Number of Ways to Split a String

Number of Ways to Split a String - Problem

Math String Medium

Given a binary string s, you can split s into 3 non-empty strings s1, s2, and s3 where s1 + s2 + s3 = s.

Return the number of ways s can be split such that the number of ones is the same in s1, s2, and s3. Since the answer may be too large, return it modulo 10⁹ + 7.

Input & Output

Example 1 — Equal Distribution

$ Input: s = "10101"

› Output: 4

💡 Note: There are 4 ways to split: "1|010|1", "1|01|01", "10|10|1", "101|0|1". Each gives 1 one per part.

Example 2 — All Zeros

$ Input: s = "0000"

› Output: 3

💡 Note: Any split gives 0 ones in each part. Possible splits: "0|0|00", "0|00|0", "00|0|0".

Example 3 — Impossible Split

$ Input: s = "101"

› Output: 0

💡 Note: 2 ones cannot be split equally among 3 parts (2 ÷ 3 ≠ integer).

Constraints

3 ≤ s.length ≤ 10⁵
s consists only of '0' and '1' characters

Visualization

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Asked in

G Google 25 a Amazon 18 M Microsoft 15

The key insight is to use mathematical analysis instead of brute force. First check if total ones are divisible by 3, then find valid cut position ranges using the positions of '1' characters. Best approach uses mathematical calculation with Time: O(n), Space: O(1).

Common Approaches

✓ Brute Force - Try All Splits

⏱️ Time: O(n³) Space: O(1)

For each possible position of first cut and second cut, check if all three parts have equal number of ones. This exhaustively checks all combinations.

Optimized - Mathematical Approach

⏱️ Time: O(n) Space: O(1)

First count total ones and check divisibility by 3. If valid, find positions where each third of ones should end, then use combinatorics to count valid arrangements.

Brute Force - Try All Splits — Algorithm Steps

Try all positions i for first cut (0 < i < n-1)
Try all positions j for second cut (i < j < n)
Count ones in each part s[0:i], s[i:j], s[j:n]
If all counts equal, increment result

Visualization

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Step-by-Step Walkthrough

Try Cut Positions

Test all combinations of first and second cut

Count Ones

Count ones in each resulting part

Check Equality

Verify all three parts have equal ones

Code -

solution.c — C

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

int countOnes(const char* str, int start, int end) {
    int count = 0;
    for (int i = start; i < end; i++) {
        if (str[i] == '1') count++;
    }
    return count;
}

int solution(char* s) {
    int n = strlen(s);
    int MOD = 1000000007;
    int result = 0;
    
    // Try all possible splits
    for (int i = 1; i < n-1; i++) {  // First cut position
        for (int j = i+1; j < n; j++) {  // Second cut position
            // Count ones in each part
            int ones1 = countOnes(s, 0, i);
            int ones2 = countOnes(s, i, j);
            int ones3 = countOnes(s, j, n);
            
            // Check if all parts have equal ones
            if (ones1 == ones2 && ones2 == ones3) {
                result++;
            }
        }
    }
    
    return result % MOD;
}

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

Time & Space Complexity

Time Complexity

⏱️

O(n³)

Two nested loops O(n²) and counting ones in each part O(n)

⚠ Quadratic Growth

Space Complexity

O(1)

Only using variables for counters and result

✓ Linear Space

28.5K Views

Medium Frequency

~25 min Avg. Time

892 Likes

Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Try All Splits — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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