[leetcode] 1653. Minimum Deletions to Make String Balanced

Description

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’​​.

分析

题目的意思是：给定一个字符串s，删除其中的字符，使得字符串平衡，即前半部分为全a，后半部分为全b。

• 这道题我参考了一下别人的解法，count_a记录的是需要删除的a的数量，count_b记录的是当前遍历的需要删除的b的数量。
• 如果当前遍历到的是a，则count_a减去1，更新res。

res=min(res,count_a+count_b)

• 如果当前遍历到的是b，则count_b加上1，更新res。

res=min(res,count_a+count_b-1)

代码

class Solution:
    def minimumDeletions(self, s: str) -> int:
        count_a=s.count('a')
        count_b=0
        res=float('inf')
        for ch in s:
            if(ch=='a'): # 如果选择当前的索引为a的结束
                count_a-=1
                res=min(res,count_a+count_b)
            elif(ch=='b'): #如果选择当前的索引为b的开始
                count_b+=1
                res=min(res,count_a+count_b-1)
        return res

参考文献

​​Python one-pass solution with comments​​