Number of Substrings With Fixed Ratio

Number of Substrings With Fixed Ratio - Problem

You are given a binary string s, and two integers num1 and num2. num1 and num2 are coprime numbers.

A ratio substring is a substring of s where the ratio between the number of 0's and the number of 1's in the substring is exactly num1 : num2.

For example, if num1 = 2 and num2 = 3, then "01011" and "1110000111" are ratio substrings, while "11000" is not.

Return the number of non-empty ratio substrings of s.

Note that:

A substring is a contiguous sequence of characters within a string.
Two values x and y are coprime if gcd(x, y) == 1 where gcd(x, y) is the greatest common divisor of x and y.

Input & Output

Example 1 — Basic Valid Substring

$ Input: s = "0110", num1 = 1, num2 = 2

› Output: 2

💡 Note: Substrings with 1:2 ratio (zeros:ones) where ones*1 = zeros*2: "011" (1 zero, 2 ones: 2*1 = 1*2) and "110" (1 zero, 2 ones: 2*1 = 1*2). Total: 2 valid substrings.

Example 2 — Single Character

$ Input: s = "0", num1 = 1, num2 = 1

› Output: 0

💡 Note: Single '0' has ratio 1:0, but we need 1:1 ratio, so no valid substrings.

Example 3 — Multiple Valid Substrings

$ Input: s = "01011", num1 = 2, num2 = 3

› Output: 1

💡 Note: Only "01011" itself has 2 zeros and 3 ones, giving exact 2:3 ratio.

Constraints

1 ≤ s.length ≤ 10⁵
s consists only of '0' and '1'
1 ≤ num1, num2 ≤ 100
gcd(num1, num2) = 1

Visualization

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Asked in

G Google 25 f Meta 18 a Amazon 15

The key insight is transforming the ratio problem into a prefix sum difference problem. Instead of checking ratios directly, we assign values to characters and use hash maps to count matching prefix sums. The optimal prefix sum approach runs in O(n) time with O(n) space.

Common Approaches

✓ Brute Force - Check All Substrings

⏱️ Time: O(n³) Space: O(1)

For each starting position, expand to all possible ending positions and count zeros and ones to verify the ratio matches num1:num2.

Prefix Sum with Hash Map

⏱️ Time: O(n) Space: O(n)

Convert the ratio problem to tracking differences between scaled counts. Use prefix sum to efficiently check all substrings.

Brute Force - Check All Substrings — Algorithm Steps

Step 1: For each starting index i from 0 to n-1
Step 2: For each ending index j from i to n-1, count zeros and ones
Step 3: Check if count0/count1 equals num1/num2 and increment result

Visualization

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Step-by-Step Walkthrough

Generate Substrings

Start from each position and expand

Count 0s and 1s

Track zeros and ones in current substring

Check Ratio

Verify if counts match required ratio

Code -

solution.c — C

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

int solution(char* s, int num1, int num2) {
    int n = strlen(s);
    int count = 0;
    
    for (int i = 0; i < n; i++) {
        int zeros = 0;
        int ones = 0;
        for (int j = i; j < n; j++) {
            if (s[j] == '0') {
                zeros++;
            } else {
                ones++;
            }
            
            // Check if ratio matches num1:num2 (zeros:ones)
            if (ones * num1 == zeros * num2) {
                count++;
            }
        }
    }
    
    return count;
}

int main() {
    char s[100001];
    int num1, num2;
    
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;  // Remove newline
    
    scanf("%d", &num1);
    scanf("%d", &num2);
    
    int result = solution(s, num1, num2);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

Triple nested structure: n starting positions × n ending positions × n substring processing

✓ Linear Growth

Space Complexity

O(1)

Only using variables to count zeros and ones

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Check All Substrings — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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