Longest String Chain

To solve this problem, we can employ dynamic programming (DP). Here’s the breakdown of the approach:

  1. Sorting: Sort the words by their lengths, since a longer word can never be a predecessor of a shorter word.

  2. DP Initialization: We’ll use a dictionary to store the maximum chain length for each word. Initialize the dictionary such that each word has a chain length of 1 (trivial chain).

  3. Building Chains: For each word in the sorted words list, try removing each character one by one, and check if the resultant word exists in our dictionary. If it does, this means we’ve found a predecessor, and we’ll update the chain length for the current word if the chain through this predecessor is longer.

Here’s how you can implement this approach:

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class Solution:
    def longestStrChain(self, words: List[str]) -> int:
        # Step 1: Sort words by their lengths
        words.sort(key=len)

        # Step 2: DP Initialization
        dp = {word: 1 for word in words}

        longest_chain = 1
        # Step 3: Building Chains
        for word in words:
            for i in range(len(word)):
                # Try removing character at position i
                predecessor = word[:i] + word[i+1:]
                if predecessor in dp:
                    dp[word] = max(dp[word], dp[predecessor] + 1)
            longest_chain = max(longest_chain, dp[word])

        return longest_chain

This algorithm builds up chains of words, incrementing the length of a chain when it finds valid predecessors. By sorting the words and iterating from shortest to longest, we ensure that we’ve always processed possible predecessors for a given word by the time we get to it.

Identifying Problem Isomorphism

“Longest String Chain” shares similarities with “Word Ladder”.

In “Longest String Chain”, we are given a list of words and we need to find the longest possible chain of words, where each next word in the chain can be formed by deleting exactly one character from the previous word.

In “Word Ladder”, we are given two words (beginWord and endWord) and a dictionary’s word list. We must change the beginWord into the endWord progressively by changing one character at a time, with each intermediate word being in the word list.

“Word Ladder” is simpler, since it merely asks for the shortest sequence, whereas “Longest String Chain” needs the longest, which could potentially require examining more possibilities.

Although the specific conditions in these two problems are different, they share a common theme of string transformation, which suggests a mapping between them. If you understand how to solve “Word Ladder”, you could potentially apply a similar strategy to “Longest String Chain” by adjusting it to the different constraints.

10 Prerequisite LeetCode Problems

For this, the following problems is a good preparation:

  1. “300. Longest Increasing Subsequence”: The concept of longest sequence is introduced in this problem which is helpful to understand the current problem.

  2. “139. Word Break”: This problem requires similar string manipulation techniques as the current problem. Understanding how to break words into sub-words can help in forming the chain in the current problem.

  3. “392. Is Subsequence”: This problem helps in understanding how to check if one string can be a subsequence of another, similar to checking if one word can be a predecessor of another.

  4. “646. Maximum Length of Pair Chain”: The approach to this problem is similar to the current problem where we try to form the longest chain possible.

  5. “72. Edit Distance”: This problem helps understand how to compare two words and the transformations required to convert one word into another.

  6. “322. Coin Change”: The dynamic programming concepts used in this problem can be helpful in solving the current problem.

  7. “1143. Longest Common Subsequence”: Understanding how to compute longest common subsequence will help in computing longest chain.

  8. “416. Partition Equal Subset Sum”: Though not directly relevant, the problem helps in understanding the concept of partitioning which is slightly similar to breaking the word chain.

  9. “91. Decode Ways”: This problem helps in understanding the different possibilities of decoding a word, similar to forming different chains.

  10. “198. House Robber”: It is a dynamic programming problem which will help in understanding how to store and use previously calculated results.

These cover dynamic programming and sequence problems which are crucial in solving “1048. Longest String Chain”.

Problem Classification

The problem falls under the domain of “Combinatorial Optimization” with a flavor of “String Manipulation” and “Graph Theory”. It involves optimizing the length of a word chain under certain constraints.

What

  1. Array of Words: A list of words made up of lowercase English letters.
  2. Predecessor Relationship: A defined relationship between two words (wordA and wordB).
  3. Word Chain: A sequence of words where each word is a predecessor of the next word in the chain.
  4. Longest Word Chain: The primary objective is to find the length of the longest possible word chain.

The problem can be further classified as:

  1. Optimization Problem: We are tasked with maximizing the length of the word chain.
  2. Graph Problem: Each word can be treated as a node, and edges can be formed if one word is a predecessor of another.
  3. String Manipulation: Since the problem deals with words and string comparison, string manipulation comes into play.
  4. Dynamic Programming: This could potentially be solved through dynamic programming, as we are essentially looking for the longest path in a graph.

Thus, the problem is multi-faceted involving several domains: optimization for finding the longest chain, graph theory for representing the relationships, and string manipulation for the basic operations on words.

Clarification Questions

  1. Input Constraints: Is the input list of words guaranteed to be non-empty?
  2. Character Set: The problem mentions lowercase English letters. Are we safe to assume no special characters or numbers will be included?
  3. Duplicate Words: Could there be duplicate words in the input list? If so, how should they be handled?
  4. Multiple Answers: Is there a possibility of multiple word chains with the same longest length? Do we need to report all of them or just one?
  5. Word Inclusion: Can a word be used more than once in a single word chain?
  6. Output Format: Is the output just the length of the longest word chain or do we also need to provide the actual chain?
  7. Insertion Point: When considering a word as a predecessor by inserting a letter, is inserting at any position valid, including the start or end of the word?
  8. Ordering of Input: Is the input list sorted in any particular order, or is it random?
  9. Performance Constraints: What is the expected time complexity? This could be important as the number of words can go up to 1000, and their length can be up to 16.
  10. Empty Words: Will there be any empty strings in the list? If so, how should they be treated?

Clarification questions aim to eliminate ambiguity and ensure a clear understanding of problem constraints and requirements.

Problem Analysis and Key Insights

  1. Predecessor Relationship: The relationship between predecessor and successor words is crucial. Understanding this helps frame the problem as a sequence-building exercise.

  2. Insertion Rule: The rule that exactly one letter can be inserted into a predecessor to form a successor word sets the boundary for our comparisons.

  3. Order Matters: The order of letters in the words matters. You can’t rearrange letters to form a successor; you can only insert a new one.

  4. Variable Word Lengths: Words can have varying lengths, ranging from 1 to 16 characters, making the problem more challenging.

