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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.
class Program {
static int maxConsecutiveAnswers(string answerKey, int k) {
int n = answerKey.Length;
int result = 0;
foreach (char x in new char[]{ 'T', 'F' }) {
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 = Math.Max(result, right - left + 1);
}
}
return result;
}
static void Main() {
string answerKey = "TFFT";
int k = 1;
Console.WriteLine(maxConsecutiveAnswers(answerKey, k));
}
}This C# code iterates over switchable characters and manages positioning with pointers, adapting metrics to track mismatch occurrences. Pivot activities govern segments' expiry and launching fidelity to k modification demands.