README.md

1653. Minimum Deletions to Make String Balanced

You are given a string s consisting only of characters 'a' and 'b'.

You can delete any number of characters in s to make s balanced. s is balanced if there is no pair of indices (i,j) such that i < j and s[i] = 'b' and s[j]= 'a'.

Return the minimum number of deletions needed to make s balanced.

Example 1:

Input: s = "aababbab"
Output: 2
Explanation: You can either:
Delete the characters at 0-indexed positions 2 and 6 ("aababbab" -> "aaabbb"), or
Delete the characters at 0-indexed positions 3 and 6 ("aababbab" -> "aabbbb").

Example 2:

Input: s = "bbaaaaabb"
Output: 2
Explanation: The only solution is to delete the first two characters.

Constraints:

• 1 <= s.length <= 105
• s[i] is 'a' or 'b'.

Solutions (Rust)

1. Solution

impl Solution {
    pub fn minimum_deletions(s: String) -> i32 {
        let mut count = s.chars().filter(|&c| c == 'a').count() as i32;
        let mut ret = count;

        for c in s.chars() {
            count += (c == 'b') as i32 - (c == 'a') as i32;
            ret = ret.min(count);
        }

        ret
    }
}