  5. No Reuse: The problem statement doesn’t explicitly allow for the reuse of words, implying each word can be used only once in a chain.

  6. Chain Length: The problem is asking for the length of the longest chain, not the chain itself, which simplifies the output requirements.

  7. Case Sensitivity: All letters are lowercase, eliminating the need to handle case sensitivity.

  8. Bounds: The problem constraints give a maximum number of words (1000) and maximum length of a word (16), helping us estimate time complexity.

These insights collectively suggest that this problem can be approached as a sequence or chain-building exercise, probably using sorting and dynamic programming for an efficient solution.

Problem Boundary

The scope of this problem is confined to:

  1. A given list of words, where each word consists of lowercase English letters.
  2. The list can have a maximum of 1000 words, and each word can have up to 16 characters.
  3. Determining predecessor-successor relationships between words based on a specific rule: you can insert exactly one letter into a predecessor word to make it a successor.
  4. Finding the length of the longest word chain possible with the given list of words.

The problem does not involve:

  1. Modifying the list of words.
  2. Case sensitivity, as all letters are already lowercase.
  3. Reuse of words in a chain.

The solution is expected to return only the length of the longest word chain, not the chain itself.

Establishing the boundary of this problem involves understanding the constraints and limits that are explicitly stated or implied:

  1. Input Constraints:

    • Number of words in the list (words.length) is between 1 and 1000.
    • Each word’s length (words[i].length) is between 1 and 16.
    • Words consist only of lowercase English letters.

  2. Functional Constraints:

    • A word is a predecessor of another word if you can insert exactly one letter without changing the order of other characters.
    • A word chain needs to have at least one word (k >= 1).

  3. Output Constraints:

    • The output is a single integer representing the length of the longest possible word chain.

  4. Operational Constraints:

    • The solution must determine predecessor-successor relationships based on the given rule.
    • Words cannot be reused in a chain.

By clearly understanding these boundaries, you can focus your problem-solving efforts within this framework, ensuring that the solution will be both accurate and efficient.

Distilling the Problem to Its Core Elements

  1. Fundamental Concept: The problem is based on the concept of Dynamic Programming and String Manipulation. Specifically, it deals with forming a chain of words where one word can be transformed into another by adding a single character while preserving the order of existing characters.

  2. Simplest Description: You have a list of words. You can pick a word and add a letter to it, anywhere you like, but the letters must stay in the same order. The aim is to create the longest chain of words where each word is a transformed version of the previous word by adding just one letter.

  3. Core Problem: Find the length of the longest chain of words that can be formed, starting from any word in the given list, where each subsequent word is an extended version of the previous one.

  4. Key Components:

    • Words in the list
    • Predecessor-successor relationship
    • Length of the longest chain

  5. Minimal Set of Operations:

    • Sort the words by length.
    • Traverse through each word and check if it can be a successor to any other smaller word.
    • Update the maximum length chain ending at each word.
    • Keep track of the length of the longest chain found so far.

By performing these operations, we can solve the problem effectively.

Visual Model of the Problem

To visualize this problem, you can think of it as a directed graph where each node represents a word, and there is a directed edge from one node to another if the word at the source can be transformed into the word at the target by adding exactly one letter without changing the order of the existing letters.

  1. Nodes: Each word in the list is a node.
  2. Directed Edges: An edge from wordA to wordB means wordA is a predecessor of wordB.
  3. Longest Path: The length of the longest chain corresponds to the length of the longest path in this graph, starting from any node.

Example Visualization:

  • Given words: [“a”, “ba”, “bda”, “bdca”]
  • Graph:
    • “a” -> “ba”
    • “ba” -> “bda”
    • “bda” -> “bdca”

The longest chain would be the path from “a” to “bdca”, and its length would be 4. You can imagine each edge as a “transformation” from one word to another by adding one letter. The objective is to find the longest sequence of such transformations.

Problem Restatement

You have a list of words made of lowercase English letters. A word can be a “predecessor” of another word if you can add just one letter to the predecessor without rearranging any letters to get the second word. Your task is to find the longest chain of words where each word is a predecessor of the next word in the chain.

Requirements:

  • Each word chain must have at least one word.
  • All words in the chain must be from the given list.
  • A single word counts as a chain of length 1.

Constraints:

  • The list of words will have a length between 1 and 1000.
  • Each word in the list will be between 1 and 16 characters long.
  • All characters in the words are lowercase English letters.

The goal is to return the length of the longest such chain.

Abstract Representation of the Problem

In abstract terms, you’re given a set of elements (words) and a specific relation (predecessor) that can exist between pairs of these elements. The relation is transitive, meaning if A is a predecessor of B, and B is a predecessor of C, then A is a predecessor of C through B. The objective is to find the longest sequence (chain) of elements where each subsequent element is related to the previous one through this predecessor relation.

In this abstraction, the specifics of what the elements are (words) and what the relation entails (inserting a single letter without rearranging) become secondary. The core of the problem focuses on identifying sequences (chains) under the given relation (predecessor).

This abstract representation lets you view the problem from a more structural standpoint, such as a graph theory problem where you need to find the longest path in a directed acyclic graph.

Terminology

Here are some specialized terms and concepts:

  1. Predecessor: In the context of this problem, one word is said to be the “predecessor” of another if you can insert exactly one letter into the predecessor to obtain the other word, without rearranging any letters. This term sets up the specific relation between words that you have to consider.

  2. Word Chain: A sequence of words where each word after the first is the successor of the previous word, i.e., the preceding word is a “predecessor” to the next one. This sequence is what you’re trying to maximize in length.

  3. Transitive Relation: In mathematics and logic, a relation is transitive if whenever an element A is related to an element B, and B is related to an element C, then A is related to C. The predecessor relationship in this problem is transitive, and understanding this can help in solving the problem more efficiently.

  4. Directed Acyclic Graph (DAG): Though the term isn’t used in the problem statement, the problem naturally maps to a DAG. Each word represents a node, and directed edges can be drawn from each “predecessor” to its “successor(s).” Understanding this graph structure can open up various algorithmic paths for solving the problem.

  5. Longest Path: Again, not explicitly mentioned, but this problem effectively asks for the longest path in the resulting DAG. This is the series of words that forms the longest word chain.

