1573

1573. Number of Ways to Split a String (Medium)

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:

• s[i] == '0' or s[i] == '1'
• 3 <= s.length <= 10^5

Related Topics:
String

Similar Questions:

• Split Array with Equal Sum (Medium)

Solution 1.

First count the number of 1s. The the count is not divisible by 3, we can't split s into 3 parts, then return 0.

If cnt == 0, what we need to do is to choose 2 out of the N - 1 gaps between the N elements to split the s, so there are combination(N - 1, 2) = (N - 1) * (N - 2) / 2 cases.

Othewise, we need to find the number of possible cases of s1 and s3 respectively.

For s1, that's the number of 0s between the cnt/3-th (1-based) and cnt/3 + 1-th 1 from the left side, plus 1. Let this be left.

For s3, that's the number of 0s between the cnt/3-th (1-based) and cnt/3 + 1-th 1 from the right side, plus 1. Let this be right.

And the answer is different combinations of left and right and thus is left * right.

// OJ: https://leetcode.com/problems/number-of-ways-to-split-a-string/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int numWays(string s) {
        long mod = 1e9+7, cnt = 0;
        for (char c : s) cnt += c == '1';  // cnt is the count of all 1s
        if (cnt % 3) return 0;  // if cnt is not divisible by 3, we can't split the string into 3 parts, return 0
        if (cnt == 0) return (long)(s.size() - 1) * (s.size() - 2) / 2 % mod; // if cnt is 0, there are (N - 1) * (N - 2) / 2 cases.
        int i = 0, c = 0, left = 0, right = 0; // left and right are the numbers of possible cases for s1 and s2 respectively
        while (c <= cnt / 3) {
            c += s[i++] == '1';
            if (c == cnt / 3) ++left;
        }
        i = s.size() - 1, c = 0;
        while (c <= cnt / 3) {
            c += s[i--] == '1';
            if (c == cnt / 3) ++right;
        }
        return (long)left * right % mod; // The answer is simply left * right
    }
};