**Three-Way String Split with Equal Ones - Microsoft Top Interview Questions**

### Problem Statement :

You are given a binary string s containing "1"s and "0"s. Return the number of ways to split the string into three non-empty parts a, b and c such that a + b + c = s and the number of "1"s in each string is the same. Mod the result by 10 ** 9 + 7. Constraints 3 < n ≤ 100,000 where n is the length of s Example 1 Input s = "11001111" Output 3 Explanation We can have "11" + "0011" + "11" "110" + "011" + "11" "1100" + "11" + "11"

### Solution :

` ````
Solution in C++ :
#define all(a) begin(a), end(a)
using ll = long long;
const int M = 1e9 + 7;
int solve(string s) {
const int n = (int)s.length();
int z = count_if(all(s), [](auto c) { return c == '1'; }); // 1st pass
if (!z) return ((ll)(n - 2) * (n - 1) / 2) % M;
if (z % 3) return 0;
z /= 3;
int k = 0;
int a, b, c, d;
for (int i = 0; i < n; ++i) { // 2nd pass
if (s[i] == '0') continue;
++k;
if (k == z) a = i;
if (k == z + 1) b = i;
if (k == 2 * z) c = i;
if (k == 2 * z + 1) {
d = i;
break;
}
}
return (ll)((b - a) * (d - c)) % M;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
long mod = (long) (1e9 + 7);
int corner_case(int n) {
long res = 0;
for (int i = 1; i <= n - 2; i++) res += n - i - 1;
return (int) (res % mod);
}
public int solve(String s) {
int total_ones = 0;
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) == '1')
total_ones++;
}
int one_third = 0;
int two_third = 0;
int one_count = 0;
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) == '1')
one_count++;
if (one_count == total_ones / 3)
one_third++;
if (one_count == 2 * total_ones / 3)
two_third++;
}
if (one_count == 0)
return corner_case(s.length());
if (one_count % 3 != 0)
return 0;
long res = one_third * two_third;
return (int) (res % mod);
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, s):
if not s:
return 1
Mod = 1000000007
o = s.count("1")
if o:
if o % 3:
return 0
i = 0
o //= 3
c = 0
while c < o:
c += s[i] == "1"
i += 1
j = i
o *= 2
while c < o:
c += s[j] == "1"
j += 1
x, y = 0, 0
while s[i] == "0":
x += 1
i += 1
while s[j] == "0":
y += 1
j += 1
return (x + 1) * (y + 1) % Mod
else:
n = len(s)
if n < 3:
return 0
return (n - 1) * (n - 2) // 2 % Mod
```

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