Find K-Length Substrings With No Repeated Characters

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class Solution:
    def numKLenSubstrNoRepeats(self, s: str, k: int) -> int:
        result = 0
        for i in range(len(s) - k + 1):
            substring = s[i:i + k]
            unique_chars = set(substring)
            if len(unique_chars) == k:
                result += 1
        return result

Identifying Problem Isomorphism

“Find K-Length Substrings With No Repeated Characters” can be mapped to “Length of the Longest Substring Without Repeating Characters”.

In both problems, the concept of substrings without repeated characters is central. The difference is in the metrics we are interested in: the former problem counts the number of such substrings with a specific length (k), while the latter problem seeks the maximum length of such substrings.

However, despite this difference, the solution strategies are very similar. Both involve traversing the given string and using a data structure (like a set or a hash map) to track the characters seen so far. Hence, they are approximately isomorphic problems.

“Length of the Longest Substring Without Repeating Characters” is simpler because it doesn’t involve counting the specific length substrings but rather just requires maintaining the longest length found so far.

10 Prerequisite LeetCode Problems

For “1100. Find K-Length Substrings With No Repeated Characters”, the following are a good preparation:

  1. “3. Longest Substring Without Repeating Characters” - This problem requires you to find the longest substring without repeating characters, which is directly related to the main problem.

  2. “76. Minimum Window Substring” - This problem is about finding a substring with certain characteristics in a string, which is a similar task to finding k-length substrings.

  3. “438. Find All Anagrams in a String” - Finding anagrams involves dealing with substrings and character counts, which can help with the main problem.

  4. “159. Longest Substring with At Most Two Distinct Characters” - This problem helps you learn how to handle substring problems where you’re limited in the diversity of characters you can use.

  5. “209. Minimum Size Subarray Sum” - It’s not about strings, but this problem’s approach of using a sliding window is highly applicable in the main problem.

  6. “424. Longest Repeating Character Replacement” - This problem involves string manipulation and window sliding, preparing you for the main problem.

  7. “567. Permutation in String” - Dealing with permutations will familiarize you with handling substrings and their characteristics, similar to the main problem.

  8. “992. Subarrays with K Different Integers” - Here, the concept of counting substrings with certain characteristics is introduced, which is related to the main problem.

  9. “340. Longest Substring with At Most K Distinct Characters” - This problem involves handling substrings with a limit on distinct characters, providing a similar challenge to the main problem.

  10. “395. Longest Substring with At Least K Repeating Characters” - This problem is a variation on finding substrings based on character frequency, helping to build intuition for the main problem.

These cover the techniques you need for the “1100. Find K-Length Substrings With No Repeated Characters” problem, including string manipulation, sliding window technique, and handling substrings with certain characteristics.

Problem Classification

The problem falls under the domain of String Manipulation and Combinatorial Search.

‘What’ Components

  1. A string s consisting of lowercase English letters.
  2. An integer k, which is the length of the substring to be considered.
  3. The goal is to find substrings of length k in the given string s that have no repeated characters.
  • Search Problem: You’re searching through all possible substrings of length k to find those with unique characters.
  • Counting Problem: The task is not just to find such substrings but to count them.
  • Constraint-based: We have constraints like 1 <= s.length <= 104 and 1 <= k <= 104.

This problem can be categorized as a Constrained Counting and Search problem within the domain of String Manipulation.

Clarification Questions

  1. Can the string s contain any special characters or is it strictly lowercase English letters?
  2. What should be returned if k is greater than the length of s? The problem states it should return 0; is that correct?
  3. Is an empty string a valid input for s? If so, what should be the output?
  4. Is it guaranteed that k will always be a positive integer?
  5. What should be the output if there are no substrings of length k with unique characters? Should it return 0?
  6. Can k be equal to 1, and if so, should the output then be equal to the length of the string s?
  7. Is performance or optimization a concern? For example, are we looking for the most efficient solution in terms of time and space complexity?
  8. What should happen if the string s contains only one unique character? Should it return 0 because there are no substrings of length k with unique characters?

These clarification questions help us understand the boundaries and special cases for the problem, allowing for a more comprehensive solution.

Problem Analysis and Key Insights

  1. Sliding Window: The problem hints at using a sliding window of size k to find substrings. You’d slide this window across the string s to check for substrings with unique characters.

  2. Count Uniqueness: Within each window, you have to verify that all characters are unique. This suggests using a data structure to keep track of character occurrences efficiently.

  3. Edge Cases: If k is larger than the string’s length or if k is 1, the problem has straightforward solutions. These are edge cases that need special handling.

  4. Character Set: The problem specifies that s consists of lowercase English letters, simplifying the types of characters we have to consider.

  5. Output Type: The output is a single integer, so we know we’ll likely be using some form of counting in our solution.

  6. Constraints: The string length constraint (1 <= s.length <= 10^4) indicates that the solution needs to be reasonably efficient, but it doesn’t have to be extremely optimized.

  7. Non-Overlapping: The problem doesn’t state that the substrings have to be non-overlapping, so overlapping substrings are permissible.

By understanding these key aspects, you get clues on potential approaches for solving the problem. You’re likely to use a sliding window, need a fast way to check for character uniqueness, and have to handle a few different edge cases.

Problem Boundary

The scope of this problem is well-defined:

  1. Input: A string s consisting of lowercase English letters and an integer k.

  2. Objective: To find and count the number of substrings of length k that have no repeated characters within them.

  3. Output: A single integer representing the count of such substrings.

  4. Constraints:

    • String length is between 1 and 10^4.
    • k is between 1 and 10^4.
    • Only lowercase English letters are involved.

