Number of Ways to Split a String

Given a binary string s (a string consisting only of '0's and '1's), we can split s into 3 non-empty strings s1, s2, s3 (s1+ s2+ s3 = s).

Return the number of ways s can be split such that the number of characters '1' is the same in s1, s2, and s3.

Since the answer may be too large, return it modulo 10^9 + 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"

Example 4:

Input: s = "100100010100110"
Output: 12

Constraints:

• 3 <= s.length <= 10^5

• s[i] is '0' or '1'.

class Solution {
    long mod = 1_000_000_007;

    public long length(String s, int index, int sum) {
        if (index == 0) {
            boolean found = false;
            long len = 0, count = 0;
            while (index < s.length()) {
                if (s.charAt(index) == '1') {
                    if (found)
                        break;
                    count++;
                    if (count == sum)
                        found = true;
                }
                if (found)
                    len++;
                index++;
            }
            return len;
        } else {
            boolean found = false;
            int len = 0, count = 0;
            while (index >= 0) {
                if (s.charAt(index) == '1') {
                    if (found)
                        break;
                    count++;
                    if (count == sum)
                        found = true;
                }
                if (found)
                    len++;
                index--;
            }
            return len;
        }
    }

    public int numWays(String s) {
        int nums = 0;
        for (char c : s.toCharArray())
            if (c == '1')
                nums++;
        if (nums % 3 != 0)
            return 0;
        if (nums == 0) {
            long len = s.length() - 2;
            return (int) ((len * (len + 1L) / 2L) % mod);
        }
        // Find out the 4 positions we need (because 2 are fixed i.e start & end of s1 & s3 resp.)
        return (int) ((length(s, 0, nums / 3) * length(s, s.length() - 1, nums / 3)) % mod);
    }
}