What is the time complexity of Replace the Substring for Balanced String?

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

What data structures are used to solve Replace the Substring for Balanced String?

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

What are the key insights for solving Replace the Substring for Balanced String?

Insight 1: The core idea is to use a sliding window approach. We aim to find the smallest window such that replacing the characters within that window will balance the entire string. Insight 2: We don't actually replace the substring. We just identify the substring that, if removed, would leave us with at most n/4 occurrences of each character. Then, the length of the substring is the answer.

Which companies ask LeetCode 1234. Replace the Substring for Balanced String in interviews?

This problem is frequently asked at Accolite.

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Replace the Substring for Balanced String - LeetCode 1234 Solution

Replace the Substring for Balanced String - Complete Solution Guide

Replace the Substring for Balanced String is LeetCode problem 1234, 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 of length n containing only four kinds of characters: 'Q' , 'W' , 'E' , and 'R' . A string is said to be balanced if each of its characters appears n / 4 times where n is the length of the string. Return the minimum length of the substring that can be replaced with any other string of the same length to make s balanced . If s is already balanced , return 0 . Example 1: Input: s = "QWER" Output: 0 Explanation: s is already balanced. Example 2: Input: s = "QQWE" Output: 1

Detailed Explanation

The problem asks us to find the minimum length of a substring within a given string `s` (containing only 'Q', 'W', 'E', 'R' characters) that needs to be replaced with any other string of the same length, such that the resulting string becomes 'balanced'. A string is considered 'balanced' if each of the four characters ('Q', 'W', 'E', 'R') appears exactly `n/4` times, where `n` is the length of the string `s`. The constraint is that `n` is always a multiple of 4. If the input string is already balanced, the function should return 0.

Solution Approach

The provided solution employs a sliding window technique. First, it calculates the required count `k = n/4` for each character. Then, it identifies the characters that have counts exceeding `k` and stores the excess counts in a `needed` counter. The sliding window expands and contracts to find the minimum window size that contains at least the number of excess characters needed to make the overall string balanced. The `formed` variable keeps track of how many of the needed characters' counts are satisfied within the current window.

Step-by-Step Algorithm

Step 1: Calculate `n = len(s)` and `k = n // 4`. `k` represents the ideal count for each character.
Step 2: Count the occurrences of each character in `s` using a `Counter` (or HashMap).
Step 3: Create a `needed` `Counter` (or HashMap) to store characters that have counts greater than `k`. The values in `needed` represent the number of extra occurrences of each character that need to be replaced.
Step 4: Initialize `left = 0`, `min_len = n`, `required = len(needed)`, and `formed = 0`. `left` is the left boundary of the sliding window, `min_len` stores the minimum window size found so far, `required` is the number of characters that have counts greater than k (number of keys in `needed`), and `formed` tracks how many of those character's needs are met within the window.
Step 5: Iterate through the string with the `right` pointer (right boundary of the sliding window).
Step 6: Update the `window_counts` `Counter` (or HashMap) with the character at the `right` pointer.
Step 7: If the character at the `right` pointer is in `needed` and its count in `window_counts` matches its count in `needed`, increment `formed`.
Step 8: While `left <= right` and `formed == required`, update `min_len` with the minimum of `min_len` and `right - left + 1`. Shrink the window from the left.
Step 9: Decrement the count of the character at the `left` pointer in `window_counts`.
Step 10: If the character at the `left` pointer is in `needed` and its count in `window_counts` is now less than its count in `needed`, decrement `formed`.
Step 11: Increment `left` to shrink the window.
Step 12: Return `min_len`.

Key Insights

Insight 1: The core idea is to use a sliding window approach. We aim to find the smallest window such that replacing the characters within that window will balance the entire string.
Insight 2: We don't actually replace the substring. We just identify the substring that, if removed, would leave us with at most n/4 occurrences of each character. Then, the length of the substring is the answer.
Insight 3: Using a `Counter` (or equivalent hash map) is crucial for efficiently tracking the frequency of each character both in the overall string and within the sliding window.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: String, Sliding Window.

Companies

Asked at: Accolite.

Replace the Substring for Balanced String | LeetCode 1234

Medium

String Sliding Window

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

Solution Code

Available in:

import java.util.HashMap;
import java.util.Map;

class Solution {
    public int balancedString(String s) {
        int n = s.length();
        int k = n / 4;

        Map<Character, Integer> counts = new HashMap<>();
        for (char c : s.toCharArray()) {
            counts.put(c, counts.getOrDefault(c, 0) + 1);
        }

