Maximize the Confusion of an Exam
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
answerKey = "TTFF", k = 24The 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
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. Running this process twice, once for target 'T' and once for 'F', ensures the maximum length substring for either character is found.
Optimal Approach
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. Track the maximum window size during this process. Return the maximum of the two runs for 'T' and 'F'.
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
| 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")) |
Step-by-Step Solution
Expand and Contract Sliding Window to Track Mismatches
| 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 |
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
| 21 | return max(longest_with_target("T"), longest_with_target("F")) |
Objective
To compute the maximum length substring by evaluating both 'T' and 'F' targets and returning the best result.
Key Insight
Since the problem allows changing characters to either 'T' or 'F', the maximum substring length must be the best achievable by targeting either character. Running the helper function twice and taking the maximum ensures completeness and correctness without complicating the sliding window logic.
Interview Quick-Check
Core Logic
Invoke the sliding window helper for both 'T' and 'F' and return the maximum result to cover all valid substrings.
Common Pitfalls & Bugs
Neglecting to check both targets can miss the optimal substring if it consists mostly of the other character.
Line Analysis
This solution has 3 Critical lines interviewers watch for.
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.
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.
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.
changes = 0Initialize the count of characters that differ from the target within the window.
Tracking mismatches is essential to know when the window violates the allowed number of changes k.
while changes > k:While the mismatch count exceeds k, shrink the window from the left.
Shrinking the window restores validity by removing characters that cause the window to violate the allowed number of changes.
left += 1Update the best window size with the current valid window length.
Tracking the maximum window size ensures the algorithm returns the longest valid substring found during iteration.
def longest_with_target(target):Define a helper function to find the longest substring with at most k changes to the target character.
Encapsulating the sliding window logic in a helper function allows reuse for both target characters and keeps the main function concise.
changes -= 1Move the left pointer forward to shrink the window.
Advancing the left pointer reduces the window size and removes characters that violate the constraint.
return bestReturn the maximum window size found for the target character.
Returning the result of the helper function provides the maximum length of a substring with at most k changes to the target.
for right in range(len(answerKey)):Iterate over the string with the right pointer expanding the window.
The right pointer explores potential window expansions, enabling the algorithm to consider all substrings efficiently.
left = 0Initialize the left pointer of the sliding window to 0.
The left pointer marks the start of the current window and will move forward to maintain the constraint on mismatches.
best = 0Initialize the variable to track the best (maximum) window size found.
This variable accumulates the maximum valid window length during the iteration.
changes += 1Blank line (ignored).
No operation.
best = max(best, right - left + 1)Blank line (ignored).
No operation.
Test Your Understanding
Why does running the sliding window twice, once for 'T' and once for 'F', guarantee finding the maximum substring length?
Because the substring must be all 'T's or all 'F's after at most k changes, checking both targets ensures that the longest valid substring for either character is found, covering all possibilities.
Related Problems
Sliding Window pattern
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