# 1573. Number of Ways to Split a String

## Description

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'.

## Solutions

• 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;
}
}
}
}

• class Solution {
public:
int numWays(string s) {
int cnt = 0;
for (char& c : s) {
cnt += c == '1';
}
int m = cnt % 3;
if (m) {
return 0;
}
const int mod = 1e9 + 7;
int n = s.size();
if (cnt == 0) {
return (n - 1LL) * (n - 2) / 2 % mod;
}
cnt /= 3;
auto find = [&](int x) {
int t = 0;
for (int i = 0;; ++i) {
t += s[i] == '1';
if (t == x) {
return i;
}
}
};
int i1 = find(cnt), i2 = find(cnt + 1);
int j1 = find(cnt * 2), j2 = find(cnt * 2 + 1);
return (1LL * (i2 - i1) * (j2 - j1)) % mod;
}
};

• 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)


• 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
}