of it.

To avoid this potential memory leak issue, the implementation of substring() was changed in Java 7 such that it copies the relevant part of the character array, resulting in a time complexity of O(n), where n is the length of the substring.

This change prevents the memory leak scenario described above. With the new implementation, when you create a substring, it doesn’t hold a reference to the original character array. Instead, it has its own copy of the characters it represents. This allows the garbage collector to free the memory of the original string if it’s no longer in scope, even if a substring of it is still in use. This change trades off some time efficiency for improved memory efficiency in certain scenarios.

why trie is faster than dp?

A Trie data structure is not inherently faster than dynamic programming (DP); the efficiency depends on the specific problem and how these methods are implemented.

In the Word Break problem, using a Trie can potentially speed up the process of looking up words in the dictionary. Here’s why:

In the DP solution, you would typically use a HashSet or similar data structure to store the dictionary words. To check if a substring of the input string is a word, you need to create a new substring and check if it’s in the HashSet. This has a time complexity of O(m) for creating the substring, where m is the length of the substring, plus O(1) for the HashSet lookup. Overall, this leads to a time complexity of O(n^2 * m) for the Word Break problem, where n is the length of the input string.

In the Trie-based solution, you can check if a substring is a word by traversing the Trie from the root to a node, one character at a time. This can be done in time proportional to the length of the substring, O(m), without having to create a new substring. This leads to a time complexity of O(n^2) for the Word Break problem.

So in this case, the Trie-based solution can be more efficient than the DP solution. However, the Trie also requires additional space to store the dictionary words, and it can be more complex to implement.

It’s also important to note that the DP solution and the Trie-based solution are not mutually exclusive. In fact, the Trie-based solution also uses DP to keep track of which substrings can be segmented. The Trie is just used to speed up the dictionary word lookup.

what if the dictionary is really large and we need to call the method search(word) multiple times

If the dictionary is large and we need to call the search method multiple times, a Trie data structure can be very efficient. Here’s why:

Space Efficiency: A Trie can be more space efficient when storing a large number of strings, especially if there are many common prefixes. This is because, in a Trie, each common prefix is stored only once.
Time Efficiency for Search: Searching for a word in a Trie takes time proportional to the length of the word, regardless of the size of the dictionary. This is very efficient compared to searching for a word in a list or an array, which can take time proportional to the size of the dictionary in the worst case.
Preprocessing Time: Building the Trie from the dictionary takes some time upfront (time complexity of O(M), where M is the total number of characters in all words in the dictionary). However, once the Trie is built, many searches can be performed quickly. If you need to perform a large number of searches, this upfront time can be well worth it.
Memory Considerations: Although Tries can be space efficient for storing the dictionary, they can still use significant memory if the dictionary is large and the words have few common prefixes. Also, each node in the Trie has an array to store its children, which can use a lot of memory if the number of possible characters is large.

In conclusion, if the dictionary is large and we need to call the search method multiple times, using a Trie can be an efficient solution, both in terms of space and time. However, the specific details can depend on the characteristics of the dictionary and the usage patterns of the search method.

Just want to clarify something, with the trie, it is O(n^2) whereas without the trie as in the original post, it is O(n^3) because constructing the substring is O(n)?

Yes, you are correct. Here’s why:

With Trie: In the case of using a Trie, the time complexity is O(n^2). This is because for each starting point in the string, you can traverse the Trie in O(m) time where m is the length of the substring (from the starting point to the end of the string). Given that you may try this for all n starting points in the string, this operation is O(n*m). But since m can be at most n (when the starting point is at the beginning of the string), the overall time complexity is O(n^2).
Without Trie: In the case without using a Trie, specifically when using a HashSet to store the dictionary words, the time complexity is O(n^3). This is because for each starting point in the string, you are creating a new substring for each possible ending point, and then checking if this substring is in the HashSet. Constructing the substring is an O(n) operation and you may do this for all n possible ending points, so this operation is O(n^2). And since you may try this for all n starting points in the string, the overall time complexity is O(n^3).

So, using a Trie can significantly reduce the time complexity for this problem by allowing efficient prefix matching without having to construct new substrings.

I found out I like to set dp array the same length as s, and it’s actually easier for me to understand what dp[i] means: whether the string ending with s.charAt(i) element can be split.

