Number of Equal Count Substrings

Number of Equal Count Substrings - Problem

You are given a 0-indexed string s consisting of only lowercase English letters, and an integer count.

A substring of s is said to be an equal count substring if, for each unique letter in the substring, it appears exactly count times in the substring.

Return the number of equal count substrings in s.

A substring is a contiguous non-empty sequence of characters within a string.

Input & Output

Example 1 — Basic Case

$ Input: s = "aaabbbccc", count = 3

› Output: 3

💡 Note: The equal count substrings are "aaa", "bbb", and "ccc". Each contains one unique character that appears exactly 3 times.

Example 2 — Mixed Characters

$ Input: s = "abababab", count = 2

› Output: 3

💡 Note: The equal count substrings are "abab" (positions 0-3), "abab" (positions 2-5), and "abab" (positions 4-7). Each contains 'a' twice and 'b' twice.

Example 3 — No Valid Substrings

$ Input: s = "abcdef", count = 2

› Output: 0

💡 Note: No substring has any character appearing exactly 2 times, so there are 0 equal count substrings.

Constraints

1 ≤ s.length ≤ 1000
1 ≤ count ≤ s.length
s consists of only lowercase English letters

Visualization

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Asked in

G Google 25 M Microsoft 20 a Amazon 15 f Facebook 12

The key insight is to check all possible substrings and verify that each unique character appears exactly count times. The sliding window approach optimizes the frequency tracking. Time: O(n²), Space: O(1)

Common Approaches

✓ Brute Force

⏱️ Time: O(n³) Space: O(1)

Generate all possible substrings, then for each substring count the frequency of each character and verify if all frequencies equal the target count.

Sliding Window

⏱️ Time: O(n²) Space: O(1)

Optimize by using a sliding window approach where we expand and contract the window while maintaining character frequencies. For each starting position, we expand the window and check validity.

Brute Force — Algorithm Steps

Step 1: Generate all possible substrings
Step 2: For each substring, count character frequencies
Step 3: Check if all frequencies equal count

Visualization

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Step-by-Step Walkthrough

Generate Substrings

Create all possible substrings

Count Frequencies

Count each character in substring

Validate

Check if all frequencies equal count

Code -

solution.c — C

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

int solution(char* s, int count) {
    int n = strlen(s);
    int result = 0;
    
    // Check all possible substrings
    for (int i = 0; i < n; i++) {
        for (int j = i; j < n; j++) {
            int freq[26] = {0};
            int uniqueChars = 0;
            
            // Count frequencies for substring s[i..j]
            for (int k = i; k <= j; k++) {
                if (freq[s[k] - 'a'] == 0) {
                    uniqueChars++;
                }
                freq[s[k] - 'a']++;
            }
            
            // Check if all frequencies equal count
            int valid = 1;
            for (int k = 0; k < 26; k++) {
                if (freq[k] > 0 && freq[k] != count) {
                    valid = 0;
                    break;
                }
            }
            
            if (valid) {
                result++;
            }
        }
    }
    
    return result;
}

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

Time & Space Complexity

Time Complexity

⏱️

O(n³)

Three nested loops: two for substring bounds and one to count frequencies

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a fixed-size frequency array for 26 letters

✓ Linear Space

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Medium Frequency

~15 min Avg. Time

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Smart Actions

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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