What is the time complexity of Longest Repeating Character Replacement?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Longest Repeating Character Replacement?

This problem uses Hash Table, String, Sliding Window. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Longest Repeating Character Replacement?

Insight 1: The problem can be solved using a sliding window approach. We need to maintain a window and keep track of the most frequent character within that window. Insight 2: If the number of characters within the window minus the count of the most frequent character is less than or equal to `k`, then the window is a valid candidate substring.

Which companies ask LeetCode 424. Longest Repeating Character Replacement in interviews?

This problem is frequently asked at ByteDance, Flipkart, MathWorks, PayU, PhonePe and 5 more companies.

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Longest Repeating Character Replacement - LeetCode 424 Solution

Longest Repeating Character Replacement - Complete Solution Guide

Longest Repeating Character Replacement is LeetCode problem 424, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given a string s and an integer k . You can choose any character of the string and change it to any other uppercase English character. You can perform this operation at most k times. Return the length of the longest substring containing the same letter you can get after performing the above operations . Example 1: Input: s = "ABAB", k = 2 Output: 4 Explanation: Replace the two 'A's with two 'B's or vice versa. Example 2: Input: s = "AABABBA", k = 1 Output: 4 Explanation: Replace the one

Detailed Explanation

The problem asks us to find the length of the longest substring in a given string `s` that contains the same letter after performing at most `k` character replacements. We can change any character in the string to any other uppercase English character. The goal is to maximize the length of a substring consisting of the same character after these replacements. For example, if `s = "ABAB"` and `k = 2`, we can replace the two 'A's with 'B's (or vice versa) to get a substring of length 4 ('BBBB' or 'AAAA').

Solution Approach

The solution uses a sliding window approach with two pointers, `left` and `right`. We expand the window by incrementing the `right` pointer and update a frequency map to keep track of the character counts within the window. We also maintain `max_freq`, which stores the maximum frequency of any character seen so far within the window. If the difference between the window size (`right - left + 1`) and `max_freq` is greater than `k`, it means we need to shrink the window by incrementing the `left` pointer. We also decrement the count of the character that `left` is pointing to. We continuously update `max_len` to keep track of the longest valid window encountered so far. The key idea is to maintain a window that, after at most `k` replacements, can be a substring of the same character.

Step-by-Step Algorithm

Step 1: Initialize `left = 0`, `max_freq = 0`, `max_len = 0`, and `counts = [0] * 26` (an array to store the frequencies of each character).
Step 2: Iterate through the string `s` using the `right` pointer (from 0 to `len(s) - 1`).
Step 3: Update the frequency count of the character at the `right` pointer: `counts[ord(s[right]) - ord('A')] += 1`.
Step 4: Update `max_freq` to be the maximum frequency encountered so far: `max_freq = max(max_freq, counts[ord(s[right]) - ord('A')])`.
Step 5: Check if the current window is valid: `if (right - left + 1) - max_freq > k`. If it's not valid, it means we need to shrink the window.
Step 6: If the window is not valid, decrement the frequency count of the character at the `left` pointer: `counts[ord(s[left]) - ord('A')] -= 1`. Increment the `left` pointer: `left += 1`.
Step 7: Update `max_len` to be the maximum window length encountered so far: `max_len = max(max_len, right - left + 1)`.
Step 8: After the loop finishes, return `max_len`.

Key Insights

Insight 1: The problem can be solved using a sliding window approach. We need to maintain a window and keep track of the most frequent character within that window.
Insight 2: If the number of characters within the window minus the count of the most frequent character is less than or equal to `k`, then the window is a valid candidate substring.
Insight 3: We don't necessarily need to shrink the window; we can just slide the right edge and maintain `max_len`. The reason is that if `(right - left + 1) - max_freq > k`, we shift the left bound; if not, we keep the left bound and update `max_len`. So if the window is invalid, we won't update `max_len`. This works because we are only concerned with the *longest* valid substring. Even if we have an invalid window now, we keep its length and shift its left pointer if it's invalid. We maintain the `max_freq` value, which may not be the accurate maximum frequency of the current window; this works because only the length of the window matters.
Insight 4: Use an array of length 26 as a hashmap to store character frequencies instead of a dictionary. Since we only use uppercase letters, this is much more efficient.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Hash Table, String, Sliding Window.

Companies

Asked at: ByteDance, Flipkart, MathWorks, PayU, PhonePe, Pocket Gems, Turing, UiPath, Yandex, Zepto.

Longest Repeating Character Replacement | LeetCode 424

Medium

Hash Table String Sliding Window

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int characterReplacement(String s, int k) {
        int left = 0;
        int max_freq = 0;
        int max_len = 0;
        int[] counts = new int[26];

        for (int right = 0; right < s.length(); right++) {
            int char_right_idx = s.charAt(right) - 'A';
            counts[char_right_idx]++;
            max_freq = Math.max(max_freq, counts[char_right_idx]);

            if ((right - left + 1) - max_freq > k) {
                int char_left_idx = s.charAt(left) - 'A';
                counts[char_left_idx]--;
                left++;
            }
            
            max_len = Math.max(max_len, right - left + 1);
        }

        return max_len;
    }
}

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Problem Info

Description

Problem 424: Longest Repeating Character Replacement - Detailed explanation and solution approaches.

