What is the time complexity of Swap For Longest Repeated Character Substring?

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

What data structures are used to solve Swap For Longest Repeated Character Substring?

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 Swap For Longest Repeated Character Substring?

Insight 1: The solution revolves around finding groups of consecutive identical characters and then strategically swapping a character to extend these groups. Insight 2: Counting the total occurrences of each character is crucial to determine if a swap is possible and to what extent a group can be extended.

Which companies ask LeetCode 1156. Swap For Longest Repeated Character Substring in interviews?

This problem is frequently asked at Nutanix.

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Swap For Longest Repeated Character Substring - LeetCode 1156 Solution

Swap For Longest Repeated Character Substring - Complete Solution Guide

Swap For Longest Repeated Character Substring is LeetCode problem 1156, 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 text . You can swap two of the characters in the text . Return the length of the longest substring with repeated characters . Example 1: Input: text = "ababa" Output: 3 Explanation: We can swap the first 'b' with the last 'a', or the last 'b' with the first 'a'. Then, the longest repeated character substring is "aaa" with length 3. Example 2: Input: text = "aaabaaa" Output: 6 Explanation: Swap 'b' with the last 'a' (or the first 'a'), and we get longest repeated character

Detailed Explanation

The problem requires finding the length of the longest substring with repeated characters in a given string `text` after performing at most one swap of two characters within the string. The input `text` consists of lowercase English letters. The goal is to return the maximum length of a substring with repeated characters achievable after the optimal swap (or no swap if it's already optimal). For instance, in "ababa", we can swap the first 'b' with the last 'a' to get "aaaba", where "aaa" has length 3. The constraints limit the length of `text` to at most 2 * 10^4.

Solution Approach

The solution first counts the occurrences of each character in the input string. Then, it identifies groups of consecutive identical characters. The core logic involves two cases: (1) Trying to extend each group by one character by swapping a suitable character from elsewhere in the string into the current group. (2) Merging two groups of the same character separated by a single character by swapping the separating character. The maximum length achieved across all scenarios is returned.

Step-by-Step Algorithm

Step 1: Count the total occurrences of each character in the input string `text` using a HashMap or Counter.
Step 2: Group consecutive identical characters in the string into a list of (character, length) pairs.
Step 3: Iterate through the list of groups and for each group, calculate the maximum possible length achievable by extending the group by one character using a swap. This is limited by the total count of that character in the string. The extended length is `min(groupLength + 1, totalCharacterCount)`.
Step 4: Iterate through the list of groups, checking if any two groups of the same character are separated by exactly one character. If they are, check if swapping the separating character can merge the groups. This is limited by `min(len1 + len3 + 1, totalCharacterCount)`
Step 5: Return the maximum length found in steps 3 and 4.

Key Insights

Insight 1: The solution revolves around finding groups of consecutive identical characters and then strategically swapping a character to extend these groups.
Insight 2: Counting the total occurrences of each character is crucial to determine if a swap is possible and to what extent a group can be extended.
Insight 3: Two main scenarios need to be considered: extending a single existing group and merging two groups of the same character separated by a single different character.
Insight 4: The total frequency of a character in the string limits the maximum length of any repeated character substring containing that character after a single swap.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

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

Companies

Asked at: Nutanix.

Swap For Longest Repeated Character Substring | LeetCode 1156

Medium

Hash Table String Sliding Window

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

Solution Code

Available in:

import java.util.HashMap;
import java.util.ArrayList;
import java.util.List;
import java.util.Map;

class Solution {
    public int maxRepOpt1(String text) {
        Map<Character, Integer> counts = new HashMap<>();
        for (char c : text.toCharArray()) {
            counts.put(c, counts.getOrDefault(c, 0) + 1);
        }