Understanding these terms and their interplay can help dissect the problem, formulating an approach that effectively navigates the relationships between words to find the longest word chain.

Problem Simplification and Explanation

This problem is about finding the longest chain of words where each word is a slightly modified version of the previous word. You start with a word and, by adding a single letter at any position, get the next word in the chain. The chain can be as long as possible, but you can only use the words provided in the initial list.

Key Concepts

  1. Starting Point: You can begin the chain with any word from the list.

  2. Adding a Letter: You add a single letter to a word to get the next word in the chain. You must use words from your initial list.

  3. Longest Chain: Your goal is to find the longest such chain possible.

How They Interact

You start with any word from your list. Then you look for another word that can be obtained by adding a single letter to the first word. You repeat this process, always checking the list of available words, trying to form the longest chain possible.

Metaphor

Imagine you are building a tower using a set of building blocks. Each block has a label on it, written with letters. You start with one block. Each time you add a new block on top, you can modify the label by adding one more letter, but the sequence of letters must stay the same. Your aim is to build the tallest tower using the blocks from your set, following these rules.

In this metaphor, each block is a word from your list, and the height of your tower is the length of the word chain you’re trying to find.

Constraints

  1. Word Length: The length of words in the list is capped at 16 letters. This provides a limit on how deep our search needs to go for each word.

  2. Limited Set of Characters: All words consist of lowercase English letters, meaning a maximum of 26 unique characters can be used. This helps in hashing or looking up words.

  3. Array Length: The list has a maximum length of 1000 words, which is not very large. This allows us to pre-process the list if necessary.

  4. Predecessor Relationship: The fact that one word can be a “predecessor” to another through the addition of a single letter provides a clear rule for linking words. This could be useful for creating a directed graph where each node is a word and edges signify predecessor relationships.

  5. No Reordering: Letters in words must remain in the same sequence, reducing the complexity in checking whether one word is a predecessor of another.

  6. Return Length, Not Sequence: The problem asks for the length of the longest word chain, not the chain itself. This simplifies the output and allows us to focus solely on counting the chain length.

  7. Start Anywhere: You can start the chain from any word, meaning we could sort the list and then traverse it to make searching for predecessors more efficient.

These characteristics and conditions could be useful for designing an efficient algorithm to solve the problem.

  1. Word Length: Create a table sorted by word lengths. You’ll quickly notice that the maximum length is capped, signaling a limit on how deep the search needs to go for each word.

  2. Limited Set of Characters: You can draw a frequency table or histogram of the characters used across all words. You’ll see that only 26 unique characters are involved.

  3. Array Length: Just noting down the number of rows in your table will tell you the maximum size of the array. The number will be 1000 or less.

  4. Predecessor Relationship: Draw a directed graph. Nodes will be the words, and an arrow from one node to another signifies that the first is a predecessor of the second. This will help you visualize the connections between words.

  5. No Reordering: In your graph, you can annotate each edge with the added letter and its position to ensure that only legal predecessor relationships are included.

  6. Return Length, Not Sequence: No special diagram needed here, but keep in mind that the solution will be a single integer (the length), not a sequence of words.

  7. Start Anywhere: In your sorted table, mark potential starting points. This will help you realize that you’re not limited to starting the chain from a specific point.

Drawing these tables and diagrams can help you visualize the inherent characteristics and constraints of the problem, making it easier to formulate an efficient algorithm.

  1. Array Length Limitation: The array has at most 1000 words. This sets an upper boundary on how big the problem can get, signaling that certain computational approaches will be feasible within this limit.

  2. Word Length Cap: Each word has a length capped at 16. This is another key constraint that limits the depth of the problem, making recursive or depth-first approaches more viable.

  3. Lowercase Letters Only: Since all the characters are lowercase English letters, there’s a maximum of 26 unique characters to consider. This simplicity could be exploited for efficient searching or hashing.

  4. Integer Output: You’re only asked for the length of the longest chain, not the chain itself. This narrows the scope of the problem and allows for optimizations, as you don’t need to store all possible chains.

  5. Words are Unique: The problem doesn’t mention duplicates, implying that each word is unique. This can simplify your data structures and eliminate the need for duplicate checks.

  6. No Reordering of Letters: The condition that a word is a predecessor only if we can add a letter without changing the order limits the complexity of the matching task. This rule can be exploited to make the search more efficient.

  7. Start Anywhere: The problem doesn’t specify a starting point, so every word can potentially be the start of a chain. This property suggests that a global search strategy might be more appropriate than trying to optimize individual chains.

Understanding these constraints allows you to make informed decisions when formulating your solution. They often indicate the types of algorithms and data structures that would be most efficient for solving the problem.

Case Analysis

Let’s dive into some examples and test cases that can help us understand the problem more clearly.

Minimal Words Case

Input: ["a", "aa"]
Output: 2
Analysis: This case has the minimum number of words. The chain [“a”, “aa”] is valid and is of length 2.

Single Character Words

Input: ["a", "b", "c"]
Output: 1
Analysis: Each word is a single letter. No word is a predecessor of any other, so the longest chain can only be a single word.

Differing Lengths

Input: ["apple", "apples", "apply", "applies", "app"]
Output: 3
Analysis: This example highlights that even if a word is similar to another, the specific requirement of adding only one letter makes [“app”, “apple”, “apples”] the longest chain, not [“app”, “apply”, “applies”].

No Chain

Input: ["apple", "orange", "banana"]
Output: 1
Analysis: No word can become another by adding a single letter. Each word is its own chain of length 1.

Mixed Lengths

Input: ["a", "ab", "abc", "def", "de"]
Output: 3
Analysis: Two separate chains are present: [“a”, “ab”, “abc”] and [“de”, “def”]. The longest has a length of 3.

Edge Case: Empty Array

Input: []
Output: 0
Analysis: The array is empty, so no chain can be formed. The answer is zero.

Edge Case: One Word

Input: ["apple"]
Output: 1
Analysis: There’s only one word. It forms a chain with itself, so the output is 1.

Edge Case: Longest Word Length

Input: ["a"*16, "b"*16]
Output: 1
Analysis: Each word has the maximum length of 16 but neither is a predecessor of the other.

Edge Case: Maximum Array Length

Input: ["a", "aa", "aaa", ..., "a"*1000]
Output: 1000
Analysis: Each word is a predecessor of the next, forming the longest possible chain. This tests the performance of the algorithm.

