leetcode - 2024. Maximize the Confusion of an Exam

Description

A teacher is writing a test with n true/false questions, with ‘T’ denoting true and ‘F’ denoting false. He wants to confuse the students by maximizing the number of consecutive questions with the same answer (multiple trues or multiple falses in a row).

You are given a string answerKey, where answerKey[i] is the original answer to the ith question. In addition, you are given an integer k, the maximum number of times you may perform the following operation:

Change the answer key for any question to ‘T’ or ‘F’ (i.e., set answerKey[i] to ‘T’ or ‘F’).
Return the maximum number of consecutive 'T’s or 'F’s in the answer key after performing the operation at most k times.

Example 1:

Input: answerKey = "TTFF", k = 2
Output: 4
Explanation: We can replace both the 'F's with 'T's to make answerKey = "TTTT".
There are four consecutive 'T's.

Example 2:

Input: answerKey = "TFFT", k = 1
Output: 3
Explanation: We can replace the first 'T' with an 'F' to make answerKey = "FFFT".
Alternatively, we can replace the second 'T' with an 'F' to make answerKey = "TFFF".
In both cases, there are three consecutive 'F's.

Example 3:

Input: answerKey = "TTFTTFTT", k = 1
Output: 5
Explanation: We can replace the first 'F' to make answerKey = "TTTTTFTT"
Alternatively, we can replace the second 'F' to make answerKey = "TTFTTTTT". 
In both cases, there are five consecutive 'T's.

Constraints:

n == answerKey.length
1 <= n <= 5 * 10^4
answerKey[i] is either 'T' or 'F'
1 <= k <= n

Solution

Similar as 1493. Longest Subarray of 1‘s After Deleting One Element, because for each try the teacher could only change answers to either T or F, so we run sliding window twice. One for changing the answers to T, one for changing the answers to F.

For each run, keep track of the number of elements are need to be changed. If the number is larger than k, then we shrink the window.

Time complexity: $o(n)$
Space complexity: $o(n)$

Code

class Solution:
    def max_consective(self, answers: str, target: str, k: int) -> int:
        """
        change character in answers to target, return the maximum length of subarray
        """
        left = -1
        ans = 0
        diff = 0
        for right in range(len(answers)):
            if answers[right] != target:
                diff += 1
            while diff > k:
                left += 1
                if answers[left] != target:
                    diff -= 1
            ans = max(ans, right - left)
        return ans

    def maxConsecutiveAnswers(self, answerKey: str, k: int) -> int:
        return max(self.max_consective(answerKey, 'T', k), self.max_consective(answerKey, 'F', k))