problem

You are given a string s.

A split is called good if you can split s into two non-empty strings sleft and sright where their concatenation is equal to s (i.e., sleft + sright = s) and the number of distinct letters in sleft and sright is 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").

 

Constraints:

  • 1 <= s.length <= 105
  • s consists of only lowercase English letters.

submission

// let do it the dumb way
impl Solution {
    pub fn num_splits(s: String) -> i32 {
        let pref: Vec<_> = s.bytes()
            .scan(
                std::collections::HashSet::new(),
                |set, byte| {
                    set.insert(byte);
                    Some(set.len())
                }
            )
            .collect();
        let mut suff: Vec<_> = s.bytes()
            .rev()
            .scan(
                std::collections::HashSet::new(),
                |set, byte| {
                    set.insert(byte);
                    Some(set.len())
                }
            )
            .collect();
        suff.reverse(); // remember to reverse
        pref.into_iter()
            .zip(suff.into_iter().skip(1)) // skip 1 to split
            .filter(|(p, s)| p == s)
            .count() as _
    }
}