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: 5

💡 Note: Substrings with at least one character appearing ≥ 2 times: "cb" (b appears 1 time - invalid), "bcb" (b appears 2 times), "abcb" (b appears 2 times), etc. Total count is 5.

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: 9

💡 Note: All substrings of length ≥ 2 are valid: "aa", "aa", "aa", "aaa", "aaa", "aaaa", plus 3 more "aa" patterns. Total = 9.

Constraints

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

Visualization

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Understanding the Visualization

Input

String s and frequency threshold k

Process

Find substrings where ≥1 character appears ≥k times

Output

Count of valid substrings

Key Takeaway

🎯 Key Insight: Use complement counting - total substrings minus those with no frequent characters

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

Approach	Time	Space	Notes
✓ Brute Force - Check Every Substring	O(n³)	O(1)	Generate all possible substrings and count character frequencies for each
Sliding Window with Complement Counting	O(n)	O(1)	Count total substrings minus substrings where no character appears k+ times

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

⚠ Quadratic Growth

Space Complexity

O(1)

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

✓ Linear Space

23.0K Views

Medium Frequency

~25 min Avg. Time

890 Likes

Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Check Every Substring — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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