Count Substrings With K-Frequency Characters I

Count Substrings With K-Frequency Characters I - 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 non-empty sequence of characters within a string.

Input & Output

Example 1 — Basic Case

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

› Output: 3

💡 Note: Substrings with at least one character appearing ≥3 times: "ccc" (position 2-4), "bccc" (position 1-4), "abccc" (position 0-4). The character 'c' appears 3+ times in each. Total: 3 substrings.

Example 2 — No Valid Substrings

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

› Output: 0

💡 Note: No character appears 4 or more times in any substring. Maximum frequency is 3 for 'a' in "aaab".

Example 3 — Single Character

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

› Output: 6

💡 Note: Substrings of length ≥2 where 'a' appears ≥2 times: "aa" appears 3 times (positions 0-1, 1-2, 2-3), "aaa" appears 2 times (positions 0-2, 1-3), "aaaa" appears 1 time (position 0-3). Total: 6 substrings.

Constraints

1 ≤ s.length ≤ 3000
1 ≤ k ≤ s.length
s consists of only lowercase English letters

Visualization

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Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is to count character frequencies incrementally rather than recalculating for each substring. The optimized approach uses frequency counting with nested loops, achieving O(n²) time complexity by updating character counts as we extend substrings. Time: O(n²), Space: O(1)

Common Approaches

✓ Brute Force - Check All Substrings

⏱️ Time: O(n³) Space: O(1)

For each possible starting position, generate all substrings and check if any character appears at least k times. This involves nested loops to generate substrings and another loop to count frequencies.

Optimized Frequency Tracking

⏱️ Time: O(n²) Space: O(1)

For each starting position, incrementally build character frequencies as we extend the substring character by character. This avoids recounting the entire substring each time and reduces redundant work.

Brute Force - Check All Substrings — Algorithm Steps

Generate all possible substrings using nested loops
For each substring, count character frequencies
Check if any character appears at least k times

Visualization

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Step-by-Step Walkthrough

Generate Substrings

Create all possible contiguous substrings

Count Frequencies

For each substring, count character occurrences

Check Condition

See if any character appears k or more times

Code -

solution.c — C

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

int solution(char* s, int k) {
    int count = 0;
    int n = strlen(s);
    
    for (int i = 0; i < n; i++) {
        for (int j = i; j < n; j++) {
            int freq[26] = {0};
            for (int l = i; l <= j; l++) {
                freq[s[l] - 'a']++;
            }
            
            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 × O(n) to count frequencies for each substring

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra space for frequency counting

✓ Linear Space

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Medium Frequency

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Check All Substrings — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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