2024. Maximize the Confusion of an Exam

Method 1

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 (adj. 连续的；连贯的) 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 (adj. 连续的；连贯的) '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.

Method 1

【O(n) time | O(1) space】

package Leetcode.SlideWindow.DynamicSlidingWindow;

/**
 * @author zhengxingxing
 * @date 2025/01/18
 */
public class MaximizeTheConfusionOfAnExam {
    public static int maxConsecutiveAnswers(String answerKey, int k) {
        // Calculate max length by considering both cases: converting to 'T' and converting to 'F'
        // Return the maximum of these two scenarios
        return Math.max(maxLength(answerKey, k, 'T'), maxLength(answerKey, k, 'F'));
    }

    private static int maxLength(String answerKey, int k, char target) {
        int maxLen = 0;        // Track the maximum length of consecutive same characters
        int left = 0;          // Left pointer of sliding window
        int count = 0;         // Count of characters that need to be changed in current window

        for (int right = 0; right < answerKey.length(); right++) {
            // Step 1: Process the current character at right pointer
            // If current character is different from target, increment the count
            // This count represents how many characters we need to change in our window
            if (answerKey.charAt(right) != target) {
                count++;
            }

            // Step 2: Maintain window validity
            // If count exceeds k, we need to shrink the window from left
            // Keep shrinking until we have a valid window (count <= k)
            while (count > k) {
                // If character at left pointer is different from target
                // Decrease count as this character will no longer be in our window
                if (answerKey.charAt(left) != target) {
                    count--;
                }
                // Move left pointer to shrink window
                left++;
            }

            // Step 3: Update result
            // Current window size is (right - left + 1)
            // Update maxLen if current window is larger
            maxLen = Math.max(maxLen, right - left + 1);
        }

        return maxLen;
    }

    public static void main(String[] args) {
        // Test cases to verify the solution
        System.out.println(maxConsecutiveAnswers("TTFF", 2));     // Output: 4
        System.out.println(maxConsecutiveAnswers("TFFT", 1));     // Output: 3
        System.out.println(maxConsecutiveAnswers("TTFTTFTT", 1)); // Output: 5
    }
}

Method 1

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