By carefully considering these edge cases and constraints, you can identify potential pitfalls and ensure your solution handles all scenarios.

Visualizing these cases can help you understand the problem better. Here are some ways to do that:

  1. Minimal Words Case:

    • Graph Nodes: “a” → “aa”
    • Chain Length: 2

  2. Single Character Words:

    • Graph Nodes: “a”, “b”, “c” (no edges)
    • Chain Length: 1

  3. Differing Lengths:

    • Graph Nodes: “app” → “apple” → “apples”, “apply” → “applies”
    • Chain Length: 3

  4. No Chain:

    • Graph Nodes: “apple”, “orange”, “banana” (no edges)
    • Chain Length: 1

  5. Mixed Lengths:

    • Graph Nodes: “a” → “ab” → “abc”, “de” → “def”
    • Chain Length: 3

  6. Empty Array:

    • Graph Nodes: (None)
    • Chain Length: 0

  7. One Word:

    • Graph Nodes: “apple” (no edges)
    • Chain Length: 1

  8. Longest Word Length:

    • Graph Nodes: “a”*16, “b”*16 (no edges)
    • Chain Length: 1

  9. Maximum Array Length:

    • Graph Nodes: “a” → “aa” → “aaa” → … → “a”*1000
    • Chain Length: 1000

Each node in the visualization represents a word, and an edge between two nodes indicates that one word can be transformed into the other by adding a single letter. No edge means no such transformation is possible. The Chain Length signifies the length of the longest chain that can be formed.

Analyzing different cases yields several key insights:

  1. Minimal Words and Single Characters: The problem is easier to solve when there are fewer words, especially single characters, as the longest chain can be found quickly.

  2. Differing Lengths: A chain can be formed by gradually adding letters to shorter words, making them longer. This emphasizes the importance of sorting the words by their length.

  3. No Chain: When words can’t be linked by adding a single character, the longest chain is 1, which is a base case.

  4. Mixed Lengths: Chains can involve words of varying lengths, so consider all possible predecessors for each word.

  5. Empty Array: An empty array means a zero-length chain, another base case.

  6. One Word: If there’s only one word, the longest chain is that single word, a special case that simplifies things.

  7. Longest Word Length: If all words are the maximum length allowed by constraints, then no chains can be formed.

  8. Maximum Array Length: In the worst case, you may have to link 1000 words, which emphasizes the need for an efficient algorithm.

  9. Sorted Order: Sorting the list of words by length can simplify the task of finding chains.

These insights help in understanding the boundaries and potential shortcuts or optimizations that can be used in solving the problem.

Identification of Applicable Theoretical Concepts

Several mathematical and algorithmic concepts can make this problem more manageable:

  1. Dynamic Programming: This problem involves subproblems that can be solved independently to construct a solution to the overall problem. Dynamic programming is particularly well-suited to store solutions to subproblems, improving efficiency.

  2. Graph Theory: The problem can be modeled as a directed graph where each word is a node, and an edge between two nodes signifies a predecessor relationship. This allows us to find the longest path in the graph.

  3. Sorting Algorithms: Sorting words by length can optimize the process of building the chain, enabling you to eliminate options quickly.

  4. Hashing: A hash table can be employed to look up words efficiently. This is beneficial for checking if a shorter word can be turned into a longer one by adding a letter.

  5. String Matching Algorithms: Given that we need to match words based on specific conditions, string matching algorithms may be useful, although the problem’s constraints are simple enough that this may not be necessary.

  6. Recursion: Recursive algorithms could also be used to explore all possible combinations of words, although this may not be the most efficient approach given the problem constraints.

  7. Greedy Algorithms: While not directly applicable, a greedy approach of always picking the longest chain at each step won’t work here. Realizing this helps you avoid incorrect strategies.

  8. Combinatorial Metrics: The number of combinations of words can be crucial for understanding the problem’s complexity. Binomial coefficients or factorial calculations could be relevant.

  9. Depth-First Search (DFS): DFS can be applied on the graph representation to find the longest path, though this requires storing intermediate results to be efficient.

These concepts provide various ways to attack the problem, and some could be combined for an efficient solution.

Simple Explanation

Imagine you have a collection of building blocks, and each block has some letters on it. You can stack these blocks on top of each other to make towers. But there’s a rule: you can only place a block on top of another block if you can add just one more letter to the word on the lower block to make the word on the higher block.

For example, if you have a block with the word “cat,” you can put a block with the word “cart” on top of it. But you can’t put a block with the word “rat” on top of “cat” because you’d have to change the ‘c’ to ‘r’, and that’s more than just adding one letter.

Your goal is to make the tallest tower you can with the blocks you have. The taller the tower, the better you’ve done.

In everyday terms, it’s like playing a special game of Scrabble where you can only add tiles to extend words already on the board by one letter at a time. You’re trying to find the longest word chain you can make.

Problem Breakdown and Solution Methodology

Approach to Solving the Problem

  1. Sort the Words by Length: Start by sorting the list of words by their length. This is similar to arranging your building blocks from smallest to largest before you start stacking them.

  2. Create a Dictionary for Chain Lengths: Initialize a dictionary to store the longest chain length ending at each word. Think of this as keeping a record of your tallest towers so far, with the top block’s word as the key.

  3. Iterate Through the Words: Loop through the sorted list. For each word, calculate the longest chain ending with that word by examining all its predecessors. This is like looking at all smaller blocks to see which ones can be extended to reach the current block.

  4. Check Predecessors and Update: For each word, generate all possible predecessors by removing one letter at a time. If any of these predecessors are in the dictionary, update the longest chain length for the current word. This is akin to seeing if you can put the current block on top of an existing tower to make it taller.

  5. Get the Longest Chain: After going through all words, find the maximum chain length from the dictionary.

Impact of Changes in Parameters

  1. Number of Words: Increasing the number of words would make our problem more computationally expensive. Our solution would still work but would take more time.

  2. Length of Words: The longer the words, the more predecessors they can have, thus increasing the number of calculations.

  3. Alphabetic Variety: If words contain a wider range of alphabets, it increases the chances of forming longer chains.

Example Case

Let’s say the input words are ["a", "ba", "bca", "bdca", "bda"].