That’s completely fine! In dynamic programming, the specific representation of the DP array can sometimes be adjusted to fit your personal preferences or the specific requirements of the problem.

If setting the DP array to the same length as s and defining dp[i] as “whether the string ending with s.charAt(i) can be split” makes more sense to you, go for it! The key is to define it in a way that allows you to correctly express the relationship between different subproblems.

Remember, the overall idea and strategy of the dynamic programming approach remain the same: you’re breaking down the problem into smaller subproblems, solving each subproblem only once, storing the results of each subproblem to avoid duplicated work, and using the stored results to build up the solution to the overall problem. As long as your definition of the DP array aligns with this strategy, it should work fine.

you would have to deal with more edge case than len+1 solution

try think of it this way , dp[i] mean whether [0,i) match , it is a common skill to use [ ) for less bug coding , rather than [ ]

Your observation is indeed correct. In many dynamic programming problems, especially those involving strings or arrays, it’s often easier and less error-prone to define the problem in terms of intervals that include the start index and exclude the end index (i.e., half-open intervals [start, end)).

The key advantage of using this [start, end) convention is that it can eliminate off-by-one errors, which are a common type of bug in programming. When the end index is excluded, it’s clear that an array of size n has valid indices from 0 to n-1, and a subarray from index i to j has j-i elements.

In the context of the Word Break problem, defining dp[i] as “whether the substring [0, i) can be segmented into words from the dictionary” aligns with this [start, end) convention. It also makes the initialization of dp[0] as true (representing an empty string that can be trivially segmented) more intuitive.

That being said, if you choose to define dp[i] as “whether the string ending with s.charAt(i) can be segmented,” it’s still workable, but as you mentioned, you might have to handle more edge cases, and the code could be more error-prone.

What is a generative way to solve string problem?

A generative approach to solving problems, including string problems, is to generate all possible solutions and then check which ones satisfy the problem’s constraints.

For example, consider the problem of finding all permutations of a string. A generative solution would involve generating all possible permutations of the characters in the string, and then possibly checking them against some constraint.

Here’s a Python example of a generative approach to finding all permutations of a string:

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def generate_permutations(string):
    if len(string) == 1:
        return [string]

    permutations = []
    for i, char in enumerate(string):
        remaining_chars = string[:i] + string[i+1:]
        for perm in generate_permutations(remaining_chars):
            permutations.append(char + perm)
            
    return permutations

This function recursively generates all permutations of a string by selecting each character in turn, generating all permutations of the remaining characters, and then appending the selected character to the front of each of these permutations.

This generative approach is simple and intuitive, but it can be inefficient for large inputs due to the large number of possible solutions that it generates. For many problems, a more efficient approach is often possible, such as dynamic programming or greedy algorithms, which construct an optimal solution incrementally while avoiding unnecessary work.

That being said, a generative approach can be a good starting point for solving a problem, and it can sometimes lead to insights that help in finding a more efficient solution.

Is it the same as exhaustive search?

Yes, a generative approach is a form of exhaustive search. In an exhaustive search, you generate all possible solutions and check each one to see if it satisfies the problem’s constraints. A generative approach is a type of exhaustive search where you systematically generate all possible solutions.

For example, if you’re solving a problem where you need to find the best arrangement of some items, an exhaustive search would involve generating all possible arrangements and checking each one.

While exhaustive search (and therefore also the generative approach) is guaranteed to find a solution if one exists, it can be very inefficient because the number of possible solutions can be very large. Therefore, exhaustive search is typically only used for small problem sizes. For larger problem sizes, other techniques like greedy algorithms, divide-and-conquer, or dynamic programming are usually more efficient.

so for dp[i] = dp[index], this means that if we split at this ‘i’ the previous word(s) must have been splittable too huh?

Yes, exactly!

In the context of the Word Break problem, dp[i] = dp[index] essentially means that if the substring ending at index i can be split (i.e., dp[i] is true), then the substring ending at index index must also be able to be split (i.e., dp[index] is true).

When we say “substring ending at index i,” we mean the substring that starts from the beginning of the string and ends at index i. Similarly, “substring ending at index index” means the substring that starts from the beginning of the string and ends at index index.

This is why we iterate over each index i from left to right, and for each i, we check all the substrings ending at i to see if they can be split. If a substring can be split, then we mark dp[i] as true.

