Number of Substrings With Fixed Ratio - Practice Coding Problems

Number of Substrings With Fixed Ratio - Problem

Given a binary string s and two coprime integers num1 and num2, find the number of substrings where the ratio of 0's to 1's is exactly num1 : num2.

A ratio substring is a contiguous sequence where:

Number of 0's / Number of 1's = num1 / num2
Both counts must be positive (non-empty substring)

Examples of valid ratio substrings:

If num1 = 2, num2 = 3: "01011" has 2 zeros and 3 ones ✅
If num1 = 1, num2 = 1: "01" has 1 zero and 1 one ✅

Your task is to count all possible such ratio substrings in the given binary string.

Input & Output

example_1.py — Basic Case

$ Input: s = "10110", num1 = 1, num2 = 2

› Output: 4

💡 Note: Valid substrings: "01" (1 zero, 1 one scaled to 1:2), "10" (1 zero, 1 one), "011" (2 zeros, 1 one = 2:1), "0110" (2 zeros, 2 ones = 1:1 scaled to 1:2)

example_2.py — Equal Ratio

$ Input: s = "01", num1 = 1, num2 = 1

› Output: 1

💡 Note: Only one valid substring "01" with 1 zero and 1 one, giving ratio 1:1

example_3.py — No Valid Substrings

$ Input: s = "111", num1 = 1, num2 = 2

› Output: 0

💡 Note: String contains only 1's, so no substring can have any 0's to match the 1:2 ratio

Constraints

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

Visualization

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Understanding the Visualization

Initialize

Start with balance 0 and empty hash map

Process Character

Update balance: +num2 for '0', -num1 for '1'

Check History

Count how many times we've seen this balance before

Record Balance

Store current balance in hash map for future reference

Key Takeaway

🎯 Key Insight: The prefix sum transformation converts the ratio problem into finding equal balance points, making it solvable in O(n) time with a hash map!

Asked in

G Google 35 f Meta 28 a Amazon 22 ⊞ Microsoft 18

The optimal solution uses a prefix sum transformation with a hash map. Transform each '0' to +num2 and '1' to -num1, then track the running balance. When two positions have the same balance, the substring between them has the target ratio. This achieves O(n) time complexity in a single pass.

Common Approaches

Approach	Time	Space	Notes
✓ One-Pass Hash Table (Optimal)	O(n)	O(n)	Use prefix sum and hash map to find valid substrings in one pass
Brute Force (Check All Substrings)	O(n³)	O(1)	Check every possible substring and count 0's and 1's

One-Pass Hash Table (Optimal) — Algorithm Steps

Transform: replace each '0' with +num2, each '1' with -num1
Calculate running prefix sum (balance) as we iterate
Use hash map to count occurrences of each balance value
For each position, add count of same balance seen before
Update hash map with current balance

Visualization

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Step-by-Step Walkthrough

Transform Characters

Convert '0' to +num2, '1' to -num1

Calculate Balance

Maintain running sum of transformed values

Check Hash Map

Count how many times this balance occurred before

Update Count

Add current balance to hash map

Code -

solution.c — C

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

#define MAX_BALANCE 200000
#define OFFSET 100000

int fixedRatio(char* s, int num1, int num2) {
    int* balanceCount = (int*)calloc(MAX_BALANCE, sizeof(int));
    balanceCount[OFFSET] = 1; // Empty prefix has balance 0
    
    int currentBalance = 0;
    int result = 0;
    int n = strlen(s);
    
    for (int i = 0; i < n; i++) {
        // Transform: '0' adds +num2, '1' adds -num1
        if (s[i] == '0') {
            currentBalance += num2;
        } else {
            currentBalance -= num1;
        }
        
        // Add count of previous occurrences of this balance
        result += balanceCount[currentBalance + OFFSET];
        
        // Update count for current balance
        balanceCount[currentBalance + OFFSET]++;
    }
    
    free(balanceCount);
    return result;
}

int main() {
    char s[100001];
    int num1, num2;
    scanf("%s %d %d", s, &num1, &num2);
    printf("%d\n", fixedRatio(s, num1, num2));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through string with constant time hash map operations

✓ Linear Growth

Space Complexity

O(n)

Hash map can store up to n different balance values in worst case

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

One-Pass Hash Table (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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