        List<Pair> groups = new ArrayList<>();
        if (text.length() > 0) {
            char currentChar = text.charAt(0);
            int currentLength = 1;

            for (int i = 1; i < text.length(); i++) {
                if (text.charAt(i) == currentChar) {
                    currentLength++;
                } else {
                    groups.add(new Pair(currentChar, currentLength));
                    currentChar = text.charAt(i);
                    currentLength = 1;
                }
            }
            groups.add(new Pair(currentChar, currentLength));
        }

        int max_len = 0;

        for (Pair pair : groups) {
            max_len = Math.max(max_len, Math.min(pair.length + 1, counts.get(pair.character)));
        }

        for (int i = 0; i < groups.size() - 2; i++) {
            Pair pair1 = groups.get(i);
            Pair pair2 = groups.get(i + 1);
            Pair pair3 = groups.get(i + 2);

            if (pair1.character == pair3.character && pair2.length == 1) {
                max_len = Math.max(max_len, Math.min(pair1.length + pair3.length + 1, counts.get(pair1.character)));
            }
        }

        return max_len;
    }

    private static class Pair {
        char character;
        int length;

        public Pair(char character, int length) {
            this.character = character;
            this.length = length;
        }
    }
}

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Problem Info

Description

Problem 1156: Swap For Longest Repeated Character Substring - Detailed explanation and solution approaches.

Topics

Hash Table String Sliding Window

Companies

Nutanix

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1156

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

5 steps

1
Step 1: Count the total occurrences of each character in the input string `text` using a HashMap or Counter.
2
Step 2: Group consecutive identical characters in the string into a list of (character, length) pairs.
3
Step 3: Iterate through the list of groups and for each group, calculate the maximum possible length achievable by extending the group by one character using a swap. This is limited by the total count of that character in the string. The extended length is `min(groupLength + 1, totalCharacterCount)`.
4
Step 4: Iterate through the list of groups, checking if any two groups of the same character are separated by exactly one character. If they are, check if swapping the separating character can merge the groups. This is limited by `min(len1 + len3 + 1, totalCharacterCount)`
5
Step 5: Return the maximum length found in steps 3 and 4.

💡

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(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The loops for counting, grouping, and checking the scenarios are all linear with respect to the length of the input string. The auxiliary data structures like the hash map (or counter) and the list of groups also have a size proportional to the input string length in the worst case. Therefore, both time and space complexity are O(n).

📖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

Sliding Window

A sliding window approach could be used, where the window expands until more than one character needs to be swapped to make it all the same, then shrinks from the left until only one swap is needed. This approach is complex to implement correctly.

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

Trade-offs: Sliding Window offers better constant space usage, but implementation is significantly harder. Main approach is clearer for understanding.

✨

Key Insights

1Insight 1: The solution revolves around finding groups of consecutive identical characters and then strategically swapping a character to extend these groups.
2Insight 2: Counting the total occurrences of each character is crucial to determine if a swap is possible and to what extent a group can be extended.
3Insight 3: Two main scenarios need to be considered: extending a single existing group and merging two groups of the same character separated by a single different character.
4Insight 4: The total frequency of a character in the string limits the maximum length of any repeated character substring containing that character after a single swap.

📚

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: Empty string. The code implicitly handles this because the loops will not execute, and 0 will be returned, which is correct.
Edge case 2: String with only one distinct character (e.g., "aaaaa"). The first loop will find one group with length 5, and the length will be returned as 5. The second loop will not execute since there is only one group.
Edge case 3: String where no swaps are needed (already optimal). The algorithm correctly identifies and returns the length of the longest repeated substring.
Edge case 4: String where no swaps are possible because the necessary character doesn't exist elsewhere (e.g., "abab"). The calculation of extending the length with a swap ensures that it is at most `counts[char]`.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to consider the total count of a character when calculating the maximum possible length after a swap. For example, if you have a group of 2 'a's but only 2 'a's in the entire string, you cannot extend the group with a swap.

✗Common mistake 2: Forgetting the edge case where no swap is needed, and the longest repeated substring is already present in the original string.

✗Common mistake 3: Not handling cases where two groups of the same character are separated by more than one character. The solution should only consider the case where they are separated by exactly one character.

✗Common mistake 4: Incorrectly implementing the grouping logic, leading to inaccurate group lengths and incorrect results.

✅

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 algorithms can use similar patterns to identify repeated sequences and reduce storage space.

🏦Real-world scenario 2: Bioinformatics, specifically in DNA sequence analysis, where finding the longest repeated sequence after a minimal alteration is important for identifying genetic mutations or similarities.

💡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 16, 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

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 13

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 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 "Swap For Longest Repeated Character Substring"?

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.

🤖 Need help with this problem?