  1. Sort by length: ["a", "ba", "bda", "bca", "bdca"]
  2. Initialize dictionary: {}
  • First word: “a” (Longest chain is 1, dictionary becomes {"a": 1})
  • Second word: “ba” (Predecessor “a” exists, so longest chain is 1+1=2, dictionary becomes {"a": 1, "ba": 2})
  • Third word: “bda” (Predecessor “ba” exists, so longest chain is 2+1=3, dictionary becomes {"a": 1, "ba": 2, "bda": 3})
  • Fourth word: “bca” (Predecessor “ba” exists, so longest chain is 2+1=3, dictionary becomes {"a": 1, "ba": 2, "bda": 3, "bca": 3})
  • Fifth word: “bdca” (Predecessor “bca” and “bda” exist, both have longest chain as 3, so new longest chain is 3+1=4, dictionary becomes {"a": 1, "ba": 2, "bda": 3, "bca": 3, "bdca": 4})

The longest chain is 4, with words ["a", "ba", "bca", "bdca"].

Inference of Problem-Solving Approach from the Problem Statement

Key Terms and Concepts

  1. Array of Words: This is the main input. We need to sort it by length to reduce the complexity of finding predecessors.

  2. Predecessor: A word that can be turned into another by adding a single character. Knowing what constitutes a predecessor helps us understand how to build word chains.

  3. Word Chain: A sequence of words where each word is a predecessor of the next. Understanding the goal of building the longest possible chain informs our approach to solving the problem.

  4. Constraints: These give us limits within which the solution must be found, helping us identify what optimizations may be needed.

  5. Length of Word Chain: This is the primary output. Knowing that we need to maximize this length guides our choice of algorithms and data structures.

Strategy and Method

  1. Array of Words: Given this is our input and we have to find chains, sorting the array helps us in optimizing predecessor search.

  2. Predecessor: We can identify a predecessor by removing a single character from a word and checking if the resulting string exists. This becomes a basic building block of our algorithm.

  3. Word Chain: The aim to build the longest word chain guides us to use dynamic programming. We maintain a record of the longest chain ending at each word to avoid redundant calculations.

  4. Constraints: Given that the number of words and their lengths are reasonably limited, our approach does not need extreme optimization. A solution with time complexity around O(N log N) or O(N^2) should be sufficient.

  5. Length of Word Chain: To maximize this, we maintain a dictionary to keep track of the longest chain ending at each word. After processing all words, the maximum value in this dictionary will be our answer.

Recognizing properties through tables or diagrams can be a useful exercise. Here’s how you can do it:

  1. Array of Words: Create a table and list the words in one column. Sort them by length in another column.

    OriginalSorted by Length
    bdaa
    bdcab
    aba
    bbda
    babdca

  2. Predecessor: Draw arrows between predecessor and successor words in the sorted table.

    Sorted by LengthPredecessor Relationship
    a
    b————————>
    ba————————>
    bda————————>
    bdca

  3. Word Chain: Create a flow diagram to visualize the longest word chain.

    a --> ba --> bda --> bdca

  4. Constraints: You could visualize constraints like word length and number of words using a bar graph or pie chart to ensure you are not exceeding limits. However, in this problem, the constraints are straightforward, so a table like the first one should suffice.

  5. Length of Word Chain: Keep a separate column in your table to mark the length of the longest chain ending at each word.

    Sorted by LengthLength of Longest Chain
    a1
    b1
    ba2
    bda3
    bdca4

By constructing these tables and diagrams, you can visually organize the problem’s properties, which can be helpful in identifying an efficient solution strategy.

How did you infer from the problem statement that this problem can be solved using ?

Simple Explanation of the Proof

I’m having trouble understanding the proof of this algorithm. Could you explain it in a way that’s easy to understand?

Stepwise Refinement

1. Approach

  1. Sort the Words: Begin by sorting the array of words by their lengths. It’s easier to find predecessors for shorter words first.

  2. Initialize Data Structures: Create a dictionary or hash map to store each word along with the length of its longest word chain.

  3. Iterate Through Sorted Words: Go through the sorted list and try to find a predecessor for each word.

  4. Check for Predecessors: For each word, remove one character at a time and check if the resultant word is in the dictionary.

  5. Update Dictionary: If a predecessor is found, update the dictionary value for the current word as one greater than its predecessor’s value.

  6. Find Longest Chain: Keep track of the longest chain found so far.

2. Granular, Actionable Steps

  1. Sort the Words:

    • Use a sort function to sort the array based on word length.

  2. Initialize Data Structures:

    • Initialize an empty dictionary.

  3. Iterate Through Sorted Words:

    • Use a loop to go through each word in the sorted array.

  4. Check for Predecessors:

    • Create all possible sub-words by removing one character at a time.
    • Check if these sub-words are keys in the dictionary.

  5. Update Dictionary:

    • If a sub-word is found, calculate the new chain length as one more than the sub-word’s chain length.
    • Update the dictionary with this new chain length if it’s longer than the existing one.

  6. Find Longest Chain:

    • Keep a variable to track the longest chain length. Update this whenever a longer chain is found.

3. Independent Parts

  • Sorting the Words: This can be done independently before diving into the problem.
  • Check for Predecessors: This is more or less an independent operation for each word.
  • Initialization: Initializing data structures is an independent step.

4. Repeatable Patterns

  • Check for Predecessors: This pattern is repeated for every word in the list.
  • Update Dictionary: Whenever a predecessor is found, the dictionary updating is a repeatable action.

By breaking down the problem into these steps, you can tackle each part independently, in sequence, while also recognizing repeatable patterns that may be optimized.

Solution Approach and Analysis

Steps

  1. Sort the Words by Length: Think of this step as arranging dominoes by size. You want to start with the smallest domino and see how you can stack up from there.

  2. Initialize a Dictionary: Imagine this as a scoreboard that keeps track of the longest chain each word can form.

  3. Iterate Over Sorted Words: This is similar to walking along the line of sorted dominoes, considering each one at a time.

  4. Find Predecessors: For each word, imagine removing one domino from the set and checking if what remains can still stand as a smaller chain. In coding terms, remove one character at a time from the word and check if the resulting smaller word is already on the scoreboard.