So dp[i] = dp[index] is a way of saying “if we can split the string up to index index, and the rest of the string from index to i is a word in the dictionary, then we can also split the string up to index i.”

Time complexity: n = length of string, m = size of dictionary First DP: O(n*m) Second DP: O(n^2)

In practice m = big set of english words, and n is some small block of text like a search result or paragraph. So n^2 will be faster than n*m. Good question to ask interviewer.

You’re correct in your time complexity analysis. The time complexity of the first DP approach, which checks each word in the dictionary for every position in the string, is indeed O(n*m). The second DP approach, which checks every possible substring for every position in the string, has a time complexity of O(n^2).

In practical terms, the size of the dictionary m can indeed be large (for example, if it includes all English words), while the length of the string n is often relatively small (like a single sentence or paragraph). In such cases, the O(n^2) approach might indeed be faster than the O(n*m) approach.

It’s worth noting that these are worst-case time complexities, and actual performance can depend on various factors, such as the specific contents of the string and the dictionary, and how the code is implemented. It’s also possible that other optimizations could be applied to improve performance further.

And you’re right, this could be a good point to discuss with an interviewer. It shows that you understand the time complexity of different approaches and can think critically about how the characteristics of the input might affect performance.

Nope, it should be First DP : - O(n * m * (avg of n)) ….O(n * m * n) Second Dp: - O(n^3)

you were missing the inner substring() method which will take O(N) time for every inner loop,making it N^3 time complexity.

You’re correct, and I apologize for the oversight. Your analysis is accurate for the typical dynamic programming approach to this problem.

The time complexity of the first dynamic programming solution is indeed O(n * m * l), where n is the length of the string, m is the size of the dictionary, and l is the average length of the words in the dictionary. This is because for each position in the string (n positions), we’re checking each word in the dictionary (m words), and comparing it to a substring of the input string, which takes time proportional to the length of the word (l).

The time complexity of the second dynamic programming solution is indeed O(n^3). This is because for each position in the string (n positions), we’re checking every possible substring that ends at that position. Checking a substring involves creating a new string, which takes time proportional to the length of the substring, leading to a worst-case time complexity of O(n^3).

Thank you for pointing out this error. It’s important to include the time complexity of the substring operation when analyzing these algorithms.

Can you give a little bit more explanation on how you come out both time complexity step by step?

I would be happy to explain the time complexity of both dynamic programming solutions in more detail.

First DP Solution: The time complexity is O(n * m * l), where n is the length of the string, m is the size of the dictionary, and l is the average length of the words in the dictionary.
Here’s how we arrive at this:
- We iterate over the entire length of the string s (n characters), so there is a factor of n.
- For each character in s, we then iterate over all the words in the dictionary (m words), so there is a factor of m.
- For each word in the dictionary, we potentially create a substring of s to compare with the dictionary word. The time complexity of creating a substring is proportional to the length of the substring. In the worst case, this can be the length of the entire string, but if we assume that the dictionary words are not too long on average, we can say this is proportional to the average length of the words in the dictionary (l).
- Hence, the overall time complexity is O(n * m * l).
Second DP Solution: The time complexity is O(n^3).
Here’s how we arrive at this:
- We iterate over the entire length of the string s (n characters), so there is a factor of n.
- For each character in s, we then iterate over all possible substrings that end at the current character. In the worst case, this can be the entire string up to the current character, so there is another factor of n.
- For each substring, we create a new string to check if it’s in the dictionary. The time complexity of creating a substring is proportional to the length of the substring. In the worst case, this can be the length of the entire string, so there is another factor of n.
- Hence, the overall time complexity is O(n^3).

Because after you processed a few words, it is likely that the match word will be found at the end of the finished part of the string, but not a really long word which begins from the beginning of the string

Yes, you’re right.

The solution you’re suggesting takes advantage of the “locality” property of the problem. If we’ve determined that a certain substring of the input string can be segmented into a sequence of words from the dictionary, we don’t need to revisit that substring again. We just need to check if the next segment, often towards the end of the processed part of the string, can be segmented.

In terms of the Word Break problem, let’s take an example. If s = "catsanddog" and wordDict = ["cat", "cats", "and", "sand", "dog"], once we’ve processed the substring “catsand”, we don’t need to check it again. We simply check if “dog” can be segmented, which it can.