  5. Edge Cases: The problem explicitly asks to consider cases where k is larger than the length of s.

The problem does not involve external elements like database interaction, file reading, or network communication. It’s a self-contained algorithmic problem.

Establishing the boundary of this problem involves understanding the limits set by the constraints and what is explicitly stated in the problem description.

  1. Input Boundaries:

    • Minimum length of s is 1, and maximum is 10^4.
    • k is an integer ranging between 1 and 10^4.

  2. Output Boundaries:

    • Output is an integer. Minimum could be 0 (if no such substrings exist), and maximum would be tied to the input string length and k.

  3. Functional Boundaries:

    • The function needs to return an integer count.
    • The function only deals with lowercase English alphabets.

  4. Operational Boundaries:

    • The problem is expected to be solved in reasonable computational time, although exact time complexity is not specified.

  5. Edge Cases:

    • Cases where k is greater than the length of s should return 0, as mentioned in the problem statement.

  6. Excluded Scenarios:

    • Uppercase letters, symbols, or numbers in the string are out of scope, as the string is specified to only contain lowercase English letters.

Understanding these boundaries helps focus on solving what is actually required without being sidetracked by what-ifs and other extraneous conditions.

Distilling the Problem to Its Core Elements

  1. Fundamental Concept:

    • The problem is essentially a sliding window problem. It explores subarrays of a given size, checking for unique elements in those subarrays.

  2. Simplest Description:

    • Imagine you have a sequence of letters. You want to slide a small window over that sequence and count how many times you see a unique set of letters within that window.

  3. Core Problem:

    • The core problem is counting the number of unique substrings of a fixed size in a given string.

  4. Key Components:

    • Input String (s)
    • Window size (k)
    • Substrings of size k
    • Uniqueness of characters in those substrings

  5. Minimal Set of Operations:

    • Initialize a counter for unique substrings.
    • Slide a window of size k over the string s.
    • Check if all characters in the window are unique.
    • If unique, increment the counter.
    • Return the counter.

By identifying these aspects, we can better focus our efforts on solving what is crucial to this problem.

Visual Model of the Problem

Visualizing this problem can be quite helpful, especially since it involves the concept of a “sliding window” over a string.

  1. String as a Sequence:

    • First, consider the string s as a sequence of characters, laid out horizontally.
    For example: h a v e f u n o n l e e t c o d e

  2. Sliding Window:

    • Then visualize a window of k size that “slides” over this sequence, one character at a time.
    Initial window (k=5): [h a v e f] u n o n l e e t c o d e
Next window: h [a v e f u] n o n l e e t c o d e

  3. Check for Uniqueness:

    • While the window slides, you could visualize a checkbox or a tick mark appearing above a window if all the characters inside it are unique.
    [h a v e f] ✓
[a v e f u] ✓

  4. Count:

    • Keep a counter next to your string to keep track of how many unique windows you’ve found.

  5. Constraints:

    • Consider marking windows that are too large (k > length of s) or too small (k < 1) as invalid.

By visualizing the problem in this manner, you can more easily understand what the problem asks for and how you might go about solving it.

Problem Restatement

The problem asks you to find all unique substrings of a given length ‘k’ in a given string ’s’. A substring is considered unique if it doesn’t have any repeated characters. The string ’s’ only contains lowercase English letters. The length of the string and ‘k’ both have to be at least 1 and can be as long as 10,000. You need to count these unique substrings and return that count. If ‘k’ is larger than the length of the string, the count should be zero because you can’t find any such substring.

Abstract Representation of the Problem

In the abstract, you have a sequence of elements, and you are interested in finding contiguous sub-sequences of a specified length that contain only unique elements. Your task is to count the number of such sub-sequences.

Key Elements:

  • A sequence of elements (in this case, characters)
  • A specified sub-sequence length
  • A criterion for sub-sequence uniqueness (no repeated elements)

Your goal is to quantify the sub-sequences that meet these criteria.

Terminology

  1. Substring: A contiguous sequence of characters within a string. Understanding what a substring is crucial for solving this problem, as we’re counting specific types of substrings.

  2. Unique Elements: In the context of this problem, it means no repetition of characters within a given substring. Knowing what “unique” means is critical because the task is to find substrings with no repeated characters.

  3. Constraints: These are limitations or conditions given for the problem, like the string length or the length of the substring. They guide the solution’s efficiency requirements.

  4. Contiguous: Elements that are next to each other without any gaps. Here, it means that the characters in the substring must come one after another in the original string.

These terms help you understand what exactly the problem is asking for and how you might approach a solution.

Problem Simplification and Explanation

Breaking it down:

  1. String: Think of this as a necklace made of letter-beads.

  2. Integer k: This represents the length of a piece of the necklace you want to cut.

  3. Substrings: These are the pieces you get after you cut the necklace. They must have ‘k’ beads each.

  4. No Repeated Characters: Each piece you cut must have different kinds of beads, no two beads should be the same in any piece.

  5. Count: Finally, you need to find out how many such unique pieces you can cut from the necklace.

Metaphor:

Imagine you have a necklace made of beads, where each bead represents a letter. You have a pair of magical scissors that can cut pieces from the necklace, but only if the piece has ‘k’ beads and each bead is unique. Your task is to figure out how many such unique pieces you can cut from your necklace.

