Find K-Length Substrings With No Repeated Characters

Find K-Length Substrings With No Repeated Characters - Problem

Given a string s and an integer k, return the number of substrings in s of length k with no repeated characters.

A substring is a contiguous sequence of characters within a string.

Example:

If s = "havefunonleetcode" and k = 5, the substring "havef" has repeated character 'e', but "onlee" has repeated 'e', while "funon" has no repeated characters.

Input & Output

Example 1 — Basic Case

$ Input: s = "havefunonleetcode", k = 5

› Output: 5

💡 Note: The valid substrings of length 5 are: "havef" (h,a,v,e,f - all unique), "vefun" (v,e,f,u,n - all unique), "efuno" (e,f,u,n,o - all unique), "etcod" (e,t,c,o,d - all unique), and "tcode" (t,c,o,d,e - all unique). Total count is 5.

Example 2 — Shorter String

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

› Output: 2

💡 Note: The substrings are: "hom" (all unique), "ome" (all unique). Both have no repeated characters, so answer is 2.

Example 3 — All Same Characters

$ Input: s = "aaa", k = 2

› Output: 0

💡 Note: All possible substrings of length 2 are "aa" which has repeated character 'a', so no valid substrings exist.

Constraints

1 ≤ s.length ≤ 10⁴
1 ≤ k ≤ 26
s consists of lowercase English letters only

Visualization

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Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is to use a sliding window approach with a hash map to efficiently track character frequencies. The optimal solution maintains a window of size k and counts unique characters in O(n) time with O(k) space. Best approach is sliding window: Time: O(n), Space: O(k)

Common Approaches

✓ One Pass Hash

⏱️ Time: N/A Space: N/A

Brute Force - Check All Substrings

⏱️ Time: O(n×k²) Space: O(1)

For each possible starting position, extract a substring of length k and check if all characters in it are unique by comparing each character with every other character in the substring.

Sliding Window with Hash Set

⏱️ Time: O(n) Space: O(k)

Maintain a sliding window of exactly k characters and use a hash set to efficiently check for duplicates. As we slide the window, we add the new character and remove the old one, updating our duplicate count accordingly.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

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

int solution(char* s, int k) {
    int n = strlen(s);
    if (k > n) {
        return 0;
    }
    
    int count = 0;
    
    // Check every possible substring of length k
    for (int i = 0; i <= n - k; i++) {
        // Check if all characters are unique in current substring
        bool hasDuplicate = false;
        for (int j = i; j < i + k; j++) {
            for (int l = j + 1; l < i + k; l++) {
                if (s[j] == s[l]) {
                    hasDuplicate = true;
                    break;
                }
            }
            if (hasDuplicate) {
                break;
            }
        }
        
        if (!hasDuplicate) {
            count++;
        }
    }
    
    return count;
}

int main() {
    char s[1000];
    int k;
    
    fgets(s, sizeof(s), stdin);
    // Remove newline if present
    int len = strlen(s);
    if (len > 0 && s[len-1] == '\n') {
        s[len-1] = '\0';
    }
    
    scanf("%d", &k);
    
    int result = solution(s, k);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

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Medium Frequency

~15 min Avg. Time

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Smart Actions

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