Longest Substring of One Repeating Character

Longest Substring of One Repeating Character - Problem

You are given a 0-indexed string s. You are also given a 0-indexed string queryCharacters of length k and a 0-indexed array of integer indices queryIndices of length k, both of which are used to describe k queries.

The i^th query updates the character in s at index queryIndices[i] to the character queryCharacters[i].

Return an array lengths of length k where lengths[i] is the length of the longest substring of s consisting of only one repeating character after the i^th query is performed.

Input & Output

Example 1 — Basic Updates

$ Input: s = "babacc", queryCharacters = "bcb", queryIndices = [1,3,3]

› Output: [3,3,4]

💡 Note: Query 1: s="baaacc" → longest "aaa" = 3. Query 2: s="babcc" → longest "cc" or "bb" = 3. Query 3: s="babbc" → longest "bbb" = 4.

Example 2 — Single Character

$ Input: s = "abcd", queryCharacters = "aa", queryIndices = [0,1]

› Output: [1,2]

💡 Note: Query 1: s="aacd" → longest single chars = 1. Query 2: s="aacd" → longest "aa" = 2.

Example 3 — All Same

$ Input: s = "aa", queryCharacters = "a", queryIndices = [0]

› Output: [2]

💡 Note: Query 1: s="aa" → longest "aa" = 2 (no change in this case).

Constraints

1 ≤ s.length ≤ 10⁵
s consists of lowercase English letters
1 ≤ k ≤ 10⁵
0 ≤ queryIndices[i] < s.length
queryCharacters[i] is a lowercase English letter

Visualization

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Understanding the Visualization

Input

String s with query characters and indices

Process

Update character and find longest repeating substring

Output

Array of lengths after each query

Key Takeaway

🎯 Key Insight: Track character segments and update only affected boundaries to avoid full string scans

Asked in

G Google 35 a Amazon 28 M Microsoft 22 f Facebook 18

The key insight is to efficiently track the longest repeating substring after each character update. The brute force approach scans the entire string after each query in O(k×n) time. The optimized approach uses segment tracking to minimize recalculations. Best approach depends on query patterns: O(k×n) time, O(n) space for segment tracking.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force	O(k × n)	O(1)	After each query, scan entire string to find longest repeating substring
Segment Tracking	O(n + k × n)	O(n)	Track contiguous segments and update only affected boundaries efficiently

Brute Force — Algorithm Steps

Update character at query index
Scan entire string from left to right
Track current run length and maximum run length
Return maximum length found

Visualization

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Step-by-Step Walkthrough

Update

Change character at query position

Scan

Check all substrings for longest repeating

Result

Return maximum length found

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

void solution(char* s, char* queryCharacters, int* queryIndices, int k, int* result) {
    int n = strlen(s);
    
    for (int i = 0; i < k; i++) {
        // Update character
        s[queryIndices[i]] = queryCharacters[i];
        
        // Find longest repeating substring
        int maxLen = 1;
        int currentLen = 1;
        
        for (int j = 1; j < n; j++) {
            if (s[j] == s[j-1]) {
                currentLen++;
                if (currentLen > maxLen) {
                    maxLen = currentLen;
                }
            } else {
                currentLen = 1;
            }
        }
        
        result[i] = maxLen;
    }
}

int main() {
    char s[100001];
    char queryCharacters[100001];
    char indicesLine[500000];
    
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    
    fgets(queryCharacters, sizeof(queryCharacters), stdin);
    queryCharacters[strcspn(queryCharacters, "\n")] = 0;
    
    fgets(indicesLine, sizeof(indicesLine), stdin);
    indicesLine[strcspn(indicesLine, "\n")] = 0;
    
    // Parse array manually
    int queryIndices[100000];
    int k = 0;
    char* token = strtok(indicesLine + 1, ",]");
    while (token != NULL) {
        queryIndices[k++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int* result = (int*)malloc(k * sizeof(int));
    solution(s, queryCharacters, queryIndices, k, result);
    
    printf("[");
    for (int i = 0; i < k; i++) {
        printf("%d", result[i]);
        if (i < k - 1) printf(",");
    }
    printf("]\n");
    
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(k × n)

k queries, each requiring O(n) scan of entire string

✓ Linear Growth

Space Complexity

O(1)

Only using variables for tracking, no extra data structures

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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