Topics

Hash Table String Sliding Window

Companies

ByteDanceFlipkartMathWorksPayUPhonePePocket GemsTuringUiPath

Stats

Difficulty:Medium

Languages:4

Companies:10

Problem #:424

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Initialize `left = 0`, `max_freq = 0`, `max_len = 0`, and `counts = [0] * 26` (an array to store the frequencies of each character).
2
Step 2: Iterate through the string `s` using the `right` pointer (from 0 to `len(s) - 1`).
3
Step 3: Update the frequency count of the character at the `right` pointer: `counts[ord(s[right]) - ord('A')] += 1`.
4
Step 4: Update `max_freq` to be the maximum frequency encountered so far: `max_freq = max(max_freq, counts[ord(s[right]) - ord('A')])`.
5
Step 5: Check if the current window is valid: `if (right - left + 1) - max_freq > k`. If it's not valid, it means we need to shrink the window.
6
Step 6: If the window is not valid, decrement the frequency count of the character at the `left` pointer: `counts[ord(s[left]) - ord('A')] -= 1`. Increment the `left` pointer: `left += 1`.
7
Step 7: Update `max_len` to be the maximum window length encountered so far: `max_len = max(max_len, right - left + 1)`.
8
Step 8: After the loop finishes, return `max_len`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(n) because the sliding window approach iterates through the string once. The inner `if` condition and updates to `left` happen at most `n` times in total. The space complexity is O(1) because the frequency map `counts` has a fixed size (26) regardless of the input string's length.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute Force

Iterate through all possible substrings and, for each substring, check if it can be made into a repeating character substring with at most `k` replacements. This would involve checking every possible substring from length 1 to `n`.

Time: O(n^3). Checking the validity of a substring of length `m` takes O(m) time, and there are O(n^2) substrings.Space: O(1).

Trade-offs: The brute force approach is straightforward but inefficient. It is not suitable for large input sizes. The sliding window approach is much more efficient in terms of time complexity.

✨

Key Insights

1Insight 1: The problem can be solved using a sliding window approach. We need to maintain a window and keep track of the most frequent character within that window.
2Insight 2: If the number of characters within the window minus the count of the most frequent character is less than or equal to `k`, then the window is a valid candidate substring.
3Insight 3: We don't necessarily need to shrink the window; we can just slide the right edge and maintain `max_len`. The reason is that if `(right - left + 1) - max_freq > k`, we shift the left bound; if not, we keep the left bound and update `max_len`. So if the window is invalid, we won't update `max_len`. This works because we are only concerned with the *longest* valid substring. Even if we have an invalid window now, we keep its length and shift its left pointer if it's invalid. We maintain the `max_freq` value, which may not be the accurate maximum frequency of the current window; this works because only the length of the window matters.
4Insight 4: Use an array of length 26 as a hashmap to store character frequencies instead of a dictionary. Since we only use uppercase letters, this is much more efficient.

📚

Key Terms Explained

For Students

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Sliding Window

A moving "window" over data to process subsets efficiently.

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: `s = "A"`, `k = 0`. The solution correctly returns 1.
Edge case 2: `s = "ABCDE"`, `k = 0`. The solution correctly returns 1.
Edge case 3: `s = "ABAB"`, `k = 2`. The solution correctly returns 4.
Edge case 4: `s = "AABABBA"`, `k = 1`. The solution correctly returns 4.
Edge case 5: Empty String: although the constraint says that the string length is between 1 and 10^5, handling an empty string would just return 0 because of the initial `max_len` value.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not understanding the sliding window approach and trying to use a brute force approach, which leads to TLE (Time Limit Exceeded).

✗Common mistake 2: Failing to update `max_freq` correctly. `max_freq` represents the maximum frequency of a character within the current window and helps determine if the window is valid.

✗Common mistake 3: Incorrectly shrinking the window. You should only shrink the window when the number of replacements needed exceeds `k`.

✗Common mistake 4: Not realizing that it's valid to not shrink the window; it is acceptable to have an invalid window temporarily as long as it doesn't affect the longest valid window found so far, because we are only interested in `max_len`.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Data compression. Finding the longest repeating sequence with a limited number of errors is useful for compressing data with slight variations or noise.

🏦Real-world scenario 2: Bioinformatics. Identifying the longest common subsequence or substring with mismatches can be used to analyze DNA sequences.

🎮Real-world scenario 3: Error Correction. Finding and correcting errors in communication signals where some characters may be corrupted.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

Hash tables trade space for time. Before using one, consider if the problem constraints allow the extra memory. In interviews, always mention this trade-off.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Don't forget to handle key collisions and null/undefined keys. Consider what happens with duplicate values.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 4, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

❓

Frequently Asked Questions

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Longest Repeating Character Replacement"?

For this medium-level Hash Table problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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