LeetCode #1573 — MEDIUM

Number of Ways to Split a String

Move from brute-force thinking to an efficient approach using math strategy.

The Problem

Problem Statement

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.

Example 1:

Input: s = "10101"
Output: 4
Explanation: There are four ways to split s in 3 parts where each part contain the same number of letters '1'.
"1|010|1"
"1|01|01"
"10|10|1"
"10|1|01"

Example 2:

Input: s = "1001"
Output: 0

Example 3:

Input: s = "0000"
Output: 3
Explanation: There are three ways to split s in 3 parts.
"0|0|00"
"0|00|0"
"00|0|0"

Constraints:

3 <= s.length <= 10⁵
s[i] is either '0' or '1'.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math

Example 1

"10101"

Example 2

"1001"

Example 3

"0000"

Core Insight

What unlocks the optimal approach

There is no way if the sum (number of '1's) is not divisible by the number of splits. So sum%3 should be 0.
Preffix s1 , and suffix s3 should have sum/3 characters '1'.
Follow up: Can you generalize the problem with numbers between [-10^9, 10^9] such the sum between subarrays s1, s2, s3 are the same?

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1573: Number of Ways to Split a String
class Solution {
    private String s;

    public int numWays(String s) {
        this.s = s;
        int cnt = 0;
        int n = s.length();
        for (int i = 0; i < n; ++i) {
            if (s.charAt(i) == '1') {
                ++cnt;
            }
        }
        int m = cnt % 3;
        if (m != 0) {
            return 0;
        }
        final int mod = (int) 1e9 + 7;
        if (cnt == 0) {
            return (int) (((n - 1L) * (n - 2) / 2) % mod);
        }
        cnt /= 3;
        long i1 = find(cnt), i2 = find(cnt + 1);
        long j1 = find(cnt * 2), j2 = find(cnt * 2 + 1);
        return (int) ((i2 - i1) * (j2 - j1) % mod);
    }

    private int find(int x) {
        int t = 0;
        for (int i = 0;; ++i) {
            t += s.charAt(i) == '1' ? 1 : 0;
            if (t == x) {
                return i;
            }
        }
    }
}

// Accepted solution for LeetCode #1573: Number of Ways to Split a String
func numWays(s string) int {
	cnt := 0
	for _, c := range s {
		if c == '1' {
			cnt++
		}
	}
	m := cnt % 3
	if m != 0 {
		return 0
	}
	const mod = 1e9 + 7
	n := len(s)
	if cnt == 0 {
		return (n - 1) * (n - 2) / 2 % mod
	}
	cnt /= 3
	find := func(x int) int {
		t := 0
		for i := 0; ; i++ {
			if s[i] == '1' {
				t++
				if t == x {
					return i
				}
			}
		}
	}
	i1, i2 := find(cnt), find(cnt+1)
	j1, j2 := find(cnt*2), find(cnt*2+1)
	return (i2 - i1) * (j2 - j1) % mod
}

# Accepted solution for LeetCode #1573: Number of Ways to Split a String
class Solution:
    def numWays(self, s: str) -> int:
        def find(x):
            t = 0
            for i, c in enumerate(s):
                t += int(c == '1')
                if t == x:
                    return i

        cnt, m = divmod(sum(c == '1' for c in s), 3)
        if m:
            return 0
        n = len(s)
        mod = 10**9 + 7
        if cnt == 0:
            return ((n - 1) * (n - 2) // 2) % mod
        i1, i2 = find(cnt), find(cnt + 1)
        j1, j2 = find(cnt * 2), find(cnt * 2 + 1)
        return (i2 - i1) * (j2 - j1) % (10**9 + 7)

// Accepted solution for LeetCode #1573: Number of Ways to Split a String
struct Solution;

const MOD: i64 = 1_000_000_007;

impl Solution {
    fn num_ways(s: String) -> i32 {
        let n = s.len();
        let s: Vec<i32> = s.chars().map(|c| if c == '1' { 1 } else { 0 }).collect();
        let m: i32 = s.iter().sum();
        if m == 0 {
            return (((n - 1) as i64 * (n - 2) as i64) / 2 % MOD) as i32;
        }
        if m % 3 != 0 {
            return 0;
        }
        let mut indexes = vec![];
        let k = m / 3;
        let mut sum = 0;
        for i in 0..n {
            if s[i] == 1 && sum % k == 0 {
                indexes.push(i);
            }
            sum += s[i];
            if s[i] == 1 && sum % k == 0 {
                indexes.push(i);
            }
        }
        (((indexes[2] - indexes[1]) as i64 * (indexes[4] - indexes[3]) as i64) % MOD) as i32
    }
}

#[test]
fn test() {
    let s = "10101".to_string();
    let res = 4;
    assert_eq!(Solution::num_ways(s), res);
    let s = "1001".to_string();
    let res = 0;
    assert_eq!(Solution::num_ways(s), res);
    let s = "0000".to_string();
    let res = 3;
    assert_eq!(Solution::num_ways(s), res);
    let s = "100100010100110".to_string();
    let res = 12;
    assert_eq!(Solution::num_ways(s), res);
}

// Accepted solution for LeetCode #1573: Number of Ways to Split a String
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1573: Number of Ways to Split a String
// class Solution {
//     private String s;
// 
//     public int numWays(String s) {
//         this.s = s;
//         int cnt = 0;
//         int n = s.length();
//         for (int i = 0; i < n; ++i) {
//             if (s.charAt(i) == '1') {
//                 ++cnt;
//             }
//         }
//         int m = cnt % 3;
//         if (m != 0) {
//             return 0;
//         }
//         final int mod = (int) 1e9 + 7;
//         if (cnt == 0) {
//             return (int) (((n - 1L) * (n - 2) / 2) % mod);
//         }
//         cnt /= 3;
//         long i1 = find(cnt), i2 = find(cnt + 1);
//         long j1 = find(cnt * 2), j2 = find(cnt * 2 + 1);
//         return (int) ((i2 - i1) * (j2 - j1) % mod);
//     }
// 
//     private int find(int x) {
//         int t = 0;
//         for (int i = 0;; ++i) {
//             t += s.charAt(i) == '1' ? 1 : 0;
//             if (t == x) {
//                 return i;
//             }
//         }
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

ITERATIVE

O(n) time

O(1) space

Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.

MATH INSIGHT

O(log n) time

O(1) space

Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.

Shortcut: Look for mathematical properties that eliminate iteration. Repeated squaring → O(log n). Modular arithmetic avoids overflow.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.