Count Number of Homogenous Substrings - Practice Coding Problems

Count Number of Homogenous Substrings - Problem

Math String Medium

Given a string s, return the number of homogenous substrings of s. Since the answer may be too large, return it modulo 10⁹ + 7.

A string is homogenous if all the characters of the string are the same.

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

Input & Output

Example 1 — Basic Case

$ Input: s = "abbccc"

› Output: 10

💡 Note: Groups: "a"(1), "bb"(3), "ccc"(6). Homogenous substrings: "a", "b", "b", "bb", "c", "c", "c", "cc", "ccc", "cccc". Total = 1+3+6 = 10

Example 2 — All Same Characters

$ Input: s = "aaaa"

› Output: 10

💡 Note: One group of 4 'a's. Using formula: 4*(4+1)/2 = 10 substrings: "a", "a", "a", "a", "aa", "aa", "aa", "aaa", "aaa", "aaaa"

Example 3 — All Different Characters

$ Input: s = "xy"

› Output: 2

💡 Note: Two groups each of size 1: "x"(1) and "y"(1). Total = 1+1 = 2 homogenous substrings

Constraints

1 ≤ s.length ≤ 10⁵
s consists of lowercase English letters

Visualization

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Understanding the Visualization

Input String

Given string with consecutive character groups

Identify Groups

Find consecutive same characters

Apply Formula

Calculate n*(n+1)/2 for each group and sum

Key Takeaway

🎯 Key Insight: Use the mathematical formula n*(n+1)/2 to count substrings in each consecutive group

Asked in

f Facebook 25 a Amazon 20 G Google 15

The key insight is that for any group of n consecutive identical characters, there are exactly n*(n+1)/2 homogenous substrings. Best approach uses mathematical formula to count groups in one pass. Time: O(n), Space: O(1)

Common Approaches

Approach	Time	Space	Notes
✓ Mathematical Approach - Count Groups	O(n)	O(1)	Count consecutive same characters and use formula n*(n+1)/2 for each group
Brute Force - Check All Substrings	O(n³)	O(1)	Generate all possible substrings and check if each is homogenous

Mathematical Approach - Count Groups — Algorithm Steps

Step 1: Iterate through string and count consecutive identical characters
Step 2: For each group of length n, add n*(n+1)/2 to the result
Step 3: Return total count modulo 10^9+7

Visualization

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Step-by-Step Walkthrough

Identify Groups

Find consecutive same characters

Apply Formula

For group size n: n*(n+1)/2 substrings

Sum Results

Add all group contributions

Code -

solution.c — C

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

int solution(char* s) {
    const int MOD = 1000000007;
    long long count = 0;
    int n = strlen(s);
    int i = 0;
    
    while (i < n) {
        int j = i;
        // Count consecutive same characters
        while (j < n && s[j] == s[i]) {
            j++;
        }
        
        // Length of current group
        long long groupLen = j - i;
        
        // Add n*(n+1)/2 substrings for this group
        count = (count + (groupLen * (groupLen + 1) / 2)) % MOD;
        
        i = j;
    }
    
    return count;
}

int main() {
    char s[100001];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;  // Remove newline
    printf("%d\n", solution(s));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through the string to count consecutive groups

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra variables for counting

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Mathematical Approach - Count Groups — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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