Key Concepts and Interactions:

  • You start with a necklace (string).
  • You have a rule for cutting pieces (substring of length ‘k’).
  • The cut pieces should have unique beads (unique characters).
  • You count such pieces (output).

The rule for cutting and the uniqueness requirement are your main constraints. They dictate how you go through the necklace and how you make each cut. Finally, you’re interested in the number of unique pieces you can create.

Constraints

  1. String Length Limit: The maximum length of the string is (10^4). This gives us room to consider algorithms that run in linear or near-linear time.

  2. Limited Alphabet: The string consists of lowercase English letters, which means there are at most 26 unique characters. This can be exploited for quick look-up and storage, possibly using an array of size 26.

  3. k-Value Constraint: (1 \leq k \leq 10^4). Since (k) could be as large as the string itself, edge cases where (k > \text{length of string}) should be quickly dismissible, resulting in a zero count.

  4. Substrings of Length k: The problem asks for substrings of exact length (k) with no repeated characters. This implies a sliding window approach could be beneficial, as we only need to check (k) characters at a time.

  5. Count of Substrings: We are asked for the number of such substrings, not the substrings themselves. This implies we can use data structures to keep track of character counts without storing the substrings.

  6. No Repeated Characters: We’re not allowed to have duplicates within a substring. This can be exploited to quickly eliminate possibilities within the sliding window by using a set to store unique characters.

  7. Lowercase English Letters: Since the input consists of lowercase English letters, data structures like arrays can be easily indexed using ASCII values, allowing quick access and manipulation.

Identifying these specific characteristics helps us recognize the possible avenues for an efficient algorithm. For example, a sliding window approach combined with an array to track character frequencies can be a powerful way to solve this problem efficiently.

  1. Linear Time Feasibility: With the maximum length of the string and (k) capped at (10^4), linear or near-linear time complexity is acceptable. This allows for a single-pass or a few passes through the string without hitting performance bottlenecks.

  2. Alphabet Limitation: The alphabet is limited to lowercase English letters, a total of 26 characters. This constraint allows us to use simple and efficient data structures like arrays for quick look-ups and modifications, which contributes to the overall performance of the algorithm.

  3. Early Exit Conditions: The constraint that (k) must be between 1 and (10^4) gives us specific conditions where we can immediately return 0 as the output (e.g., when (k > \text{length of string})).

  4. Memory Efficiency: The limited size of the string and alphabet makes it possible to use extra memory for data structures without worrying about exceeding memory limits.

  5. No Repeated Characters: This constraint gives us the idea of using a set or an array to track unique characters in each substring of length (k), allowing us to quickly identify valid substrings.

  6. Fixed Substring Length: Because we’re looking for substrings of a fixed length (k), we can efficiently slide a window of size (k) across the string to examine each candidate substring, without having to look at substrings of other lengths.

Understanding these constraints can guide us in choosing the most suitable data structures and algorithms for solving the problem efficiently. For instance, a sliding window approach seems to naturally fit within these constraints.

Case Analysis

Test Cases

Case 1: Minimal Input (Edge Case)

  • Input: s = “a”, k = 1
  • Output: 1
  • Reasoning: The string has only one character, and ( k = 1 ). There is one substring (“a”) that satisfies the condition.

Case 2: k > String Length (Edge Case)

  • Input: s = “abc”, k = 4
  • Output: 0
  • Reasoning: ( k ) is greater than the length of the string, so it’s impossible to find any substring.

Case 3: k = String Length

  • Input: s = “abcd”, k = 4
  • Output: 1
  • Reasoning: ( k ) equals the length of the string, and all characters are unique, so there is only one possible substring.

Case 4: Repeated Characters

  • Input: s = “aabb”, k = 2
  • Output: 0
  • Reasoning: All 2-character substrings will contain repeated characters, hence no valid substrings.

Case 5: Varied Substring Counts

  • Input: s = “abcabc”, k = 3
  • Output: 4
  • Reasoning: The valid substrings are “abc”, “bca”, “cab”, “abc”. Notice that “abc” can appear more than once as long as they are from different positions in the string.

Case 6: All Characters Same (Edge Case)

  • Input: s = “aaaa”, k = 2
  • Output: 0
  • Reasoning: All substrings of length ( k ) have repeated characters.

Case 7: Multiple Unique Substrings

  • Input: s = “xyzabcxyz”, k = 3
  • Output: 6
  • Reasoning: The valid substrings are “xyz”, “yza”, “zab”, “abc”, “bcx”, “cxy”, all of which have unique characters.

Case 8: Single Repeating Character Breaks Multiple Substrings

  • Input: s = “abcda”, k = 4
  • Output: 1
  • Reasoning: Even though you could form two substrings of length 4 (“abcd”, “bcda”), the second one contains a repeating character ‘a’.

Key Insights

  1. Edge Cases: The edge cases primarily revolve around ( k ) being greater than or equal to the length of the string and the presence of repeating characters in the string.

  2. Early Exit: Some cases allow for early exits, like when ( k ) is greater than the string length.

  3. Counting Uniqueness: The substrings must have unique characters, so the use of data structures for quick look-up of repeating characters is crucial.

  4. Fixed Length: Substrings are of fixed length ( k ), which simplifies the problem and indicates that a sliding window approach could be useful.

These test cases cover a range of scenarios and help ensure that the implemented solution would be robust and comprehensive.

