Given an array A of 0s and 1s, we may change up to K values from 0 to 1.
Return the length of the longest (contiguous) subarray that contains only 1s. 
Example 1:
Input: A = [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: A = [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.
Note:
1 <= A.length <= 20000
0 <= K <= A.length
A[i] is 0 or 1
class Solution:
    def longestOnes(self, A: List[int], K: int) -> int:
        longest = start = end = zeroes = 0
        while end < len(A):
            num = A[end]

            if not num:
                zeroes += 1
            end += 1
            while zeroes > K:
                if A[start] == 0:
                    zeroes -= 1
                start += 1
            longest = max(longest, end - start)
        return longest 

class Solution:
    def longestOnes(self, A: List[int], K: int) -> int:
        longest = start = end = zeroes = 0
        while end < len(A):
            num = A[end]

            if not num:
                zeroes += 1
            end += 1
            while zeroes > K:
                if A[start] == 0:
                    zeroes -= 1
                start += 1
            longest = max(longest, end - start)
        return longest 
