LeetCode 2024 - Maximize the Confusion of an Exam

This problem asks us to maximize the length of a contiguous block of identical answers in a true/false exam answer key. The input string answerKey contains only two characters, 'T' and 'F', representing the answers for each question.

leetcodemediumstringbinary-searchsliding-windowprefix-sum

LeetCode Problem 2024

Difficulty: 🟡 Medium
Topics: String, Binary Search, Sliding Window, Prefix Sum

Solution

Problem Understanding

This problem asks us to maximize the length of a contiguous block of identical answers in a true/false exam answer key. The input string answerKey contains only two characters, 'T' and 'F', representing the answers for each question. We are also given an integer k, which represents the maximum number of modifications we are allowed to make.

A modification means changing any character to either 'T' or 'F'. Since there are only two possible characters, every modification effectively flips one character into the opposite value when needed.

The goal is to determine the largest possible number of consecutive identical characters after performing at most k changes.

For example, if we have "TFFT" and k = 1, we can change one 'T' into 'F' and obtain "FFFT" or "TFFF". In either case, the longest consecutive block has length 3.

The constraints are important:

  • 1 <= n <= 5 * 10^4
  • The string contains only 'T' and 'F'
  • 1 <= k <= n

The input size can be as large as fifty thousand characters, which immediately tells us that quadratic algorithms will likely be too slow. An O(n^2) approach may require billions of operations in the worst case, so we need a linear or near linear solution.

Several edge cases are important:

  • The entire string may already consist of the same character.
  • k may be large enough to change every opposite character.
  • The optimal answer may involve turning characters into 'T' or into 'F', so we must consider both possibilities.
  • Very small inputs such as length 1 must still work correctly.
  • Long alternating patterns like "TFTFTFTF" can expose mistakes in window management logic.

Approaches

Brute Force Approach

A straightforward approach is to examine every possible substring and determine whether it can be converted into all 'T' or all 'F' using at most k changes.

For every starting index, we expand the substring one character at a time. While expanding, we count how many 'T' and 'F' characters exist inside the current substring. A substring can become all identical if the smaller count between the two characters is at most k.

For example:

  • Substring "TFFT" contains:

  • 2 'T'

  • 2 'F'

  • We would need to flip the smaller group, which costs 2 operations.

If that cost is less than or equal to k, the substring is valid.

This approach is correct because it checks every possible substring and directly verifies whether it satisfies the condition. However, it is too slow because there are O(n^2) substrings, and each may require additional work to evaluate.

With n = 50,000, this becomes impractical.

Optimal Sliding Window Approach

The key observation is that we do not need to evaluate every substring independently.

Suppose we want to find the longest substring that can become all 'T'. That means the substring may contain at most k 'F' characters, because those are the characters we would need to flip.

Similarly, if we want the substring to become all 'F', then it may contain at most k 'T' characters.

This immediately suggests a sliding window strategy.

We maintain a window [left, right] and expand it to the right. Inside the window, we count how many characters would need to be flipped. If the number exceeds k, the window is invalid, so we move the left boundary forward until the window becomes valid again.

Since each character enters and leaves the window at most once, the algorithm runs in linear time.

We run the sliding window twice:

  1. Longest window transformable into all 'T'
  2. Longest window transformable into all 'F'

The answer is the maximum of the two results.

Approach Time Complexity Space Complexity Notes
Brute Force O(n²) O(1) Checks every substring individually
Optimal Sliding Window O(n) O(1) Expands and shrinks a valid window dynamically

Algorithm Walkthrough

Step 1: Define a helper function

We create a helper function that computes the longest valid substring that can be converted into a chosen target character.

For example:

  • Target 'T' means we can tolerate at most k 'F' characters.
  • Target 'F' means we can tolerate at most k 'T' characters.

Step 2: Initialize the sliding window

We maintain:

  • left, the left boundary of the window
  • max_length, the best answer seen so far
  • changes_needed, the number of characters inside the window that do not match the target

Initially:

  • left = 0
  • changes_needed = 0

Step 3: Expand the window

We iterate right from 0 to n - 1.

