README.md

1525. Number of Good Ways to Split a String

You are given a string s, a split is called good if you can split s into 2 non-empty strings p and q where its concatenation is equal to s and the number of distinct letters in p and q are the same.

Return the number of good splits you can make in s.

Example 1:

Input: s = "aacaba"
Output: 2
Explanation: There are 5 ways to split "aacaba" and 2 of them are good. 
("a", "acaba") Left string and right string contains 1 and 3 different letters respectively.
("aa", "caba") Left string and right string contains 1 and 3 different letters respectively.
("aac", "aba") Left string and right string contains 2 and 2 different letters respectively (good split).
("aaca", "ba") Left string and right string contains 2 and 2 different letters respectively (good split).
("aacab", "a") Left string and right string contains 3 and 1 different letters respectively.

Example 2:

Input: s = "abcd"
Output: 1
Explanation: Split the string as follows ("ab", "cd").

Example 3:

Input: s = "aaaaa"
Output: 4
Explanation: All possible splits are good.

Example 4:

Input: s = "acbadbaada"
Output: 2

Constraints:

• s contains only lowercase English letters.
• 1 <= s.length <= 10^5

Solutions (Rust)

1. Solution

impl Solution {
    pub fn num_splits(s: String) -> i32 {
        let mut p_count = vec![];
        let mut q_count = vec![];
        let mut count = 0i32;

        for ch in s.bytes().take(s.len() - 1) {
            count |= 1 << (ch - b'a');
            p_count.push(count.count_ones());
        }
        count = 0;
        for ch in s.bytes().rev().take(s.len() - 1) {
            count |= 1 << (ch - b'a');
            q_count.push(count.count_ones());
        }

        p_count
            .iter()
            .zip(q_count.iter().rev())
            .filter(|(x, y)| x == y)
            .count() as i32
    }
}