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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.
1#include <stdio.h>
2#include <string.h>
3
4int maxConsecutiveAnswers(char *answerKey, int k) {
5 int maxLen = 0;
6 int left = 0, right = 0, maxT = 0, maxF = 0;
7 int len = strlen(answerKey);
8
9 for (; right < len; ++right) {
10 if (answerKey[right] == 'T')
11 maxT++;
12 else
13 maxF++;
14
15 while (maxT > k && maxF > k) {
16 if (answerKey[left] == 'T')
17 maxT--;
18 else
19 maxF--;
20 left++;
21 }
22 maxLen = maxLen > right - left + 1 ? maxLen : right - left + 1;
23 }
24 return maxLen;
25}
26
27int main() {
28 char answerKey[] = "TTFTTFTT";
29 int k = 1;
30 printf("%d\n", maxConsecutiveAnswers(answerKey, k));
31 return 0;
32}The function maxConsecutiveAnswers applies the sliding window technique twice: once assuming we change all 'F' to 'T' and once the reverse. It uses two counters, maxT and maxF, to track the number of changes made. As soon as the number of required changes exceeds the allowed k, the window is shrunk from the left. The maximum length of such a valid window is tracked and returned.
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.