Whenever answerKey[right] is not equal to the target character, we increment changes_needed.

This means the current window requires one more modification.

Step 4: Shrink the window if invalid

If changes_needed > k, the current window cannot be converted into the target character within the allowed operations.

We repeatedly move left forward until the window becomes valid again.

Whenever a removed character was contributing to changes_needed, we decrement the counter.

Step 5: Update the best answer

After restoring validity, the current window satisfies the constraint.

We compute:

window_length = right - left + 1

and update max_length.

Step 6: Solve for both characters

We run the helper function twice:

  • Once targeting 'T'
  • Once targeting 'F'

The final answer is the larger of the two results.

Why it works

The sliding window always maintains the largest valid window ending at the current right index.

A window is valid if it requires at most k modifications to become entirely one character. Whenever the window becomes invalid, shrinking from the left is guaranteed to restore validity because removing characters can only reduce the required number of changes.

Since every valid candidate window is considered during expansion, and no larger valid window is skipped, the algorithm correctly finds the maximum possible length.

Python Solution

class Solution:
    def maxConsecutiveAnswers(self, answerKey: str, k: int) -> int:
        def longest_window(target: str) -> int:
            left = 0
            changes_needed = 0
            max_length = 0

            for right in range(len(answerKey)):
                if answerKey[right] != target:
                    changes_needed += 1

                while changes_needed > k:
                    if answerKey[left] != target:
                        changes_needed -= 1
                    left += 1

                max_length = max(max_length, right - left + 1)

            return max_length

        return max(longest_window('T'), longest_window('F'))

The implementation follows the sliding window strategy directly.

The helper function longest_window computes the longest substring that can become entirely equal to the chosen target character. The variable changes_needed tracks how many characters inside the current window differ from the target.

As the right pointer expands the window, we include new characters and update the mismatch count when necessary.

Whenever the mismatch count exceeds k, the window becomes invalid. We then move the left pointer forward until the number of required changes is once again within the allowed limit.

At every step, the current window represents the largest valid substring ending at the current right position, so we update max_length.

Finally, we compute the best result for both possible target characters and return the larger value.

Go Solution

func maxConsecutiveAnswers(answerKey string, k int) int {
    longestWindow := func(target byte) int {
        left := 0
        changesNeeded := 0
        maxLength := 0

        for right := 0; right < len(answerKey); right++ {
            if answerKey[right] != target {
                changesNeeded++
            }

            for changesNeeded > k {
                if answerKey[left] != target {
                    changesNeeded--
                }
                left++
            }

            windowLength := right - left + 1
            if windowLength > maxLength {
                maxLength = windowLength
            }
        }

        return maxLength
    }

    tResult := longestWindow('T')
    fResult := longestWindow('F')

    if tResult > fResult {
        return tResult
    }

    return fResult
}

The Go implementation mirrors the Python solution closely.

One notable difference is that Go strings are indexed as bytes, so the target parameter uses type byte instead of a string character. Since the input contains only ASCII characters 'T' and 'F', byte indexing is completely safe and efficient.

Go does not provide a built in max function for integers, so the comparison is written manually.

No additional slices or maps are needed, keeping the memory usage constant.

Worked Examples

Example 1

Input: answerKey = "TTFF", k = 2

We try converting everything into 'T'.

Right Character Changes Needed Left Window Length Max
0 T 0 0 T 1 1
1 T 0 0 TT 2 2
2 F 1 0 TTF 3 3
3 F 2 0 TTFF 4 4

The entire string is valid because only two changes are needed.

Final answer:

4

Example 2

Input: answerKey = "TFFT", k = 1

We try converting everything into 'F'.

Right Character Changes Needed Left Window Length Max
0 T 1 0 T 1 1
1 F 1 0 TF 2 2
2 F 1 0 TFF 3 3
3 T 2 0 TFFT Invalid 3

Now the window is invalid because two changes are needed.

We shrink from the left:

  • Remove index 0
  • changes_needed becomes 1
  • New window is "FFT"

Length remains 3.

Final answer:

3

Example 3

Input: answerKey = "TTFTTFTT", k = 1

We target 'T'.

