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 :



title-img 




                        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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