  5. Update the Scoreboard (Dictionary): If you find that a smaller chain can stand on its own (the smaller word exists in the list), update the score of the current domino set (the current word) to be one point more than the smaller set (the predecessor).

  6. Keep Track of the Longest Chain: Think of this as keeping an eye on the tallest stack of dominoes you’ve built so far.

Impact of Changing Parameters

  1. Increasing Number of Words: The more words you have, the more time it will take to sort and iterate through them.

  2. Longer Word Lengths: The longer each word is, the more time it will take to find all its possible predecessors.

Example Case

Let’s consider the first example case: words = [“a”, “b”, “ba”, “bca”, “bda”, “bdca”].

  1. Sort the Words by Length: [“a”, “b”, “ba”, “bda”, “bca”, “bdca”]

  2. Initialize a Dictionary: {}

  3. Iterate Over Sorted Words: Start with “a”.

  4. Find Predecessors for “a”: No predecessors as it’s the first word.

    • Update the dictionary: {“a”: 1}

  5. Find Predecessors for “b”: No predecessors.

    • Update the dictionary: {“a”: 1, “b”: 1}

  6. Find Predecessors for “ba”: “b” and “a” are predecessors.

    • Update the dictionary: {“a”: 1, “b”: 1, “ba”: 2}

  7. Find Predecessors for “bda”: “ba” is a predecessor.

    • Update the dictionary: {“a”: 1, “b”: 1, “ba”: 2, “bda”: 3}

  8. Find Predecessors for “bca”: “ba” is a predecessor.

    • Update the dictionary: {“a”: 1, “b”: 1, “ba”: 2, “bda”: 3, “bca”: 3}

  9. Find Predecessors for “bdca”: “bca” and “bda” are predecessors, but “bda” has a longer chain.

    • Update the dictionary: {“a”: 1, “b”: 1, “ba”: 2, “bda”: 3, “bca”: 3, “bdca”: 4}

The longest chain is 4, as per the dictionary.

Identify Invariant

The invariant in this problem is the relationship between a word and its predecessor. Specifically, the condition that a word is a predecessor of another if we can insert exactly one letter into it without changing the order of the existing letters. This relationship doesn’t change and holds consistently across the problem space, serving as a stable rule for constructing the word chains.

In simple terms, think of it as the rule of the game that stays the same, no matter how the words change or how long they get. Like in a game of dominoes, the rule that you can only connect pieces with matching numbers remains the same, regardless of how many pieces you have.

Identify Loop Invariant

What is the loop invariant in this problem?

Is invariant and loop invariant the same for this problem?

Thought Process

Thought Process and Steps

  1. Understand the Problem: The first step is to fully understand what constitutes a predecessor and a word chain. A predecessor is a word that can become another word with the addition of a single letter, without changing the order of the existing letters. A word chain is a sequence of such predecessors.

  2. Sort the Words: Given that we want the longest chain, sorting the words by length will allow us to build upon smaller words to find longer ones.

  3. Use Dynamic Programming: A Dynamic Programming (DP) approach can help here. Specifically, we can use a hash table to keep track of the longest chain that ends at each word.

  4. Iterate and Build: For each word, look at its possible predecessors (by removing one letter at a time) and update the longest chain that can be formed with this word as the end.

  5. Global Maximum: Maintain a variable to keep track of the longest chain found so far.

Code

Here’s how to implement the above logic in Python:

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def longestStrChain(words):
    # Sort the words by length.
    words.sort(key=len)
    
    # Initialize DP table.
    dp = {}
    
    # Initialize the maximum length variable.
    max_len = 1
    
    # Iterate through the words.
    for word in words:
        dp[word] = 1  # A single word is trivially a chain of length 1.
        
        # Check all possible predecessors.
        for i in range(len(word)):
            prev_word = word[:i] + word[i + 1:]
            
            # If the predecessor exists in DP table, update the chain length.
            if prev_word in dp:
                dp[word] = max(dp[word], dp[prev_word] + 1)
        
        # Update the global maximum length.
        max_len = max(max_len, dp[word])
        
    return max_len

Cues and Insights

  1. Predecessor Definition: The definition of what constitutes a predecessor is crucial. It informs us that we can derive it by removing a letter from the word.

  2. Length Matters: The need to find the “longest” word chain suggests sorting by length and building up from shorter to longer words.

  3. Dynamic Programming: The idea that we can build upon previously calculated longest chains for smaller words is a hint that DP is a good fit for this problem.

By following these steps and insights, we can effectively solve the problem in an efficient manner.

Establishing Preconditions and Postconditions

Parameters

  1. Inputs: The method takes a single input parameter words.
  2. Types: This parameter is a list of strings.
  3. Representation: The parameter represents an array of words consisting of lowercase English letters. The array contains potential elements of word chains.

Preconditions

  1. State of the Program: No specific state of the program is needed before this method is called.
  2. Constraints: The input list should have a length between 1 and 1000, and individual words should have lengths between 1 and 16. All letters in the words must be lowercase English letters.
  3. Specific State: None required.

Method Functionality

  1. Expected Action: The method is expected to find the length of the longest word chain possible from the given list of words.
  2. Interaction: The method reads the input list, sorts the words, applies dynamic programming to find chain lengths, and identifies the longest chain.

Postconditions

  1. State of the Program: The program’s state remains unchanged, except for any variables local to the method.
  2. Return Value: The return value is an integer representing the length of the longest word chain.
  3. Side Effects: There are no side effects on global program state.

Error Handling

  1. Response to Unmet Preconditions: The method doesn’t explicitly check preconditions. If the input doesn’t meet the constraints, the behavior is undefined.
  2. Exception Handling: The method doesn’t throw any exceptions or return special values for incorrect inputs. However, it’s assumed that the input will always meet the defined constraints.

Problem Decomposition

Problem Understanding

The problem asks for the longest chain of words where each word is a “predecessor” of the next. A word is a predecessor if you can insert exactly one letter into it to form the next word, without rearranging any other letters.

Initial Breakdown

  1. Sorting: Sort the words by length, so we consider shorter words first.
  2. Identify Predecessors: Determine if one word is a predecessor of another.
  3. Find Chains: Use dynamic programming to find the longest chain starting from each word.
  4. Return Longest Chain: After going through all words, return the longest chain found.

Subproblem Refinement

  1. Sorting

    • Iterate through the list and sort it.