Visualization helps to understand the problem space better. Here’s a simple way to visualize each of the test cases:

Case 1: Minimal Input (Edge Case)

  • Visual: a
  • Substrings: [a]
  • Count: 1

Case 2: k > String Length (Edge Case)

  • Visual: abc -> k = 4
  • Substrings: None
  • Count: 0

Case 3: k = String Length

  • Visual: abcd
  • Substrings: [abcd]
  • Count: 1

Case 4: Repeated Characters

  • Visual: aabb
  • Substrings: None
  • Count: 0

Case 5: Varied Substring Counts

  • Visual: abcabc
  • Substrings: [abc, bca, cab, abc]
  • Count: 4

Case 6: All Characters Same (Edge Case)

  • Visual: aaaa
  • Substrings: None
  • Count: 0

Case 7: Multiple Unique Substrings

  • Visual: xyzabcxyz
  • Substrings: [xyz, yza, zab, abc, bcx, cxy]
  • Count: 6

Case 8: Single Repeating Character Breaks Multiple Substrings

  • Visual: abcda
  • Substrings: [abcd]
  • Count: 1

Visualization Tips

  1. Arrows: You can use arrows to mark the (k)-length substrings you are examining, e.g., “ab**->c<-**d” for a 1-length substring around ‘c’.

  2. Highlighting: Use highlighting or bolding to emphasize unique substrings that meet the problem’s requirements.

  3. Count: Keep a running tally to understand how many substrings meet the condition.

  4. Color Coding: Use different colors for valid and invalid substrings to easily distinguish between them.

This visual representation provides a quick way to comprehend the various test cases and how they meet or fail the problem’s conditions.

Identification of Applicable Theoretical Concepts

Certainly. This problem intersects with several mathematical and algorithmic concepts that can make it more manageable:

Sliding Window Technique

The problem asks for contiguous substrings of length (k). The sliding window technique is ideal for working with such problems, allowing us to check each substring in (O(1)) time by simply adding and removing characters from the window.

Hashing

Hashing is another concept that can be applied here. We can use a hash set to store characters and efficiently check for duplicates within a given substring.

Combinatorics

The problem could also be viewed as a combinatorial problem, where you’re selecting (k) characters from (n) without replacement, though this is more of a stretch and might not lead to efficient solutions.

String Metrics

We’re interested in substrings without repeating characters. This is related to metrics that measure string diversity or entropy, though in a simplified way.

Greedy Algorithm

We could also frame this as a greedy algorithm problem. We are essentially trying to maximize the number of unique substrings that meet the condition. Every time we slide the window to a new position, we make the “greedy” choice of adding a new character and removing an old one to maintain a window of size (k).

Graph Theory

Theoretically, you could build a graph where each vertex is a unique character, and an edge exists if two characters are adjacent in a substring. A valid substring would correspond to a simple path of length (k). However, this approach is likely to be overcomplicated for this problem.

Time Complexity

Understanding of (O(n)) time complexity will come in handy, as we aim to solve this problem in linear time.

By leveraging these mathematical and algorithmic concepts, we can aim for a more efficient solution. Most prominently, the sliding window technique combined with hashing can be very effective.

Simple Explanation

Let’s say you have a bag of letter blocks, like the ones children use to learn the alphabet. You’re asked to make small rows of these blocks, each with exactly 5 blocks in a row. But there’s a catch: each row must have 5 different letters; no letter should repeat in the same row.

You can keep making these rows one after the other. For example, with the blocks spelling out “havefunonleetcode,” you could make rows like “havef,” “avefu,” “vefun,” and so on. The question is, how many different rows can you make following this rule?

So, this problem is basically asking the same thing but with letters in a string instead of blocks. You need to find out how many such “rows” or “substrings” you can form with no repeating letters.

Problem Breakdown and Solution Methodology

Certainly. The overall approach to solving this problem can be broken down into a series of steps. Let’s think of it like a conveyor belt in a factory that produces toy blocks. Each block has a letter on it, and you are in charge of quality control.

Steps to Solve the Problem

  1. Initialize Counters and Set: At the start of your shift, you need to set up a counter for the number of valid rows you find, and an empty “inspection area” to examine each row of blocks.

    • Metaphor: Think of this as setting your workstation with a tally counter and an empty table space.

  2. Slide the Window: Imagine a 5-block window sliding over the conveyor belt to look for valid rows. Your job is to inspect this window.

    • Metaphor: The sliding window is like your inspection area on the table. You’re checking to see if each window holds a valid product—blocks with unique letters.

  3. Check Validity: As each window slides in, check if it contains all unique letters.

    • Metaphor: You look at the blocks in the inspection area. If you find duplicates, it’s a bad product.

  4. Count if Valid: If the row is valid (no repeating letters), increase your valid row counter by 1.

    • Metaphor: Each time you find a valid product, you click your tally counter.

  5. Move to Next: Slide the window one block to the right and repeat steps 3 and 4.

    • Metaphor: You slide the blocks on the table to make space for a new one from the conveyor belt.

  6. Return the Count: At the end of your shift, or when the conveyor belt is empty, your tally counter shows the total number of valid rows.

Changes in Problem’s Parameters

  1. Change in k: If k changes (the length of each row or substring), the size of your inspection window changes too.

    • Metaphor: Your table space gets bigger or smaller based on the new row size.

  2. Change in String Length: A shorter or longer string is like a shorter or longer conveyor belt.

    • Metaphor: A longer belt gives you more products to inspect, possibly increasing the number of valid products.

Example

Let’s use the example: s = “havefunonleetcode”, k = 5.

