Find Special Substring of Length K

Find Special Substring of Length K - Problem

String Easy

You are given a string s and an integer k. Determine if there exists a substring of length exactly k in s that satisfies the following conditions:

The substring consists of only one distinct character (e.g., "aaa" or "bbb").
If there is a character immediately before the substring, it must be different from the character in the substring.
If there is a character immediately after the substring, it must also be different from the character in the substring.

Return true if such a substring exists. Otherwise, return false.

Input & Output

Example 1 — Valid Substring

$ Input: s = "aabbbcc", k = 3

› Output: true

💡 Note: Substring "bbb" at position 2-4 has length 3, all same character 'b', with different characters before (s[1]='a') and after (s[5]='c').

Example 2 — No Valid Substring

$ Input: s = "aaabbbccc", k = 4

› Output: false

💡 Note: No substring of length 4 with identical characters exists. Longest runs are of length 3.

Example 3 — Boundary Violation

$ Input: s = "aaaaa", k = 3

› Output: false

💡 Note: While "aaa" exists, any 3-character substring would have matching characters before or after, violating boundary conditions.

Constraints

1 ≤ s.length ≤ 1000
1 ≤ k ≤ s.length
s consists of lowercase English letters only

Visualization

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Asked in

G Google 25 a Amazon 20 M Microsoft 15

The key insight is to find runs of consecutive identical characters and check if any valid k-length substring can be isolated. Best approach uses single pass tracking for O(n) time complexity. Time: O(n), Space: O(1)

Common Approaches

✓ Brute Force - Check All Substrings

⏱️ Time: O(n×k) Space: O(1)

For each starting position, extract substring of length k, verify it has one character, then check if boundaries are different.

Optimized - Single Pass Tracking

⏱️ Time: O(n) Space: O(1)

Scan the string once, counting consecutive identical characters. When count reaches k, check isolation boundaries.

Brute Force - Check All Substrings — Algorithm Steps

Iterate through all possible starting positions
For each position, check if substring of length k has one character
Verify boundary conditions (before/after characters are different)

Visualization

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Step-by-Step Walkthrough

Start

Check substring at position 0

Validate

Check if characters are same and boundaries differ

Result

Return true if valid substring found

Code -

solution.c — C

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

bool solution(char* s, int k) {
    int n = strlen(s);
    if (k > n) return false;
    
    for (int i = 0; i <= n - k; i++) {
        // Check if all characters in substring are the same
        char ch = s[i];
        bool allSame = true;
        for (int j = 1; j < k; j++) {
            if (s[i + j] != ch) {
                allSame = false;
                break;
            }
        }
        
        if (!allSame) continue;
        
        // Check before character
        if (i > 0 && s[i-1] == ch) continue;
        
        // Check after character
        if (i + k < n && s[i+k] == ch) continue;
        
        return true;
    }
    
    return false;
}

int main() {
    char s[1001];
    int k;
    
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;  // Remove newline
    scanf("%d", &k);
    
    bool result = solution(s, k);
    printf(result ? "true\n" : "false\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n×k)

For each of n-k+1 positions, we check k characters for uniformity

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables, no extra data structures

✓ Linear Space

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~15 min Avg. Time

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Input & Output

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Related Problems

Common Approaches

Brute Force - Check All Substrings — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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