  2. Identify Predecessors

    • Take two words and compare them letter by letter.
    • Check if one can be transformed into the other by inserting a single letter.

  3. Find Chains

    • For each word, find all its predecessors.
    • Use dynamic programming to update the longest chain that ends at this word.

  4. Return Longest Chain

    • Return the maximum value in the dynamic programming array.

Task Identification

  1. Compare Two Words: Repeated task of identifying if one word is a predecessor of another.

Task Abstraction

  1. Compare Two Words: Abstracted enough to take two strings and return a boolean indicating if one is a predecessor of the other.

Method Naming

  1. Sorting: sortWordsByLength
  2. Identify Predecessors: isPredecessor
  3. Find Chains: findLongestChain
  4. Return Longest Chain: returnMaxChainLength

Subproblem Interactions

  1. Sorting must be done first to prepare for chain building.
  2. Identify Predecessors is an independent task used by Find Chains.
  3. Find Chains relies on Identify Predecessors and works after Sorting.
  4. Return Longest Chain is the final step and depends on Find Chains.

From Brute Force to Optimal Solution

Brute-Force Solution

  1. Start with each word: For every word in the list, we try to find its longest chain by starting from that word and looking for valid successors in the list.
  2. Search for Successors: For each word, iterate through all other words to find if they are successors.
  3. Recursively Build Chain: For each successor, again repeat steps 1 and 2 to build the chain.
  4. Find Longest: Keep track of the longest chain you’ve found.

Python3 Code Snippet for Brute-force

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def is_predecessor(word1, word2):
    count = 0
    j = 0
    for i in range(len(word1)):
        while j < len(word2) and word1[i] != word2[j]:
            j += 1
        if j == len(word2):
            return False
        j += 1
    return True

def longest_chain(words, current_word):
    max_chain = 1
    for word in words:
        if is_predecessor(current_word, word):
            max_chain = max(max_chain, 1 + longest_chain(words, word))
    return max_chain

words = ["a", "ab", "abc", "abcd"]
result = 0
for word in words:
    result = max(result, longest_chain(words, word))

print("Longest chain length:", result)

Inefficiencies of the Brute-Force Approach

  1. Time Complexity: For each word, we go through all other words to see if they can be its successor. This makes the time complexity O(N^2), but since we’re doing this recursively, it could be even worse, approximating O(N!).
  2. Repeated Work: We may check the longest chain for the same word multiple times if it appears as a successor for different words.

Optimization Steps

  1. Sort the words by length: Sorting will allow us to consider only shorter words as potential predecessors.
  2. Use Dynamic Programming (DP): Store the longest chain for each word so we don’t recompute it.

Optimized Python3 Code Snippet

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def longest_chain(words):
    word_dict = {}
    for word in words:
        word_dict[word] = 1

    words.sort(key=len)
    max_chain = 1
    for word in words:
        for i in range(len(word)):
            prev_word = word[:i] + word[i+1:]
            if prev_word in word_dict:
                word_dict[word] = max(word_dict[word], word_dict[prev_word] + 1)
        max_chain = max(max_chain, word_dict[word])

    return max_chain

words = ["a", "ab", "abc", "abcd"]
print("Longest chain length:", longest_chain(words))

Reasoning Behind Each Decision

  1. Sorting: By sorting the words, we only have to consider shorter words when searching for predecessors.
  2. Dynamic Programming: We store the longest chain length ending at each word, so we don’t recompute it.

Complexity Analysis

  1. Time Complexity: Sorting takes O(NlogN). The loops take O(N * M), where M is the maximum length of any word. Overall, O(NlogN + N * M).
  2. Space Complexity: We’re storing a longest chain value for each word, making it O(N).

The optimized solution is significantly better than the brute-force approach.

Code Explanation and Design Decisions

  1. Initial Parameters:

    • words: This is a list of words given as input. It’s crucial as we want to find the longest chain of words where each word is a predecessor to the next word in the chain.

  2. Primary Loop:

    • The main loop iterates through each word in the sorted words list.
    • Each iteration represents the calculation of the longest chain ending at that word.
    • This advances the solution by updating the longest chain for each word, thereby making it easier to compute the longest chain for words that come after.

  3. Conditions/Branches:

    • There’s an inner loop that removes one character at a time from the current word to find its predecessor.
    • The condition if prev_word in word_dict: checks if this predecessor word exists.
    • This is logical because we are only interested in valid chains, and a word can be part of a chain only if its predecessor exists in the given list.

  4. Updates/Modifications:

    • word_dict[word] = max(word_dict[word], word_dict[prev_word] + 1) updates the longest chain ending at the current word.
    • This is necessary as we need to keep track of the longest chain for each word so that we don’t have to recalculate it.

  5. Invariant:

    • The invariant here is the word_dict that always contains the longest chain length ending at each word up to the current point of the loop.
    • It helps by ensuring that we don’t repeat calculations and always have the correct information needed to compute the longest chain for any word.

  6. Final Output:

    • The final output is the length of the longest valid chain of words, represented by the variable max_chain.
    • This satisfies the problem’s requirement of finding the longest such chain within the given list of words.

Coding Constructs

  1. High-Level Strategies:

    • The code employs Dynamic Programming to keep track of the longest chain ending at each word, reducing repeated calculations.
    • It also sorts the list of words to ensure that we can compute the longest chain efficiently.

  2. Purpose to a Non-Programmer:

    • Imagine you have a box of magnetic letters, and you’re trying to make the longest possible chain of words. Each word in the chain has to be made by removing just one letter from the previous word. The code does this but much faster than we could do it by hand.

  3. Logical Elements:

    • Sorting Mechanism: to sort words by length.
    • Data Storage: a dictionary or hash table to store the longest chain for each word.
    • Loop Iteration: to go through each word.
    • Conditional Checks: to verify if a word can be a successor to another.

  4. Algorithmic Approach in Plain English:

    • First, sort all the words by their length.
    • Start with the shortest word and find the longest chain that ends with this word.
    • Use this information to find the longest chain for the next word, and the next, all the way to the longest word.
    • Along the way, keep updating and storing the longest chain information for each word.
    • In the end, find out which word has the longest chain leading to it.