  1. Start with your counter at 0 and the inspection area empty.
  2. Slide the first window: “havef”. It’s valid (no repeating letters). Counter = 1.
  3. Next window: “avefu”. It’s also valid. Counter = 2.
  4. And so on. When you reach the end, your counter is at 6, which is the answer.

By following these steps, we can solve the problem efficiently, making sure we count all possible valid rows of blocks (or substrings).

Inference of Problem-Solving Approach from the Problem Statement

  1. Substring: This refers to a contiguous set of characters within a string. Knowing that we are looking for substrings helps guide us toward using a “sliding window” strategy to examine each possible substring of length ( k ).

  2. Length k: The problem specifies that we are interested in substrings of a certain length ( k ). This length defines the width of our sliding window and ensures we only consider substrings of this size.

  3. No Repeated Characters: This is a constraint that narrows down which substrings we are interested in. Knowing this, we can use a data structure like a set to easily identify and ignore substrings with repeated characters.

  4. Return the Number: The problem asks for a numerical answer, so we know we’ll need a counter variable to keep track of the number of valid substrings we find.

  5. Constraints: The problem’s constraints on the string length (( 1 \leq \text{s.length} \leq 10^4 )) and ( k ) (( 1 \leq k \leq 10^4 )) tell us that the solution must be efficient. This informs our choice of using a sliding window strategy, which is ( O(n) ), instead of a brute-force approach that would be less efficient.

These key terms guide the method and strategy for solving the problem. For example, the “sliding window” approach naturally emerges from the need to examine substrings of a specific length ( k ), while the constraint of “no repeated characters” informs the use of a set for quick look-up.

  1. Substring and Length k: Create a table where each row represents a possible substring of length ( k ). This table can help you visualize the “sliding window” moving across the string.

    Substring 1Substring 2Substring 3…
    havefavenuvefun…

  2. No Repeated Characters: In another column next to each substring, use checkboxes or symbols to indicate whether the substring has no repeated characters or not. This could be thought of as a filter for the substrings we are interested in.

    SubstringNo Repeats?
    havef✓
    avenux
    vefun✓

  3. Return the Number: At the bottom of the table, you could have a row that counts the number of checked substrings, representing the number of substrings with no repeated characters.

  4. Constraints: Visualize the constraints by marking the table’s size limitations based on ( s.length ) and ( k ). If ( k ) is greater than ( s.length ), indicate that it’s not possible to find any substring.

By setting up tables like these, you can visualize the problem’s properties and constraints, making it easier to come up with a solution.

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. Stepwise Refinement of Approach

    • Initialize Counter: Start by initializing a counter to zero. This counter will keep track of the number of valid substrings.

    • Iterate Through String: Loop through the given string, examining each substring of length ( k ).

    • Check for Uniqueness: Within each substring, check if all the characters are unique.

    • Update Counter: If a substring meets the criteria, increment the counter.

    • Output the Counter: After examining all possible substrings, output the counter’s value.

  2. Granular, Actionable Steps

    • Initialization

      1. Create a counter variable set to zero.
      2. Create a “window” data structure (like an array or hash table) to store characters of the current substring.

    • Main Loop

      1. Start from the first character and establish a “window” of ( k ) characters.
      2. Check for duplicate characters within this “window.”
      3. Move the “window” one character to the right.
      4. Repeat steps 2-3 until the end of the string is reached.

    • Uniqueness Check

      1. For each character in the “window,” mark its occurrence.
      2. If any character appears more than once, the substring is invalid.

    • Counter Update

      1. If the substring is valid, increment the counter by 1.

    • Final Output

      1. Return the counter’s value.

  3. Independent Parts

    • The operation of checking for duplicate characters within a given “window” can be isolated as an independent function.
    • The counter can be updated independently of the other operations.

  4. Repeatable Patterns

    • The “sliding window” technique is a repeatable pattern, moving one character at a time from start to end.
    • Checking for uniqueness within a “window” is another repeatable operation, performed each time the “window” moves.

By breaking down the problem in this way, we can approach the solution step-by-step, ensuring that each part contributes to solving the overall problem.

Solution Approach and Analysis

Approach to Solving the Problem

  1. Initialization

    • Think of this step like setting the odometer in your car to zero before a road trip. You initialize a counter to zero that will keep track of the number of unique substrings.

  2. Sliding Window Setup

    • Imagine a “window” that captures a portion of the string, much like using a magnifying glass to look at a section of a road map. This window will always cover (k) characters.

  3. Sliding Window Movement

    • Move the “window” one character at a time, from the beginning to the end of the string. It’s like moving your magnifying glass along the road map to examine different sections.

  4. Uniqueness Check

    • Within each “window,” see if all characters are unique. This is akin to checking if all the landmarks in a section of your road map are distinct.

  5. Counter Update

    • Every time a “window” passes the uniqueness check, increment the counter. It’s like adding a point to your score in a game each time you find a unique set of landmarks.

  6. Return the Counter

    • At the end, the counter will tell you the total number of unique substrings of length (k).

How Parameters Affect the Solution

  • Length of the string ((s.length)): The longer the string, the more “windows” you’ll have to check. This increases the time it takes to find the answer.

  • Size of (k): A larger (k) means fewer windows to check but requires more time to check each window for uniqueness.

