LeetCode #2024 — MEDIUM

Maximize the Confusion of an Exam

Move from brute-force thinking to an efficient approach using binary search strategy.

The Problem

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 i^th 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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Binary Search · Sliding Window

Example 1

"TTFF"
2

Example 2

"TFFT"
1

Example 3

"TTFTTFTT"
1

Core Insight

What unlocks the optimal approach

Can we use the maximum length at the previous position to help us find the answer for the current position?
Can we use binary search to find the maximum consecutive same answer at every position?

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2024: Maximize the Confusion of an Exam
class Solution {
    private char[] s;
    private int k;

    public int maxConsecutiveAnswers(String answerKey, int k) {
        s = answerKey.toCharArray();
        this.k = k;
        return Math.max(f('T'), f('F'));
    }

    private int f(char c) {
        int l = 0, cnt = 0;
        for (char ch : s) {
            cnt += ch == c ? 1 : 0;
            if (cnt > k) {
                cnt -= s[l++] == c ? 1 : 0;
            }
        }
        return s.length - l;
    }
}

// Accepted solution for LeetCode #2024: Maximize the Confusion of an Exam
func maxConsecutiveAnswers(answerKey string, k int) int {
	f := func(c byte) int {
		l, cnt := 0, 0
		for _, ch := range answerKey {
			if byte(ch) == c {
				cnt++
			}
			if cnt > k {
				if answerKey[l] == c {
					cnt--
				}
				l++
			}
		}
		return len(answerKey) - l
	}
	return max(f('T'), f('F'))
}

# Accepted solution for LeetCode #2024: Maximize the Confusion of an Exam
class Solution:
    def maxConsecutiveAnswers(self, answerKey: str, k: int) -> int:
        def f(c: str) -> int:
            cnt = l = 0
            for ch in answerKey:
                cnt += ch == c
                if cnt > k:
                    cnt -= answerKey[l] == c
                    l += 1
            return len(answerKey) - l

        return max(f("T"), f("F"))

// Accepted solution for LeetCode #2024: Maximize the Confusion of an Exam
impl Solution {
    pub fn max_consecutive_answers(answer_key: String, k: i32) -> i32 {
        let n = answer_key.len();
        let k = k as usize;
        let s: Vec<char> = answer_key.chars().collect();

        let f = |c: char| -> usize {
            let mut l = 0;
            let mut cnt = 0;
            for &ch in &s {
                cnt += if ch == c { 1 } else { 0 };
                if cnt > k {
                    cnt -= if s[l] == c { 1 } else { 0 };
                    l += 1;
                }
            }
            n - l
        };

        std::cmp::max(f('T'), f('F')) as i32
    }
}

// Accepted solution for LeetCode #2024: Maximize the Confusion of an Exam
function maxConsecutiveAnswers(answerKey: string, k: number): number {
    const n = answerKey.length;
    const f = (c: string): number => {
        let [l, cnt] = [0, 0];
        for (const ch of answerKey) {
            cnt += ch === c ? 1 : 0;
            if (cnt > k) {
                cnt -= answerKey[l++] === c ? 1 : 0;
            }
        }
        return n - l;
    };
    return Math.max(f('T'), f('F'));
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

Shrinking the window only once

Wrong move: Using `if` instead of `while` leaves the window invalid for multiple iterations.

Usually fails on: Over-limit windows stay invalid and produce wrong lengths/counts.

Fix: Shrink in a `while` loop until the invariant is valid again.