2024. Maximize the Confusion of an Exam LeetCode Solution
In this guide, you will get 2024. Maximize the Confusion of an Exam LeetCode Solution with the best time and space complexity. The solution to Maximize the Confusion of an Exam problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
Problem Statement of Maximize the Confusion of an Exam
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.
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.
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
Complexity Analysis
Time Complexity: O(n)
Space Complexity: O(2) = O(1)
2024. Maximize the Confusion of an Exam LeetCode Solution in C++
class Solution {
public:
int maxConsecutiveAnswers(string answerKey, int k) {
int ans = 0;
int maxCount = 0;
vector<int> count(2);
for (int l = 0, r = 0; r < answerKey.length(); ++r) {
maxCount = max(maxCount, ++count[answerKey[r] == 'T']);
while (maxCount + k < r - l + 1)
--count[answerKey[l++] == 'T'];
ans = max(ans, r - l + 1);
}
return ans;
}
}; /* code provided by PROGIEZ */
2024. Maximize the Confusion of an Exam LeetCode Solution in Java
class Solution {
public int maxConsecutiveAnswers(String answerKey, int k) {
int ans = 0;
int maxCount = 0;
int[] count = new int[2];
for (int l = 0, r = 0; r < answerKey.length(); ++r) {
maxCount = Math.max(maxCount, ++count[answerKey.charAt(r) == 'T' ? 1 : 0]);
while (maxCount + k < r - l + 1)
--count[answerKey.charAt(l++) == 'T' ? 1 : 0];
ans = Math.max(ans, r - l + 1);
}
return ans;
}
} // code provided by PROGIEZ
2024. Maximize the Confusion of an Exam LeetCode Solution in Python
class Solution:
def maxConsecutiveAnswers(self, answerKey: str, k: int) -> int:
ans = 0
maxCount = 0
count = collections.Counter()
l = 0
for r, c in enumerate(answerKey):
count[c == 'T'] += 1
maxCount = max(maxCount, count[c == 'T'])
while maxCount + k < r - l + 1:
count[answerKey[l] == 'T'] -= 1
l += 1
ans = max(ans, r - l + 1)
return ans # code by PROGIEZ
