Count Substrings With K-Frequency Characters II

Count Substrings With K-Frequency Characters II - Problem

Given a string s and an integer k, return the total number of substrings of s where at least one character appears at least k times.

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

Input & Output

Example 1 — Basic Case

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

› Output: 2

💡 Note: Substrings with at least one character appearing ≥ 2 times: "bcb" (b appears 2 times) and "abcb" (b appears 2 times). Total count is 2.

Example 2 — No Valid Substrings

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

› Output: 0

💡 Note: No character appears 2 or more times in any substring, so result is 0.

Example 3 — All Same Characters

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

› Output: 6

💡 Note: All substrings of length ≥ 2 are valid since 'a' appears at least 2 times: "aa" (3 occurrences), "aaa" (2 occurrences), "aaaa" (1 occurrence). Total = 6.

Constraints

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

Visualization

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Asked in

G Google 15 M Microsoft 12 a Amazon 8

The key insight is to use complement counting - calculate total substrings minus those where no character appears k+ times. Best approach uses sliding window to efficiently count invalid substrings. Time: O(n), Space: O(1)

Common Approaches

✓ Brute Force - Check Every Substring

⏱️ Time: O(n³) Space: O(1)

For each possible starting position, extend the substring one character at a time and maintain frequency counts. Check if any character reaches frequency k or more.

Sliding Window with Complement Counting

⏱️ Time: O(n) Space: O(1)

Use the complement approach: calculate total possible substrings, then subtract substrings where NO character appears k or more times using sliding window technique.

Brute Force - Check Every Substring — Algorithm Steps

For each starting position i
Extend substring from i to j, maintaining character counts
If any character count >= k, increment result

Visualization

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Step-by-Step Walkthrough

Generate

Create all possible substrings

Count

Count character frequencies in each

Validate

Check if any character appears k+ times

Code -

solution.c — C

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

int solution(char* s, int k) {
    int n = strlen(s);
    int count = 0;
    
    for (int i = 0; i < n; i++) {
        int freq[26] = {0};
        for (int j = i; j < n; j++) {
            freq[s[j] - 'a']++;
            
            // Check if any character appears k or more times
            for (int f = 0; f < 26; f++) {
                if (freq[f] >= k) {
                    count++;
                    break;
                }
            }
        }
    }
    
    return count;
}

int main() {
    char s[1001];
    int k;
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    scanf("%d", &k);
    
    int result = solution(s, k);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

O(n²) substrings, each requiring O(n) to check frequency counts

✓ Linear Growth

Space Complexity

O(1)

Only using a fixed-size frequency array for 26 lowercase letters

✓ Linear Space

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Medium Frequency

~25 min Avg. Time

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Related Problems

Common Approaches

Brute Force - Check Every Substring — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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