Leetcode problem 108 - Maximize the Confusion of an Exam

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

/**
 * @param {string} answerKey
 * @param {number} k
 * @return {number}
 */
var maxConsecutiveAnswers = function (answerKey, k) {
  // Initialize counters for the number of consecutive 'T' and 'F'
  let countT = 0;
  let countF = 0;

  // Initialize the current index for the sliding window
  let current = 0;

  // Initialize the result to store the maximum number of consecutive answers
  let result = 0;

  // Iterate through the answerKey using a sliding window approach
  for (let i = 0; i < answerKey.length; i++) {
    // Update counters based on the current character
    if (answerKey[i] === "T") {
      countT++;
    }
    if (answerKey[i] === "F") {
      countF++;
    }

    // Check if the number of flips needed is exceeded (more than k)
    if (Math.min(countT, countF) > k) {
      // Adjust the window by moving it to the right and decrementing counters
      if (answerKey[current] === "T") {
        countT--;
      } else {
        countF--;
      }
      current++;
    }

    // Update the result with the maximum number of consecutive answers
    result = Math.max(result, countT + countF);
  }

  // Return the final result representing the maximum number of consecutive answers
  return result;
};

The time complexity of this code is O(n), where n is the length of the answerKey. This is because we iterate through the answerKey once, performing constant time operations for each character.

The space complexity of this code is O(1), as we only use a constant amount of extra space to store the counters and the current index.