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This approach uses the sliding window technique to maintain a window of the answerKey where the number of modifications does not exceed k. We attempt to maximize the length of this window by evaluating either 'T' or 'F' transformations separately and then taking the maximum length between both transformations.
Time Complexity: O(n), where n is the length of the answerKey string. We traverse the string once.
Space Complexity: O(1), as we use constant space related to variables.
1public class MaxConsecutiveAnswers {
2 public static int maxConsecutiveAnswers(String answerKey, int k) {
3 int maxLen = 0, left = 0, countT = 0, countF = 0;
4
5 for (int right = 0; right < answerKey.length(); right++) {
6 char currentChar = answerKey.charAt(right);
7 if (currentChar == 'T') countT++;
8 else countF++;
9
10 while (Math.min(countT, countF) > k) {
11 if (answerKey.charAt(left) == 'T') countT--;
12 else countF--;
13 left++;
14 }
15 maxLen = Math.max(maxLen, right - left + 1);
16 }
17 return maxLen;
18 }
19
20 public static void main(String[] args) {
21 String answerKey = "TTFF";
22 int k = 2;
23 System.out.println(maxConsecutiveAnswers(answerKey, k));
24 }
25}The Java implementation creates a similar sliding window by maintaining pointers left and right. The resolution of window shrinkage occurs once more than k changes are needed. It employs counting when handling both 'T' and 'F' changes distinctly.
In this approach, we utilize two pointers to explore the string sequence, adjusting pointers as we evaluate possibilities involving updates utilizing k. The purpose is to ensure the longest alteration window without exceeding k modifications. It proceeds by specifically examining alternate possibilities and then concluding with the optimal maximum length.
Time Complexity: O(n), based on the two-pointer traversal.
Space Complexity: O(1), engaging a minimal variable set, bypassing auxiliary data structures.
#include <string>
using namespace std;
int maxConsecutiveAnswers(string answerKey, int k) {
int n = answerKey.length(), result = 0;
char directions[] = {'T', 'F'};
for (char x : directions) {
int count = 0, left = 0;
for (int right = 0; right < n; ++right) {
if (answerKey[right] != x) count++;
while (count > k) {
if (answerKey[left] != x) count--;
left++;
}
result = max(result, right - left + 1);
}
}
return result;
}
int main() {
string answerKey = "TFFT";
int k = 1;
cout << maxConsecutiveAnswers(answerKey, k) << endl;
return 0;
}This C++ implementation iterates each conversion target ('T' and 'F'), maintaining two pointers to span over the sequence length. Within loops when limits surpass, it retracts the left pointer to create valid windows, simultaneously evaluating interval length maximization concerning left and right boundary positions.