Example Cases

  1. Example 1: ( s = “havefunonleetcode”, k = 5 )

    • Windows: “havef”, “avefu”, “vefun”, “efuno”, “funon”, “unonl”, “onlet”, “letco”, “etcod”, “tcode”
    • Unique Substrings: “havef”, “avefu”, “vefun”, “efuno”, “etcod”, “tcode” (Counter = 6)

  2. Example 2: ( s = “home”, k = 5 )

    • No window of size 5 can be formed. (Counter = 0)

This breakdown should help you understand the problem’s structure and how each step contributes to the solution.

Identify Invariant

The invariant in this problem is the size of the “sliding window,” which is (k). No matter where the sliding window is positioned within the string (s), the window size always remains constant at (k) characters. This constant window size is what allows us to systematically scan the string for substrings with no repeated characters.

Identify Loop Invariant

The loop invariant in this problem would be that, at the beginning of each iteration of the loop, the sliding window of size (k) contains no duplicate characters. This must be true before the loop starts, remain true after each iteration, and still be true when the loop terminates.

This loop invariant enables us to keep track of unique substrings of length (k) as we go through each iteration. It also helps to ensure that we’re consistently following the problem’s requirement to find substrings with no repeated characters.

In this specific problem, the loop invariant and the invariant can be considered the same. Both refer to the property that a sliding window of size (k) contains no duplicate characters as you iterate through the string. This property must hold before, during, and after the loop execution to ensure that you are correctly identifying substrings of length (k) with no repeated characters.

Thought Process

Certainly. Let’s start by identifying the key cues in the problem statement and the insights they provide.

Key Cues and Insights:

  1. Substrings of length k: You’re looking for substrings that are exactly (k) characters long.
  2. No repeated characters: These substrings should not have any repeated characters.
  3. Return the number of such substrings: The final output is a count, not the substrings themselves.

These cues point to a few possible approaches:

  1. Sliding window: Given we need contiguous substrings of fixed length (k).
  2. Hashing or Set: To quickly check for repeated characters.

Thought Process:

  1. Initialization: Initialize a set and a count variable.
  2. Iterate through the string: Use a sliding window of size (k) to iterate through the string (s).
  3. Character Checks: For each window, check for duplicate characters. If none, increment the count.
  4. Final Output: Return the count as the answer.

Python Code:

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def count_substrings(s, k):
    # Initialize variables
    unique_substrings = set()
    n = len(s)
    
    # Base condition
    if k > n:
        return 0
    
    # Initialize sliding window
    char_set = set(s[:k])
    
    # Special case for first window
    if len(char_set) == k:
        unique_substrings.add(s[:k])
    
    # Sliding window logic
    for i in range(1, n - k + 1):
        # Remove the outgoing character from set
        char_set.remove(s[i - 1])
        
        # Add the incoming character into set
        char_set.add(s[i + k - 1])
        
        # Check if all characters in the window are unique
        if len(char_set) == k:
            unique_substrings.add(s[i:i + k])
    
    return len(unique_substrings)

# Example usage
print(count_substrings("havefunonleetcode", 5))  # Output: 6
print(count_substrings("home", 5))  # Output: 0

The above code efficiently solves the problem using a sliding window approach, with a time complexity of (O(n)) and space complexity of (O(k)).

Establishing Preconditions and Postconditions

Parameters:

  1. Inputs: The method takes two parameters:

    • s (type: str): A string representing the sequence of characters to analyze.
    • k (type: int): An integer specifying the length of the substrings to consider.

  2. Types:

    • s is a string.
    • k is an integer.

  3. Context:

    • s represents the text in which we’re finding substrings.
    • k defines the exact length of each substring we are interested in.

Preconditions:

  1. State: No specific state required for the program.

  2. Constraints:

    • ( 1 \leq \text{len}(s) \leq 10^4 )
    • ( 1 \leq k \leq 10^4 )

  3. Program State: No specific state requirements.

Method Functionality:

  1. Expected to Do: The method counts the number of unique substrings in s that are of length k and have no repeating characters.

  2. Interaction: It iterates through the input string s using a sliding window approach. It maintains a set to check the uniqueness of each window of size k.

Postconditions:

  1. State: No changes to the state of the program or parameter values.

  2. Return Value: An integer is returned representing the count of unique substrings of length k with no repeating characters.

  3. Side Effects: None.

Error Handling:

  1. Preconditions Not Met: The method accounts for the case where k is greater than the length of s by returning 0.

  2. Exception Handling: No exceptions are thrown; the function simply returns an integer value as specified.

Problem Decomposition

Problem Understanding:

  1. In Own Words: The problem asks to find how many unique substrings of length k exist in a given string s, such that these substrings have no repeated characters.

  2. Key Components:

    • Input string s.
    • Length k for the substrings.
    • No repeated characters in each substring.

  3. Requirements:

    • Count of unique substrings.

Initial Breakdown:

  1. Major Parts:
    • Reading and validating inputs.
    • Iterating through the string to find substrings.
    • Checking each substring for uniqueness of characters.
    • Counting the valid substrings.

