Maximize the Confusion of an Exam - Solution & Explanation
Problem Statement
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., setanswerKey[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.length1 <= n <= 5 * 104answerKey[i]is either'T'or'F'1 <= k <= n
Approach Overview
Problem Overview: You receive a string representing exam answers where each character is 'T' or 'F'. You may change at most k answers. The goal is to maximize the length of a consecutive block of identical answers after performing those changes.
Approach 1: Sliding Window Technique (O(n) time, O(1) space)
The optimal strategy uses a sliding window. Treat the problem as finding the longest substring where you can flip at most k characters to make all values the same. Run the window twice: once assuming the final sequence should be all 'T', and once for 'F'. Expand the right pointer while counting mismatches. When mismatches exceed k, move the left pointer forward and shrink the window. The window size always represents the longest valid segment you can transform with ā¤ k flips.
This works because the window always maintains a valid state where the number of required flips is within the allowed limit. Each character is visited at most twice (entering and leaving the window), giving linear complexity. This approach is the standard pattern used in many string optimization problems involving limited modifications.
Approach 2: Two Pointer Technique (O(n) time, O(1) space)
The two pointer method is conceptually the same as sliding window but framed more explicitly with left and right indices. Move the right pointer forward to expand the candidate segment and maintain counts of 'T' and 'F'. When the number of characters needing conversion exceeds k, advance the left pointer to restore validity. The current window length represents a segment that can be converted into all identical answers.
This formulation emphasizes pointer movement and state maintenance rather than a formal window abstraction. It is useful when practicing general pointer techniques used in binary search-style range problems and substring optimization tasks.
Recommended for interviews: The sliding window solution is what most interviewers expect. It demonstrates that you can transform the problem into a longest-valid-substring pattern and maintain constraints efficiently in O(n) time. A brute-force approach that checks all substrings would take O(nĀ²) time and quickly becomes impractical. Showing the transition from brute force thinking to a sliding window optimization demonstrates strong problem-solving skills.
Approach 1: Sliding Window Technique
This approach uses the sliding window technique to maintain a window of the answerKey where the number of modifications does not exceed k. We attempt to maximize the length of this window by evaluating either 'T' or 'F' transformations separately and then taking the maximum length between both transformations.
The function maxConsecutiveAnswers applies the sliding window technique twice: once assuming we change all 'F' to 'T' and once the reverse. It uses two counters, maxT and maxF, to track the number of changes made. As soon as the number of required changes exceeds the allowed k, the window is shrunk from the left. The maximum length of such a valid window is tracked and returned.
Complexity
Time Complexity: O(n), where n is the length of the answerKey string. We traverse the string once.
Space Complexity: O(1), as we use constant space related to variables.
Approach 2: Two Pointer Technique
In this approach, we utilize two pointers to explore the string sequence, adjusting pointers as we evaluate possibilities involving updates utilizing k. The purpose is to ensure the longest alteration window without exceeding k modifications. It proceeds by specifically examining alternate possibilities and then concluding with the optimal maximum length.
Here we loop over each character possibility ('T' and 'F') and use a simple two-pointer technique to determine the length of the segment to convert. As pointers move, each change is applied and the window is adjusted to stay under the threshold for changes.
Complexity
Time Complexity: O(n), based on the two-pointer traversal.
Space Complexity: O(1), engaging a minimal variable set, bypassing auxiliary data structures.
Approach 3: Sliding Window
We design a function f(c), which represents the longest length of consecutive characters under the condition that at most k characters c can be replaced, where c can be 'T' or 'F'. The answer is max(f('T'), f('F')).
We iterate through the string answerKey, using a variable cnt to record the number of characters c within the current window. When cnt > k, we move the left pointer of the window one position to the right. After the iteration ends, the length of the window is the maximum length of consecutive characters.
Time complexity is O(n), where n is the length of the string. Space complexity is O(1).
Similar problems:
Complexity Comparison
| Approach | Complexity |
|---|---|
| Sliding Window Technique | Time Complexity: O(n), where n is the length of the answerKey string. We traverse the string once. |
| Two Pointer Technique | Time Complexity: O(n), based on the two-pointer traversal. |
| Sliding Window | ā |
Detailed Complexity Analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Brute Force (Check All Substrings) | O(nĀ²) | O(1) | Conceptual baseline to understand the problem constraints |
| Sliding Window | O(n) | O(1) | Best general solution for longest valid substring with limited modifications |
| Two Pointer Technique | O(n) | O(1) | When implementing substring expansion/shrinking logic using explicit pointer control |
Video Solution
Maximize the Confusion of an Exam | 3 Approaches | MICROSOFT | Leetcode-2024 | Explanation ā¢ codestorywithMIK ā¢ 12,179 views views
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