Maximize the Confusion of an Exam

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StringsSliding WindowsArraysTwo Pointers

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

Try coding it in LeetCode

Maximize the Confusion of an Exam

Medium

Asked by:

6 views

Topics:

StringsSliding WindowsArraysTwo Pointers

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

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible values for 'k' (the maximum number of allowed changes), and what is the maximum length of the answerKey string?
Does the answerKey string only contain 'T' and 'F' characters, or can it contain other characters as well?
If it's impossible to achieve a higher confusion than the original string (even with k changes), what should I return?
If there are multiple substrings of the same maximum length that maximize confusion, is it acceptable to return any one of them?
Is 'k' always non-negative, and is the answerKey string guaranteed to be non-empty?

Brute Force Solution

Approach

The brute force strategy for this problem involves trying every single possible chunk of answers and figuring out the longest possible chunk we can make confusing enough. We will examine every possible section of the answer key to find the biggest confusing part.

Here's how the algorithm would work step-by-step:

Start by looking at the first answer, then the first two answers, then the first three answers, and so on, until you reach the entire answer key.
For each of these chunks of answers, count how many trues and falses there are.
If the count of trues or falses is greater than the allowed amount of changes we can make, move on to the next chunk.
If the number of changes we need to make to make a confusing exam is less than or equal to the limit, remember the size of this chunk.
Now, repeat this process, but this time start at the second answer. Then the third answer, and so on, until you reach the last answer.
Keep track of the size of the largest chunk we found that did not exceed the changes that we were allowed to make.
The size of the biggest allowable chunk we found in all these tries is our answer.

Code Implementation

def maximize_confusion_brute_force(answer_key, allowed_changes):
    maximum_length = 0
    number_of_answers = len(answer_key)

    for start_index in range(number_of_answers):
        for end_index in range(start_index, number_of_answers):
            # Consider every sub-array of the answer key
            sub_array = answer_key[start_index:end_index + 1]
            true_count = sub_array.count('T')
            false_count = sub_array.count('F')

            # Checks if we can make the entire subarray T or F
            if min(true_count, false_count) <= allowed_changes:

                current_length = end_index - start_index + 1
                # Keep track of the longest valid sub-array
                maximum_length = max(maximum_length, current_length)

    return maximum_length

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through all possible subarrays of the input string of length n. The outer loop determines the starting position of the subarray, and the inner loop determines the ending position. In the worst case, for each starting position, we iterate through almost all remaining elements to form a subarray, resulting in a nested loop structure. The number of operations can be approximated by n * (n+1) / 2. Therefore, the time complexity is O(n²).

Space Complexity

O(1) – The provided algorithm uses a constant amount of extra space. It primarily involves iterating through subarrays and counting true and false values within those subarrays. These counts are stored in a fixed number of variables, independent of the input string's length, N. Thus, no auxiliary data structures that scale with the input size are required, resulting in constant space complexity.

Optimal Solution

Approach

The problem asks us to find the longest possible segment in a series of 'T's and 'F's where we're allowed to change a certain number of 'T's to 'F's or 'F's to 'T's. The trick is to slide a window across the series, expanding it when possible and shrinking it when necessary, to find the longest valid segment.

Here's how the algorithm would work step-by-step:

Imagine a window that starts at the beginning of the series.
Expand the window to the right, one step at a time.
While expanding, count how many 'T's we would need to change to 'F's or how many 'F's we would need to change to 'T's to make the whole window the same letter.
If the number of changes we need is more than the number of changes we're allowed, shrink the window from the left until the number of required changes is within our allowed limit again.
Keep track of the length of the longest valid window we've seen so far.
Repeat this process of expanding and shrinking the window until we reach the end of the series.
The length of the longest valid window we found is our answer.

Code Implementation

def maximize_the_confusion_of_an_exam(exam_string, max_allowed_changes):
    window_start = 0
    maximum_length = 0
    true_count = 0
    false_count = 0

    for window_end in range(len(exam_string)):
        if exam_string[window_end] == 'T':
            true_count += 1
        else:
            false_count += 1

        # Shrink the window if changes exceed limit.
        while min(true_count, false_count) > max_allowed_changes:
            if exam_string[window_start] == 'T':
                true_count -= 1
            else:
                false_count -= 1

            window_start += 1

        # Update max length with the current window size.
        maximum_length = max(maximum_length, window_end - window_start + 1)

    return maximum_length

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the exam string using a sliding window approach. The window expands to the right and may shrink from the left. Each character in the exam string is visited at most twice: once by the right pointer (expanding the window) and once by the left pointer (shrinking the window). Therefore, the number of operations is proportional to the length of the exam string, n, resulting in a time complexity of O(n).

Space Complexity

O(1) – The algorithm uses a sliding window approach with a fixed number of variables to track the window's boundaries and counts. It does not create any auxiliary data structures like arrays, lists, or hash maps whose size depends on the input size N (the length of the answerKey). Therefore, the auxiliary space used remains constant irrespective of the input size, resulting in a space complexity of O(1).

Edge Cases

Case	How to Handle
Empty input string	Return 0 as no flip can increase contiguous block size if there is no characters
k is 0	Return the length of the longest contiguous block of either 'T' or 'F' without any flips
k is greater than the length of the input string	Return the length of the input string as we can flip all characters to maximize contiguous block size
Input string contains only 'T' or only 'F'	Return the length of the string if k is greater or equal to 0; otherwise return length - k if k is less than 0 which should never happen
Input string with very large length	Ensure the algorithm's time and space complexity remains efficient (O(n)) for large inputs
k is a negative number	Throw IllegalArgumentException, since flipping a negative number of characters doesn't make sense in the context of this problem.
String contains characters other than T and F	Throw IllegalArgumentException or filter out invalid chars, depending on the specification.
Maximum value of k reaches Integer.MAX_VALUE	No special action needed as sliding window approach operates based on window conditions not the magnitude of k.