23. Maximize the Confusion of an Exam / Max Consecutive Ones III

A teacher is writing a test with n true/false questions, with 'T' denoting true and 'F' denoting false. He wants to confuse the students by maximizing the number of consecutive questions with the same answer (multiple trues or multiple falses in a row).

You are given a string answerKey, where answerKey[i] is the original answer to the ith question. In addition, you are given an integer k, the maximum number of times you may perform the following operation:

• Change the answer key for any question to 'T' or 'F' (i.e., set answerKey[i] to 'T' or 'F').

Return the maximum number of consecutive 'T's or 'F's in the answer key after performing the operation at most k times.

Example 1:

Input: answerKey = "TTFF", k = 2
Output: 4
Explanation: We can replace both the 'F's with 'T's to make answerKey = "TTTT".
There are four consecutive 'T's.

Example 2:

Input: answerKey = "TFFT", k = 1
Output: 3
Explanation: We can replace the first 'T' with an 'F' to make answerKey = "FFFT".
Alternatively, we can replace the second 'T' with an 'F' to make answerKey = "TFFF".
In both cases, there are three consecutive 'F's.

Example 3:

Input: answerKey = "TTFTTFTT", k = 1
Output: 5
Explanation: We can replace the first 'F' to make answerKey = "TTTTTFTT"
Alternatively, we can replace the second 'F' to make answerKey = "TTFTTTTT". 
In both cases, there are five consecutive 'T's.

Solution: (Sliding Window)

Finding a window with max no true having k false and vice versa

class Solution
{
public:
    int findMaxCons(string s, int k, char val)
    {

        int start = 0;
        int count = 0;

        int maxLen = INT_MIN;

        for (int i = 0; i < s.length(); i++)
        {
            if (s[i] == val)
            {
                count++;
            }

            while (count > k)
            {

                if (s[start] == val)
                {
                    count--;
                }
                start++;
            }

            maxLen = max(maxLen, i - start);
        }
        
        return maxLen+1;
    }

    int maxConsecutiveAnswers(string answerKey, int k)
    {
        
        return max(findMaxCons(answerKey, k, 'T'), findMaxCons(answerKey, k, 'F'));
    }
};

Time Complexity: O(n)

Max Consecutive Ones III

Given a binary array nums and an integer k, return the maximum number of consecutive 1's in the array if you can flip at most k 0's.

Example 1:

Input: nums = [1,1,1,0,0,0,1,1,1,1,0], k = 2
Output: 6
Explanation: [1,1,1,0,0,1,1,1,1,1,1]
Bolded numbers were flipped from 0 to 1. The longest subarray is underlined.

Example 2:

Input: nums = [0,0,1,1,0,0,1,1,1,0,1,1,0,0,0,1,1,1,1], k = 3
Output: 10
Explanation: [0,0,1,1,1,1,1,1,1,1,1,1,0,0,0,1,1,1,1]
Bolded numbers were flipped from 0 to 1. The longest subarray is underlined.

Solution:

class Solution
{
public:
    int longestOnes(vector<int> &nums, int k)
    {

        int start = 0;
        int count = 0;
        int maxLen = INT_MIN;

        for (int i = 0; i < nums.size(); i++)
        {

            if (nums[i] == 0)
            {
                count++;
            }

            while (count > k)
            {
                if (nums[start] == 0)
                {
                    count--;
                }
                start++;
            }
            
            maxLen = max(maxLen, i - start + 1);
        }

        return maxLen;
    }
};