This locality property is what makes the problem a good candidate for a dynamic programming solution. By remembering (or memoizing) the results of previous computations, we can avoid unnecessary work and improve efficiency. This makes it possible to solve the problem in O(n^2) time, where n is the length of the string, which is a significant improvement over the naive O(2^n) solution that involves checking all possible partitions of the string.

In your solution, you are building up the solution from smaller problems (substrings of the original string), which is the key idea behind dynamic programming.

Reverse the second loop, the program will get faster,like this:

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for(int j = i - 1; j >= 0; --j) {
    if(f[j] && wordDict.contains(s.substring(j, i))) {
        f[i] = true;
        break;
    }
}

Reversing the second loop can indeed potentially speed up the program for certain inputs, especially for cases where the valid words are likely to be found at the end of the substring rather than at the beginning.

The reason this works is because you are checking the most recently processed parts of the string first, and these are more likely to match the words in the dictionary, especially if the words are short and the string has repeated patterns.

Here’s what the code does:

It starts at the current index i and checks backwards (j = i - 1; j >= 0; --j).
For each index j, it checks if the substring ending at i and starting at j is a word in the dictionary (wordDict.contains(s.substring(j, i))).
It also checks if the substring ending at j was valid (f[j]). This is the dynamic programming part, where we use previous results to build up the solution.
If both conditions are met, it marks the substring ending at i as valid (f[i] = true) and breaks out of the loop.

This is still an O(n^2) solution in terms of time complexity, but it could run faster in practice due to the reasons explained above. However, the actual speedup will depend on the specific inputs to the problem.

I’m wondering if the Complexity is still O(n^2) if the Substring method is not O(1) - I understand that the Substring method in recent version of Java is O(n).

You’re correct in pointing out that the complexity of the substring method has changed in more recent versions of Java. Prior to Java 7, substring had a time complexity of O(1), but since Java 7, it creates a new string and thus has a time complexity of O(n), where n is the size of the substring.

In your specific scenario, where substring is used inside nested loops, the time complexity would indeed increase.

The outer loop runs n times, and for each iteration i of the outer loop, the inner loop can run up to i times. The substring method is called within the inner loop, taking time proportional to the length of the substring, which is up to n.

So, the overall time complexity of the method becomes O(n^3), not O(n^2), due to the additional factor of n introduced by the substring method.

This shows the importance of being aware of the time complexities of built-in methods when doing complexity analysis. It’s also a reminder that improvements in language implementations, like the change in the substring method in Java 7, can have significant impacts on the performance of algorithms.

Q&A

Approach using Dynamic Programming: We will utilize a boolean array dp of size n+1 where n is the length of the string s. The idea is to maintain a DP table such that dp[i] is True if the substring s[0:i] can be segmented into dictionary words, and False otherwise.

Initialize dp[0] as True because an empty string can always be segmented (base case).
For each index i from 1 to n, check every possible word ending at i and starting at some j before i. If the substring s[j:i] is a valid word (exists in the wordDict) and dp[j] is True (meaning the substring s[0:j] can be segmented), then set dp[i] to True.
If by the end, dp[n] is True, it means the entire string s can be segmented using words from the wordDict.

Let’s understand with an example: s = "leetcode" wordDict = ["leet", "code"]

Create a boolean array dp of size 9 (length of s + 1).
dp[0] = True
Start with i=1. Since no word in wordDict ends at 1, move on.
At i=4 (corresponding to substring “leet”), we find that “leet” is in the wordDict and dp[0] is True, so we set dp[4] to True.
At i=8 (corresponding to substring “code”), since “code” is in the wordDict and dp[4] is True, we set dp[8] to True.

By the end, dp[8] or dp[n] is True, so the string can be segmented.

Python code:

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def wordBreak(s: str, wordDict: List[str]) -> bool:
    n = len(s)
    dp = [False] * (n + 1)
    dp[0] = True
    wordSet = set(wordDict)
    
    for i in range(1, n + 1):
        for j in range(i):
            if dp[j] and s[j:i] in wordSet:
                dp[i] = True
                break

    return dp[n]

The inner loop iterates over all previous positions to check for valid segmentations, and the outer loop iterates over the length of the string s. The DP table is updated based on the segments found in the word dictionary. By the end, if the entire string can be segmented, dp[n] will be True.

This is a top-down approach to the problem, where you’re building the solution by checking smaller and smaller pieces of the string.