  5. Key Steps or Operations:

    • Sorting the list of words: to streamline the chain-building process.
    • Looping through sorted words: to build up the chain information.
    • Updating the longest chain in a dictionary: to keep track of the chain lengths and avoid recalculating.
    • Fetching the longest chain from the dictionary: to get the answer.

  6. Algorithmic Patterns:

    • Dynamic Programming: for storing intermediate results in a dictionary.
    • Iteration: for looping through the sorted list of words.
    • Lookup: for fast retrieval of stored information from the dictionary.
    • Sorting: for arranging words by length to facilitate the chain-building process.

Language Agnostic Coding Drills

  1. Distinct Coding Concepts:

    • Variable Declaration and Initialization: Storing simple types like integers and strings.
    • Array Handling: Creating and manipulating arrays/lists.
    • Sorting Algorithm: Implementing or using a built-in sort function.
    • Looping Constructs: Using loops to iterate through data structures.
    • Conditional Statements: Making decisions based on conditions.
    • Dictionary/Hash Table: Creating and updating key-value pairs.
    • Dynamic Programming: Storing intermediate computed values for reuse.

  2. List of Concepts in Order of Increasing Difficulty:

    • Variable Declaration and Initialization: Basic and foundational.
    • Array Handling: Requires understanding of data structures but is fundamental.
    • Looping Constructs: Basic control flow, slightly more complex than array handling.
    • Conditional Statements: Introduces logic but still fundamental.
    • Sorting Algorithm: Requires an understanding of algorithms, slightly more advanced.
    • Dictionary/Hash Table: Advanced data structure, requires knowledge of key-value mapping.
    • Dynamic Programming: Most complex, requires a good grasp of optimization and previous concepts.

  3. Problem-Solving Approach Using These Drills:

    • Understand the Problem: Read the problem statement to understand what you’re asked to do. Identify input, output, and constraints.

    • Initialize Variables: Start by declaring variables that will hold data, like the list of words or the longest chain length.

    • Handle the Array of Words: Load the words into an array for easier manipulation.

    • Sort the Words: Use a sorting algorithm to sort the array of words by length. This is critical for the dynamic programming approach.

    • Initialize Dictionary: Create a dictionary to store the length of the longest chain for each word.

    • Start Looping Through Words: Use a loop to iterate through each word in the sorted list.

      • Use Conditional Statements: Within the loop, use conditions to check if one word can be formed by removing a character from another.

      • Update Dictionary: If a condition is met, update the value in the dictionary to reflect the new longest chain.

    • Apply Dynamic Programming: Use the dictionary to refer to already computed longest chains for smaller words to build up the chain for larger words without recalculating.

    • Extract Final Answer: At the end of the loop, examine the dictionary to find the longest chain.

    • Return or Display Result: Finally, the longest chain value is either returned as output or displayed.

Each of these drills contributes to the overall solution. For example, sorting enables efficient dynamic programming, and dynamic programming saves time by avoiding recalculations. Together, they build up to solve the problem in an efficient manner.

Targeted Drills in Python

  1. Python Coding Drills for General Concepts:

    • Variable Declaration and Initialization

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      x = 5
name = "Alice"

    • Array Handling

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      arr = [1, 2, 3, 4]
arr.append(5)
first_element = arr[0]

    • Looping Constructs

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      for i in range(5):
    print(i)

    • Conditional Statements

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      if x > 5:
    print("x is greater than 5")
else:
    print("x is not greater than 5")

    • Sorting Algorithm

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      arr = [3, 1, 4, 1, 5, 9]
sorted_arr = sorted(arr)

    • Dictionary/Hash Table

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      my_dict = {}
my_dict['key'] = 'value'
value = my_dict.get('key', 'default_value')

    • Dynamic Programming

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      memo = {}
def fib(n):
    if n in memo:
        return memo[n]
    if n <= 1:
        return n
    memo[n] = fib(n-1) + fib(n-2)
    return memo[n]

  2. Problem-Specific Coding Drills:

    • Chain Length Calculation
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      def chain_length(word, word_dict):
    if word in word_dict:
        return word_dict[word]
    # Logic to calculate chain length
    return length
      This drill is essential for our problem as calculating chain length is the core part of the problem. It would be repeatedly used for each word.

  3. Integrating the Pieces

    • Initialize variables and arrays.
    • Load the words into an array and sort them.
    • Create an empty dictionary to store the chain lengths.
    • Loop through each sorted word.
      • Use conditional statements to check chain possibilities.
      • Use the chain_length() function to compute or retrieve the length of the longest chain.
      • Store the computed chain length in the dictionary.
    • Once all words have been processed, find the longest chain length from the dictionary.

By performing these drills, we build a strong foundation in Python and problem-specific routines. Integrating them according to the described plan will solve the initial problem effectively.

Q&A

Similar Problems

Here are 10 problems that has similar underlying concepts:

  1. Longest Substring Without Repeating Characters
    Similarity: Utilizes a hash table to keep track of characters. The concept of maintaining an optimal value is similar to storing chain lengths.

  2. Coin Change
    Similarity: Dynamic Programming is used to store previously computed minimum coins needed, which is akin to storing chain lengths.

  3. Word Break
    Similarity: Uses Dynamic Programming and dictionary to check if a word can be formed by words in a given list, similar to forming chains.

  4. Longest Palindromic Substring
    Similarity: Both problems require scanning through a string to find a solution and may employ dynamic programming for optimization.

  5. 3Sum
    Similarity: Involves sorting and iterating through an array to find specific conditions, similar to finding chainable words.

  6. Maximum Subarray
    Similarity: Both problems look for a condition-based sequence of elements, whether it’s an array of numbers or a chain of words.

  7. Unique Paths
    Similarity: Requires you to find the number of unique paths from the start to the end, which can be solved through Dynamic Programming.

  8. Climbing Stairs
    Similarity: A classic dynamic programming problem where you calculate the number of ways to reach the top, similar to calculating chain lengths.

  9. Letter Combinations of a Phone Number
    Similarity: This problem also requires generating sequences (combinations of letters) based on conditions, similar to generating word chains.

  10. Minimum Path Sum
    Similarity: Uses Dynamic Programming to find the minimum path sum from top left to bottom right of a grid. The grid can be thought of as a network of chained elements, like the words in our original problem.

Each of these problems involves techniques like Dynamic Programming, sorting, or hash table manipulations, which were pivotal in solving our original problem.