Leetcode 2024: Maximize the Confusion of an Exam
You are given a string
answerKey where each character represents a question’s answer (either ‘T’ for True or ‘F’ for False), and an integer k. You are allowed to change at most k answers. Your task is to find the maximum number of consecutive ‘T’s or ‘F’s in the string after performing at most k changes.Problem
Approach
Steps
Complexity
Input: The input consists of a string `answerKey` where each character is either 'T' or 'F', and an integer `k` representing the maximum number of changes allowed.
Example: answerKey = "FFTT", k = 2
Constraints:
• 1 <= answerKey.length <= 50000
• 1 <= k <= answerKey.length
• answerKey[i] is either 'T' or 'F'.
Output: Return the maximum number of consecutive 'T's or 'F's that can be achieved after performing at most `k` changes.
Example: Output: 4
Constraints:
• The answer must be the maximum consecutive occurrence of either 'T' or 'F'.
Goal: The goal is to find the longest subsequence of consecutive 'T's or 'F's that can be formed after changing at most `k` characters.
Steps:
• 1. Use a sliding window approach where you iterate over the string while maintaining a window of consecutive characters that are either 'T' or 'F'.
• 2. Track the number of changes you make and adjust the window size such that the number of changes does not exceed `k`.
• 3. Update the result as you move through the string to ensure the maximum length of consecutive 'T's or 'F's is obtained.
Goal: Constraints for input values and operations.
Steps:
• 1 <= answerKey.length <= 50000
• 1 <= k <= answerKey.length
• answerKey[i] is either 'T' or 'F'.
Assumptions:
• The string `answerKey` contains only the characters 'T' and 'F'.
• The number of changes is limited by the integer `k`.
• Input: answerKey = "FFTT", k = 2
• Explanation: We can replace both 'F's with 'T's, forming the string 'TTTT', which gives us 4 consecutive 'T's.
• Input: answerKey = "TFTF", k = 1
• Explanation: We can replace one 'F' with 'T', forming either 'TTTF' or 'TTFT', which gives us 3 consecutive 'T's.
• Input: answerKey = "FTFTFT", k = 1
• Explanation: We can replace one 'F' with 'T', resulting in either 'TTFTFT' or 'FTTTFT', both giving us 3 consecutive 'T's.
Approach: We can solve this problem efficiently using a sliding window approach to track the maximum number of consecutive 'T's or 'F's that can be formed with at most `k` changes.
Observations:
• The problem can be reduced to finding the longest subsequence of 'T's or 'F's after changing at most `k` elements.
• By applying a sliding window, we can efficiently find the longest possible subsequence that meets the constraints.
Steps:
• 1. Initialize two sliding windows: one for 'T' and one for 'F'.
• 2. As you iterate over the string, track the number of changes made to convert the current window into all 'T's or all 'F's.
• 3. Adjust the window size such that the number of changes does not exceed `k`.
• 4. Track the maximum length of any valid window during the process.
Empty Inputs:
• If the input string is empty, return 0.
Large Inputs:
• Ensure the solution efficiently handles the maximum input size (up to 50000 characters).
Special Values:
• If `k` is 0, no changes can be made, and the answer should be the longest sequence of consecutive 'T's or 'F's in the original string.
Constraints:
• Ensure that the solution does not exceed time limits for large inputs.
int maxConsecutiveAnswers(string key, int k) {
int ans = 1;
int fidx = 0, tidx = 0, fcnt = 0, tcnt = 0;
for(int i = 0; i < key.size(); i++) {
if(key[i] == 'F') fcnt++;
else tcnt++;
while(fcnt > k) {
if(key[tidx] == 'F') fcnt--;
tidx++;
}
ans = max(ans, i - tidx + 1);
while(tcnt > k) {
if(key[fidx] == 'T') tcnt--;
fidx++;
}
ans = max(ans, i - fidx + 1);
}
return ans;
}
1 : Function Definition
int maxConsecutiveAnswers(string key, int k) {
This defines the function 'maxConsecutiveAnswers', which takes a string 'key' and an integer 'k'. It returns the maximum length of a subsequence where at most 'k' characters can be changed to either 'F' or 'T'.
2 : Variable Initialization
int ans = 1;
This initializes an integer variable 'ans' to store the maximum length of consecutive answers. The initial value is set to 1 since the minimum answer will be 1.
3 : Variable Initialization
int fidx = 0, tidx = 0, fcnt = 0, tcnt = 0;
These variables are initialized to keep track of the indices and counts for 'F' and 'T' characters. 'fidx' and 'tidx' are indices for the start of the window, while 'fcnt' and 'tcnt' are the counts of 'F' and 'T' characters within the window.
4 : Loop Start
for(int i = 0; i < key.size(); i++) {
This loop iterates through each character in the string 'key' to explore different possible subsequences.
5 : Condition Check
if(key[i] == 'F') fcnt++;
This checks if the current character is 'F'. If it is, the count of 'F' characters within the current window ('fcnt') is incremented.
6 : Condition Check
else tcnt++;
If the current character is 'T', the count of 'T' characters within the current window ('tcnt') is incremented.
7 : Window Adjustment
while(fcnt > k) {
This loop ensures that the number of 'F' characters within the window does not exceed 'k'. If the count of 'F' exceeds 'k', the window is adjusted.
8 : Window Adjustment
if(key[tidx] == 'F') fcnt--;
This checks if the character at the leftmost index ('tidx') is 'F'. If it is, the count of 'F' characters ('fcnt') is decreased.
9 : Window Adjustment
tidx++;
This moves the leftmost index of the window ('tidx') to the right, effectively shrinking the window from the left.
10 : Update Answer
ans = max(ans, i - tidx + 1);
This updates the value of 'ans' by comparing the current answer with the length of the current window (i - tidx + 1). It ensures that 'ans' always holds the maximum length of a valid subsequence.
11 : Window Adjustment
while(tcnt > k) {
This loop ensures that the number of 'T' characters within the window does not exceed 'k'. If the count of 'T' exceeds 'k', the window is adjusted.
12 : Window Adjustment
if(key[fidx] == 'T') tcnt--;
This checks if the character at the leftmost index ('fidx') is 'T'. If it is, the count of 'T' characters ('tcnt') is decreased.
13 : Window Adjustment
fidx++;
This moves the leftmost index of the window ('fidx') to the right, shrinking the window from the left.
14 : Update Answer
ans = max(ans, i - fidx + 1);
This updates 'ans' again by comparing it with the length of the current window (i - fidx + 1).
15 : Return Statement
return ans;
This returns the final result, which is the length of the longest valid subsequence where at most 'k' characters can be changed.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The sliding window approach iterates through the string once, resulting in a time complexity of O(n).
Best Case: O(1)
Worst Case: O(1)
Description: The space complexity is O(1) as the solution only requires a few variables to track the window and counts.
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