Leetcode 2024. Maximize the Confusion of an Exam [Python]

维护一段连续的subarray，窗口内只有要么k个T或者k个F，并计算此时窗口的大小，更新全局最大值。

class Solution:
    def maxConsecutiveAnswers(self, answerKey: str, k: int) -> int:
        rare = 0
        count_T = 0
        count_F = 0
        res = 0
        for front in range(len(answerKey)):
            if answerKey[front] == 'T':
                count_T += 1
            else:
                count_F += 1
            while count_F > k and count_T > k:
                if answerKey[rare] == 'F':
                    count_F -= 1
                else:
                    count_T -= 1
                rare += 1
            res = max(res, front - rare + 1)
        return res
                    
                

标签：count,rare,Exam,Python,Confusion,int,answerKey,res,front
来源： https://blog.csdn.net/sinat_30403031/article/details/122155122