Subproblem Refinement:

  1. Reading Inputs: Ensure the string and length are within the defined bounds.

  2. Iterating through String: Use a sliding window approach.

  3. Checking Uniqueness: Utilize a data structure to quickly determine if a substring contains repeated characters.

  4. Counting Substrings: Keep a counter to tally the valid substrings.

Task Identification:

  1. Input Validation: Ensure k is less than or equal to the length of s.

  2. Sliding Window: Move a window of size k across s.

  3. Uniqueness Check: A set can be used for O(1) lookups.

  4. Counter Update: Increment counter if the substring is valid.

Task Abstraction:

  1. Input Validation: General enough but specific to this problem.

  2. Sliding Window: Reusable concept.

  3. Uniqueness Check: Abstract but still specific to ensure no repeated characters.

  4. Counter Update: Very generic.

Method Naming:

  1. Input Validation: validate_inputs()

  2. Sliding Window: sliding_window()

  3. Uniqueness Check: check_uniqueness()

  4. Counter Update: update_counter()

Subproblem Interactions:

  1. Order:

    • Validate Inputs
    • Initialize Counter
    • Sliding Window
      • Uniqueness Check
      • Counter Update

  2. Dependencies:

    • Input validation must be done first.
    • Uniqueness Check is dependent on each window from the Sliding Window.
    • Counter Update is dependent on the outcome of the Uniqueness Check.

From Brute Force to Optimal Solution

Brute-Force Solution

  1. Iterate through all substrings of length k in the string s.
  2. For each substring, use a set to check if all characters are unique.
  3. If unique, increment the counter.
  4. Return the counter.

Here is some sample Python code to demonstrate the brute-force solution:

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def count_unique_substrings(s, k):
    count = 0
    for i in range(0, len(s) - k + 1):
        substring = s[i:i+k]
        if len(set(substring)) == k:
            count += 1
    return count

Inefficiencies:

  1. Time Complexity: For each of the n - k + 1 substrings, we are creating a set, which takes O(k) time. So the time complexity is O((n - k + 1) * k) = O(n * k).
  2. Space Complexity: For each substring, a new set is created leading to O(k) extra space.

Optimized Solution

We can optimize this problem by using the Sliding Window Technique.

  1. Initialize a counter to zero and create an empty set.
  2. Use a window of size k and slide it across s.
  3. For the first window, check if all characters are unique. If so, increment the counter.
  4. For each subsequent window, remove the first character of the previous window and add the new character. Check for uniqueness and update the counter accordingly.

Here’s the code:

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def count_unique_substrings_optimized(s, k):
    count = 0
    char_set = set()
    
    for i in range(len(s)):
        char_set.add(s[i])
        
        if i >= k:
            char_set.remove(s[i - k])
        
        if i >= k - 1 and len(char_set) == k:
            count += 1
            
    return count

Optimizations:

  1. Time Complexity: This approach needs only one pass through the string, so the time complexity is O(n).
  2. Space Complexity: The set will at most contain k characters, so the space complexity is O(k).

Summary of Optimizations:

  1. Time complexity reduced from O(n * k) to O(n) by using a sliding window.
  2. Space complexity remains the same at O(k), but the code is more efficient and elegant.

Code Explanation and Design Decisions

1. Initial Parameters

  • s: This is the input string where we look for unique substrings of length k. It’s the main data source for our problem.
  • k: This parameter specifies the length of the substrings that we consider. It guides how we form substrings from s.

2. Primary Loop

The primary loop iterates over each character in the string s. Each iteration represents an opportunity to form a new substring of length k that might be unique. As we advance in the loop, we shift our k-length window one character to the right.

3. Conditions or Branches

There are two significant conditions:

  • if i >= k: This condition ensures we don’t try to remove an element from char_set before it has k elements. It aligns with the constraint that we only consider substrings of length k.
  • if i >= k - 1 and len(char_set) == k: Checks whether the current window’s characters are all unique. If so, the counter is incremented.

4. Updates or Modifications

  • char_set.add(s[i]): Adds the new character at the window’s end.
  • char_set.remove(s[i - k]): Removes the character that just slid out of the window. These modifications keep char_set aligned with the current window of length k.

5. Invariant

The invariant here is the char_set, which always contains the characters from the current k-length window in the string s. It helps us quickly check whether all characters in the current window are unique, which aligns with our problem’s objective of finding such unique substrings.

6. Final Output Significance

The final output, count, represents the number of unique substrings of length k in the input string s. It directly answers the problem’s query, adhering to all its constraints and requirements.

Coding Constructs

1. High-Level Strategies

The code primarily uses a sliding window technique coupled with a set data structure for quick look-up. These strategies make it efficient to find unique substrings of a given length k in a string s.

2. Non-Programmer Explanation

Imagine you’re reading a book and using a transparent ruler to highlight k letters at a time. You only care about the highlighted sections where all the letters are different, and you want to count how many such unique sections there are as you move through the book one letter at a time.

3. Logical Elements

  • Iteration: Looping through each character in the string.
  • Conditionals: Checking specific criteria (like set size matching k) to make decisions.
  • Data Structure: Using a set for quick look-up of characters.

4. Algorithmic Approach in Plain English

Start from the first character and form a substring of length k. Check if all the characters are unique. Move one character forward and form a new substring of the same length, again checking for uniqueness. Repeat this until you reach the end of the string. Keep a count of the unique substrings found.

5. Key Steps or Operations

  • Forming a k-length substring starting from the current position.
  • Checking the uniqueness of characters in the substring.
  • Keeping a count of such unique substrings.

6. Algorithmic Patterns

  • Sliding Window: The code uses a fixed-size window that slides over the string.
  • Quick Lookup: Utilizes a set for constant-time look-up to check uniqueness.
  • Counter: Maintains a running total of unique substrings satisfying the conditions.

Language Agnostic Coding Drills

1. Distinct Concepts in the Code

  1. Variable Declaration and Initialization: Creating variables to hold temporary and final values.
  2. Looping Through a Sequence: Iterating over the elements in the string.
  3. String Slicing: Extracting a part of the string based on an index range.
  4. Data Structure (Set): Using a set to hold unique elements.
  5. Conditional Statements: If-else constructs for decision-making.
  6. Set Operations: Adding and removing elements to/from a set.
  7. Counters: Keeping track of counts or sums.
  8. Output Formatting: Producing the final output in the required format.

