LeetCode 2024: Maximize the Confusion of an Exam ·

Table of Contents

Meta Data
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Difficulty: medium First Attempt: 2026-01-10

Submissions: Accepted (first try)

Approach
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Making a stretch of answers all the same needs at least as many changes as the rarer of T and F. So a window is valid ⇔ min(count of T, count of F) <= k.

The rarer count only grows or stays as the window grows, so the condition is monotonic: use the “longest” sliding window and shrink the left when invalid.

My Solution
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class Solution:
    def maxConsecutiveAnswers(self, answerKey: str, k: int) -> int:
        cnt = Counter()
        left = best = 0
        for right, c in enumerate(answerKey):
            cnt[c] += 1
            while min(cnt["T"], cnt["F"]) > k:     # invalid only if both exceed k
                cnt[answerKey[left]] -= 1
                left += 1
            best = max(best, right - left + 1)
        return best

My original counted only T, with the condition count_true > k and (length - count_true) > k, which means the same thing.

Complexity
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Time O(n), space O(1).

Mistakes to Avoid
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Running two passes, one for “all T” and one for “all F”, also works, but one window looking at the rarer letter is enough.

Interview Explanation
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To make a stretch identical, the fewest changes is the rarer of T and F, so a window is valid when that count is <= k. The condition is monotonic in the window size, so I slide a window, shrink the left when invalid, and track the longest length. Time O(n).