Right Character Changes Needed Left Window Length Max
0 T 0 0 T 1 1
1 T 0 0 TT 2 2
2 F 1 0 TTF 3 3
3 T 1 0 TTFT 4 4
4 T 1 0 TTFTT 5 5
5 F 2 0 TTFTTF Invalid 5

Shrink window:

  • Move left until only one 'F' remains.

Eventually the window becomes valid again.

The maximum valid length found is 5.

Final answer:

5

Complexity Analysis

Measure Complexity Explanation
Time O(n) Each pointer moves across the string at most once
Space O(1) Only a few integer variables are used

The algorithm is linear because both the left and right pointers only move forward. Even though there is a nested while loop, each character is removed from the window at most once, so the total number of operations remains proportional to n.

The memory usage is constant because no auxiliary arrays, hash maps, or other data structures proportional to the input size are allocated.

Test Cases

solution = Solution()

# Provided examples
assert solution.maxConsecutiveAnswers("TTFF", 2) == 4  # convert both F to T
assert solution.maxConsecutiveAnswers("TFFT", 1) == 3  # longest block after one flip
assert solution.maxConsecutiveAnswers("TTFTTFTT", 1) == 5  # optimal window length

# Single character
assert solution.maxConsecutiveAnswers("T", 1) == 1  # minimum input size

# Already uniform
assert solution.maxConsecutiveAnswers("TTTT", 2) == 4  # no changes needed
assert solution.maxConsecutiveAnswers("FFFFF", 3) == 5  # already optimal

# Alternating pattern
assert solution.maxConsecutiveAnswers("TFTFTFTF", 2) == 5  # expand through flips

# Large k
assert solution.maxConsecutiveAnswers("TFTF", 4) == 4  # can convert everything

# k equals 1
assert solution.maxConsecutiveAnswers("FTFTTTFF", 1) == 4  # one strategic flip

# Window shrinking required
assert solution.maxConsecutiveAnswers("TFFFTT", 1) == 4  # must shrink invalid window

# Entire string convertible
assert solution.maxConsecutiveAnswers("FTFTFT", 3) == 6  # all characters can match

# Long repeated segments
assert solution.maxConsecutiveAnswers("TTTTFFFFTTTT", 2) == 6  # bridge two regions partially
Test Why
"TTFF", 2 Verifies converting all characters into one value
"TFFT", 1 Tests choosing the optimal single flip
"TTFTTFTT", 1 Confirms longest valid window tracking
"T", 1 Minimum input size
"TTTT", 2 Already uniform string
"FFFFF", 3 Uniform string with unused operations
"TFTFTFTF", 2 Alternating pattern stresses window logic
"TFTF", 4 Large k allows full conversion
"FTFTTTFF", 1 Tests selective optimal flipping
"TFFFTT", 1 Forces repeated shrinking of the window
"FTFTFT", 3 Entire string becomes identical
"TTTTFFFFTTTT", 2 Large grouped segments with limited flips

Edge Cases

One important edge case occurs when the string already contains all identical characters. For example, "TTTTTT" requires no modifications at all. A buggy implementation might still attempt unnecessary shrinking or incorrectly count modifications. The sliding window solution handles this naturally because changes_needed never exceeds k, so the window simply expands across the entire string.

Another critical case is an alternating pattern such as "TFTFTFTF". These inputs frequently expose errors in window maintenance because the mismatch count changes often. The implementation correctly handles this by incrementing changes_needed whenever a non target character enters the window and decrementing it whenever such a character leaves the window.

A third important edge case happens when k is large enough to convert the entire string. For example, if answerKey = "TFTF" and k = 4, every character can be converted into the same value. The algorithm correctly expands the window to the full string length because the mismatch count never exceeds the allowed limit.

Another subtle case involves repeated shrinking of the window. Consider "TFFFTT" with k = 1. As the window expands, it eventually contains too many mismatches and must repeatedly move the left pointer forward. Incorrect implementations sometimes shrink only once instead of continuing until the window becomes valid again. The while changes_needed > k loop guarantees correctness by fully restoring validity before continuing.