Maximize the Confusion of an Exam

class Solution {
public:
    int getMaxAnswerStreak(string answerKey, int k) {
        int maxStreak = 0;
        unordered_map<char, int> countLetters;
            
        for (int r = 0; r < answerKey.size(); r++) {
            countLetters[answerKey[r]]++;
            int smallest = min(countLetters['T'], countLetters['F']);
                
            if (smallest <= k) {
                maxStreak++;
            } else {
                countLetters[answerKey[r - maxStreak]]--;
            }
        }
    
        return maxStreak;
    }
};

The problem "Maximize the Confusion of an Exam" is efficiently tackled using a C++ implementation that employs the sliding window technique alongside a frequency counting strategy. Here’s a breakdown of the solution:

• Initialize maxStreak to zero to keep track of the maximum number of consecutive answers that match a certain criterion.
• Utilize an unordered_map named countLetters to monitor the frequency of each character ('T' for true and 'F' for false) within the current sliding window.
• Traverse the answerKey using a loop, dynamically adjusting the sliding window:
  • Add the current character to countLetters.
  • Determine the count of the less frequent answer within the window using the min function on the values from countLetters.
  • If the counts of the less frequent answer is less than or equal to k, increase maxStreak to include the current character.
  • If not, modify the window by decrementing the count of the character that falls out of the left side of the current window length to maintain the allowed number of transformations denoted by k.
• Continue this process until you've evaluated all possible substrings of answerKey.
• Return maxStreak, which now holds the maximum length of a contiguous segment of answers where one can change up to k answers to either all 'T' or all 'F', thereby maximizing the confusion or difficulty of the exam.

This approach ensures a swift traversal through the string while dynamically managing and evaluating the possibilities of maximizing confusion under the stated constraints.

class Solution {
    public int longestSequence(String responses, int k) {
        int lengthMax = 0;
        Map<Character, Integer> frequency = new HashMap<>();
            
        for (int end = 0; end < responses.length(); end++) {
            frequency.put(responses.charAt(end), frequency.getOrDefault(responses.charAt(end), 0) + 1);
            int minimum = Math.min(frequency.getOrDefault('T', 0), frequency.getOrDefault('F', 0));
                
            if (minimum <= k) {
                lengthMax++;
            } else {
                frequency.put(responses.charAt(end - lengthMax), frequency.get(responses.charAt(end - lengthMax)) - 1);
            }
        }
    
        return lengthMax;
    }
}

The provided Java solution focuses on finding the maximum length of a substring in a string where you can make at most k substitutions to convert it either entirely to 'T's or entirely to 'F's. The function longestSequence uses a sliding window approach with a hashmap to maintain the frequency of each character within the window.

Here’s a breakdown of how the solution operates:

1. Initialize lengthMax to 0 to store the maximum length of the substring that meets the condition.
2. Create a frequency hashmap to track the number of occurrences of each character in the current sliding window.
3. Iterate through each character in the responses string using an index end.
4. Update the frequency of the current character in the hashmap.
5. Calculate minimum, the smaller frequency between 'T' and 'F' within the current window.
6. If minimum is less than or equal to k, the substring can be made uniform by replacing these minimum characters, and thus, increase lengthMax.
7. If not, adjust the window by reducing the frequency of the character at the start of the window (end - lengthMax) and continue.
8. Finally, the function returns lengthMax, which represents the length of the longest valid substring found.

This effectively tracks the longest stretch where you can replace up to k characters to make the substring uniform, using an optimal approach through a sliding window mechanism combined with frequent character tracking.

class Solution:
    def maximumConsecutiveAnswers(self, answers: str, maxFlips: int) -> int:
        longest_sequence = 0
        char_freq = collections.Counter()
            
        for end in range(len(answers)):
            char_freq[answers[end]] += 1
            least_frequent = min(char_freq['T'], char_freq['F'])
                
            if least_frequent <= maxFlips:
                longest_sequence += 1
            else:
                char_freq[answers[end - longest_sequence]] -= 1
    
        return longest_sequence

The provided Python3 code defines a function maximumConsecutiveAnswers that determines the maximum number of consecutive 'T' or 'F' answers you can obtain in a string answers by flipping at most maxFlips characters. Here is a summary of how this function works:

• Initialize longest_sequence to store the length of the longest sequence achievable.
• Use a Counter from the collections module to track the frequency of 'T' and 'F' answers.
• Iterate through each character in the answers string using a loop.
• Update the frequency of the currently processed character.
• Calculate the frequency of the least frequent character between 'T' and 'F'.
• If the number of least frequent characters is less than or equal to maxFlips, increment longest_sequence.
• If not, decrease the count of the character at the start of the current longest sequence to adjust the window size.
• Finally, longest_sequence is returned, which gives the maximum number of consecutive characters that can be achieved by flipping at most maxFlips characters.

Adapt this approach by modifying variables and conditions to suit specific needs or different variations of the problem.