        Map<Character, Integer> needed = new HashMap<>();
        for (Map.Entry<Character, Integer> entry : counts.entrySet()) {
            if (entry.getValue() > k) {
                needed.put(entry.getKey(), entry.getValue() - k);
            }
        }

        if (needed.isEmpty()) {
            return 0;
        }

        int left = 0;
        int min_len = n;
        int required = needed.size();
        int formed = 0;
        Map<Character, Integer> window_counts = new HashMap<>();

        for (int right = 0; right < n; right++) {
            char char_right = s.charAt(right);
            window_counts.put(char_right, window_counts.getOrDefault(char_right, 0) + 1);

            if (needed.containsKey(char_right) && window_counts.get(char_right).equals(needed.get(char_right))) {
                formed += 1;
            }

            while (left <= right && formed == required) {
                min_len = Math.min(min_len, right - left + 1);

                char char_left = s.charAt(left);
                window_counts.put(char_left, window_counts.get(char_left) - 1);

                if (needed.containsKey(char_left) && window_counts.get(char_left) < needed.get(char_left)) {
                    formed -= 1;
                }

                left += 1;
            }
        }

        return min_len;
    }
}

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Problem Info

Description

Problem 1234: Replace the Substring for Balanced String - Detailed explanation and solution approaches.

Topics

String Sliding Window

Companies

Accolite

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1234

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

12 steps

1
Step 1: Calculate `n = len(s)` and `k = n // 4`. `k` represents the ideal count for each character.
2
Step 2: Count the occurrences of each character in `s` using a `Counter` (or HashMap).
3
Step 3: Create a `needed` `Counter` (or HashMap) to store characters that have counts greater than `k`. The values in `needed` represent the number of extra occurrences of each character that need to be replaced.
4
Step 4: Initialize `left = 0`, `min_len = n`, `required = len(needed)`, and `formed = 0`. `left` is the left boundary of the sliding window, `min_len` stores the minimum window size found so far, `required` is the number of characters that have counts greater than k (number of keys in `needed`), and `formed` tracks how many of those character's needs are met within the window.
5
Step 5: Iterate through the string with the `right` pointer (right boundary of the sliding window).
6
Step 6: Update the `window_counts` `Counter` (or HashMap) with the character at the `right` pointer.
7
Step 7: If the character at the `right` pointer is in `needed` and its count in `window_counts` matches its count in `needed`, increment `formed`.
8
Step 8: While `left <= right` and `formed == required`, update `min_len` with the minimum of `min_len` and `right - left + 1`. Shrink the window from the left.
9
Step 9: Decrement the count of the character at the `left` pointer in `window_counts`.
10
Step 10: If the character at the `left` pointer is in `needed` and its count in `window_counts` is now less than its count in `needed`, decrement `formed`.
11
Step 11: Increment `left` to shrink the window.
12
Step 12: Return `min_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 linear because we iterate through the string using a sliding window. The space complexity is constant because the number of distinct characters is limited to four, making the size of the counters independent 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, create a modified string with the substring replaced by other characters. Check if the resulting string is balanced. This approach is extremely inefficient.

Time: O(n^3)Space: O(n)

Trade-offs: The brute force approach is simple to understand but has a high time complexity, making it unsuitable for large input strings. Sliding window is significantly faster.

✨

Key Insights

1Insight 1: The core idea is to use a sliding window approach. We aim to find the smallest window such that replacing the characters within that window will balance the entire string.
2Insight 2: We don't actually replace the substring. We just identify the substring that, if removed, would leave us with at most n/4 occurrences of each character. Then, the length of the substring is the answer.
3Insight 3: Using a `Counter` (or equivalent hash map) is crucial for efficiently tracking the frequency of each character both in the overall string and within the sliding window.

📚

Key Terms Explained

For Students

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: The string is already balanced. In this case, the `needed` `Counter` will be empty, and the function will immediately return 0.
Edge case 2: A single character needs to be replaced. The sliding window will quickly find a substring of length 1 that satisfies the balancing requirement.
Edge case 3: String consists of just one repeated character, like 'QQQQ'. In this case, the entire string except for n/4 occurences needs to be replaced. The code correctly identifies the entire string (minus those n/4) as the minimum substring.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where the string is already balanced (returning 0).

✗Common mistake 2: Incorrectly updating the `formed` variable, leading to the window not shrinking or expanding correctly.

✗Common mistake 3: Not accounting for the required number of occurrences of each character (`k = n/4`).

✗Common mistake 4: Using a naive nested loop which leads to a Time Limit Exceeded.

✅

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: Resource balancing in a distributed system. Imagine 'Q', 'W', 'E', 'R' represent different types of tasks. You might want to identify the smallest set of tasks that, if reassigned, would balance the load across all task types.

🏦Real-world scenario 2: DNA sequencing analysis. If 'Q','W','E','R' represented different nucleotide bases, this algorithm could be adapted to find regions that disrupt the balance of bases, suggesting potential mutations or anomalies. This adapted algorithm would be useful if you want to find the smallest region of nucleotides needing edits to 'balance' the DNA sequence.

💡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

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

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

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

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: November 2, 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

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.

How do I determine the window size in sliding window problems?

For fixed-size windows (like max sum of k elements), window size is given. For variable-size windows (like smallest subarray with sum >= target), expand right pointer to meet condition, then shrink left pointer to find optimal. Track the best result during shrinking.

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 "Replace the Substring for Balanced String"?

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

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