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: 6

💡 Note: Substrings with at least one character appearing ≥3 times: "ccc", "cccc", "bccc", "bcccc", "abccc", "abcccc". The character 'c' appears 3+ times in each.

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: 9

💡 Note: All substrings of length ≥2 are valid: "aa"(×3), "aaa"(×2), "aaaa"(×1), plus longer ones starting from different positions. Total: 9 substrings.

Constraints

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

Visualization

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Understanding the Visualization

Input

String s="abccc" and threshold k=3

Process

Find all substrings where at least one character appears ≥k times

Output

Count of valid substrings: 6

Key Takeaway

🎯 Key Insight: Count character frequencies incrementally to avoid redundant recalculation

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

Approach	Time	Space	Notes
✓ Optimized Frequency Tracking	O(n²)	O(1)	Build frequency map incrementally while expanding each substring
Brute Force - Check All Substrings	O(n³)	O(1)	Generate all possible substrings and count character frequencies in each

Optimized Frequency Tracking — Algorithm Steps

For each starting position, initialize empty frequency map
Extend substring one character at a time
Update frequency incrementally and check condition

Visualization

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Step-by-Step Walkthrough

Start Position

Fix starting index and initialize frequency map

Extend & Update

Add one character and update its frequency

Check & Count

If any character reaches k frequency, count this substring

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++) {
        int freq[26] = {0};
        for (int j = i; j < n; j++) {
            char c = s[j];
            freq[c - 'a']++;
            
            if (freq[c - 'a'] >= k) {
                count++;
            }
        }
    }
    
    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(n²)

Two nested loops to generate all substrings, with O(1) frequency updates

⚠ Quadratic Growth

Space Complexity

O(1)

Hash map stores at most 26 characters (constant space)

✓ Linear Space

8.5K Views

Medium Frequency

~25 min Avg. Time

234 Likes

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Optimized Frequency Tracking — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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