【leetcode】1653. Minimum Deletions to Make String Balanced

题目如下：

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'​​.

解题思路：要使得s是一个Balanced的字符串，有两种情况，一个是把'a'全部删除，另一个是前半部分是'a'后半部分是'b'的字符串。第二种情况的关键在于找出Balanced后的字符串的最后一个'a'，假设其下标为index，那么删除操作的次数就是index前面'b'的个数加上index后面'a'的个数。

代码如下：

class Solution(object):
    def minimumDeletions(self, s):
        """
        :type s: str
        :rtype: int
        """
        count_a = s.count('a')
        count_b = 0
        res = len(s) - count_a

        for i in range(len(s)):
            if s[i] == 'a':
                count_a -= 1
                continue
            res = min(res,count_b + count_a)
            count_b += 1
        return res