2. Concepts in Order of Increasing Difficulty

  1. Variable Declaration and Initialization: Easiest, as it’s a foundational skill.
  2. Looping Through a Sequence: Also foundational but involves control flow.
  3. String Slicing: Slightly more complex due to the understanding of indices.
  4. Conditional Statements: Requires logical reasoning for decisions.
  5. Output Formatting: May involve multiple steps to get the output in the desired form.
  6. Counters: Introduces the need for maintaining state during computation.
  7. Data Structure (Set): Involves understanding a specialized data structure and its properties.
  8. Set Operations: Most complex, requires understanding how to efficiently manipulate sets.

3. Problem-Solving Approach

  1. Understand the Problem: First, grasp what the problem asks for—unique substrings of a given length k in a string s.

  2. Variable Initialization: Initialize variables like counters or sets that you’ll use for storing intermediate and final values.

  3. Loop Through String: Use a loop to iterate through the string s from the first character to the last.

  4. Extract Substring: Use string slicing to pull out a substring of length k starting from your current loop index.

  5. Check Uniqueness: Utilize a set to check the uniqueness of the characters in the substring.

  6. Conditional Checks: Add conditional statements to decide whether the substring meets your criteria. If yes, you may want to update a counter.

  7. Update Set: Based on the loop’s progress, update the set efficiently to reflect the characters in the new substring without having to create a new set from scratch.

  8. Counter: Keep a counter updated for each unique substring found.

  9. Format Output: Finally, ensure that the output— the count of unique substrings— is produced in a format that satisfies the problem’s requirements.

By mastering each of these drills, you can assemble them to form the complete solution. Each drill plays a specific role, and when they work in tandem, they efficiently solve the problem at hand.

Targeted Drills in Python

1. Python-based Coding Drills for General Concepts

Variable Declaration and Initialization

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# Declare an integer variable and initialize it
my_int = 0

# Declare a string variable and initialize it
my_str = "hello"

Looping Through a Sequence

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# Loop through a list of numbers
for num in [1, 2, 3, 4]:
    print(num)

String Slicing

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# Slice a string
text = "hello"
print(text[1:4])  # Outputs "ell"

Conditional Statements

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# Simple if-else statement
x = 5
if x > 4:
    print("Greater")
else:
    print("Not greater")

Output Formatting

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# Formatting a string output
name = "John"
print(f"Hello, {name}!")

Counters

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# Counter for counting events
counter = 0
for i in range(5):
    counter += 1

Data Structure (Set)

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# Create a set and add an element
my_set = set()
my_set.add(1)

Set Operations

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# Add and remove elements from a set
my_set = {1, 2}
my_set.add(3)  # {1, 2, 3}
my_set.remove(1)  # {2, 3}

2. Problem-Specific Concepts and Drills

Substring Uniqueness Check

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# Check if all characters in a string are unique
def is_unique(s):
    return len(s) == len(set(s))

This is crucial for our problem because we need to check the uniqueness of each substring.

3. Integrating the Drills to Solve the Problem

  1. Initialize Variables: Start by declaring a counter variable to keep track of the number of unique substrings of length k.

  2. Loop Through String: Implement a for loop that goes through the string, s.

  3. Extract Substring: In each iteration, slice a substring from s using string slicing.

  4. Check Uniqueness: Utilize the is_unique() function to determine if the substring’s characters are unique.

  5. Conditional Checks: Use an if statement to decide whether to increment the counter or not based on the result from is_unique().

  6. Counter Update: If the substring is unique, increment the counter.

  7. Format Output: Finally, print the value of the counter variable as the output.

By following these steps in this sequence, we can combine all the drills into a complete solution to solve our problem.

Q&A

Similar Problems

Here are 10 problems that use similar underlying concepts:

  1. Two Sum

    • Similarity: Uses a set or hash table to find complementary elements, much like how we used sets to check for unique elements.

  2. Valid Anagram

    • Similarity: Checks for character uniqueness and counts, similar to how we used set operations to verify substring uniqueness.

  3. Palindrome Permutation

    • Similarity: Requires you to work with substrings and character frequency counts, similar to our focus on substrings and unique characters.

  4. Longest Substring Without Repeating Characters

    • Similarity: Directly related as it also deals with finding unique substrings within a string.

  5. Remove Duplicates from Sorted Array

    • Similarity: Requires the identification and removal of duplicates, somewhat similar to how we had to identify unique substrings.

  6. Subarray Sum Equals K

    • Similarity: Involves traversing a sequence to find a specific property (a sum in this case), similar to our traversal of substrings to find unique ones.

  7. Minimum Window Substring

    • Similarity: Requires an understanding of substrings and how to manipulate them efficiently, much like our original problem.

  8. Valid Parentheses

    • Similarity: Uses a stack to ensure each opening bracket has a closing bracket, comparable to how we used a set to check for unique characters in a substring.

  9. Find the Duplicate Number

    • Similarity: Requires a technique to find duplicates in an array, which can be conceptually linked to finding unique substrings.

  10. Permutations

    • Similarity: Works with the idea of unique arrangements, much like how we were looking at unique substrings.

Each of these problems is chosen due to their requirement for understanding sequences, substrings, unique elements, or some combination of these, which were key aspects in solving our original problem.