2024. Maximize the Confusion of an Exam

Problem

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 * 104

• answerKey[i] is either 'T' or 'F'

• 1 <= k <= n

Solution

/**
 * @param {string} answerKey
 * @param {number} k
 * @return {number}
 */
var maxConsecutiveAnswers = function(answerKey, k) {
    var max = 1;
    var numF = answerKey[0] === 'F' ? 1 : 0;
    var numT = answerKey[0] === 'T' ? 1 : 0;
    var left = 0;
    var right = 0;
    while (left <= right && right <= answerKey.length - 1) {
        if (Math.min(numF, numT) <= k) {
            max = Math.max(max, right - left + 1);
            right += 1;
            answerKey[right] === 'T' ? numT++ : numF++;
        } else {
            answerKey[left] === 'T' ? numT-- : numF--;
            left += 1;
        }
    }
    return max;
};

Explain:

Sliding window.

If we slice a sub string from the answer string, we could either change T to F or F to T to make this sub string consecutive.

Because we can only change k times, which means min(num(T), num(F)) should less or equal than k.

If that sub string is valid, we could move right index forward and update the max length of consecutive sub string, otherwise we move left index forward.

Complexity:

• Time complexity : O(n).
• Space complexity : O(1).