Maximize the Confusion of an Exam

Q: What is the time complexity of Maximize the Confusion of an Exam?

O(n). Each character is visited at most twice: once when the right pointer expands the window and once when the left pointer shrinks it, resulting in linear time.

Q: What pattern does Maximize the Confusion of an Exam use?

The Pattern: Sliding Window (Variable Size with Constraint) The Heuristic: When asked to find the longest substring satisfying a constraint on character counts or changes, use a sliding window that expands greedily and contracts only when the constraint is violated, maintaining a running count of violations to efficiently track valid windows.

Medium Sliding Window

Problem

Given a string answerKey consisting of 'T' and 'F' characters and an integer k, return the maximum length of a substring containing only 'T' or only 'F' after changing at most k characters.

1 ≤ answerKey.length ≤ 10⁵
answerKey[i] is either 'T' or 'F'
0 ≤ k ≤ answerKey.length

Example

Input: answerKey = "TTFF", k = 2

Output: 4

The brute-force approach would check every substring and count how many characters need to be changed to make it all 'T' or all 'F', which is O(n^2) and inefficient. The optimal approach uses a sliding window to find the longest substring where the number of characters different from the target ('T' or 'F') is at most k. For example, considering target 'T', the window expands while the number of 'F's inside is ≤ k. When it exceeds k, the window shrinks from the left. The same is done for target 'F'. The maximum length found between these two runs is the answer.

Approach

Straightforward Solution

A brute-force approach would check all substrings and count mismatches, resulting in O(n^2) time complexity, which is too slow for large inputs.

Core Observation

The problem reduces to finding the longest substring where at most k characters differ from a chosen target character ('T' or 'F'). This is a classic sliding window problem where the window maintains a count of mismatches and adjusts boundaries to satisfy the constraint.

Path to Optimal

Preview

Recognizing that the problem asks for the longest substring with at most k mismatches to a target character suggests a sliding window approach. By maintaining a window and counting mismatches, the window can expand greedily and shrink when the mismatch count exceeds k…

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Optimal Approach

Preview

Implement a helper function that uses two pointers to maintain a sliding window over the string, counting how many characters differ from the target. Expand the right pointer, increment mismatch count if needed, and shrink the left pointer when mismatches exceed k…

Full step-by-step walkthrough on Pro →

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Time

O(n)

Each character is visited at most twice: once when the right pointer expands the window and once when the left pointer shrinks it, resulting in linear time.

Space

O(1)

Only a fixed number of variables are used to track counts and pointers, independent of input size.

Pattern Spotlight

Sliding Window (Variable Size with Constraint)

When asked to find the longest substring satisfying a constraint on character counts or changes, use a sliding window that expands greedily and contracts only when the constraint is violated, maintaining a running count of violations to efficiently track valid windows.

Solution

Python

1class Solution:
  def maxConsecutiveAnswers(self, answerKey: str, k: int) -> int:
      def longest_with_target(target):
          left = 0
          changes = 0
          best = 0
7
          for right in range(len(answerKey)):
              if answerKey[right] != target:
                  changes += 1
11
              while changes > k:
                  if answerKey[left] != target:
                      changes -= 1
                  left += 1
16
              best = max(best, right - left + 1)
18
          return best
20
      return max(longest_with_target("T"), longest_with_target("F"))

Step-by-Step Solution

Expand and Contract Sliding Window to Track Mismatches

3def longest_with_target(target):
  left = 0
  changes = 0
  best = 0
  for right in range(len(answerKey)):
      if answerKey[right] != target:
          changes += 1
      while changes > k:
          if answerKey[left] != target:
              changes -= 1
          left += 1
      best = max(best, right - left + 1)
  return best

Objective

To find the longest substring where at most k characters differ from the target character by dynamically adjusting window boundaries.

Key Insight

The sliding window maintains a count of characters that differ from the target. Expanding the right pointer greedily increases the window size, while shrinking the left pointer when mismatches exceed k restores validity. This approach efficiently finds the longest valid substring in O(n) time by avoiding redundant checks and leveraging the problem's constraint.

Interview Quick-Check

Core Logic

Maintain a window with two pointers and a mismatch count; expand right pointer and increment mismatches if needed; shrink left pointer when mismatches exceed k to restore validity.

State & Boundaries

The window is valid as long as the mismatch count is ≤ k; the loop invariant ensures the window always satisfies this constraint.

Common Pitfalls & Bugs

Forgetting to decrement mismatch count when moving the left pointer past a mismatched character leads to incorrect window validity checks.

Complexity

Each character is processed at most twice, ensuring O(n) time complexity.

Combine Results from Both Target Characters to Find Maximum

To compute the maximum length substring by evaluating both 'T' and 'F' targets and returning the best result.

1 more step with full analysis available on Pro.

Line Analysis

This solution has 3 Critical lines interviewers watch for.

Line 9 Critical

if answerKey[right] != target:

Increment mismatch count if the current character differs from the target.

Incrementing the mismatch count exactly when the character differs from the target is critical to correctly enforcing the constraint on allowed changes.

Line 13 Critical

if answerKey[left] != target:

If the character at the left pointer differs from the target, decrement mismatch count.

Failing to decrement here would cause the algorithm to incorrectly believe the window is invalid, leading to incorrect window sizes.

Line 21 Critical

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

Return the maximum length found by targeting either 'T' or 'F'.

This line is critical because the problem requires the maximum length substring of either 'T' or 'F', and only by comparing both can the correct answer be found.

Full line-by-line criticality + rationale for all 14 lines available on Pro.

Test Your Understanding

Why does running the sliding window twice, once for 'T' and once for 'F', guarantee finding the maximum substring length?

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1	`class Solution:`
2	`def maxConsecutiveAnswers(self, answerKey: str, k: int) -> int:`
3	`def longest_with_target(target):`
4	`left = 0`
5	`changes = 0`
6	`best = 0`
7
8	`for right in range(len(answerKey)):`
9	`if answerKey[right] != target:`
10	`changes += 1`
11
12	`while changes > k:`
13	`if answerKey[left] != target:`
14	`changes -= 1`
15	`left += 1`
16
17	`best = max(best, right - left + 1)`
18
19	`return best`
20
21	`return max(longest_with_target("T"), longest_with_target("F"))`

3	`def longest_with_target(target):`
4	`left = 0`
5	`changes = 0`
6	`best = 0`
8	`for right in range(len(answerKey)):`
9	`if answerKey[right] != target:`
10	`changes += 1`
12	`while changes > k:`
13	`if answerKey[left] != target:`
14	`changes -= 1`
15	`left += 1`
17	`best = max(best, right - left + 1)`
19	`return best`