I found an interesting problem on https://leetcode.com/problems/max-consecutive-ones-iii/:
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

I tried it in Python, but could not do it. Can you solve it?
class Solution:
    def longestOnes(self, A: List[int], K: int) -> int:
        i = 0
        j = 0
        
        numZeros = 0
        
        bestSize = 0
        while j < len(A):
            if A[j] == 1:
                pass
            elif A[j] == 0:
                numZeros += 1
                while numZeros > K:
                    if i >= 0 and A[i] == 0:
                        numZeros -= 1
                    i += 1
            bestSize = max(bestSize, j - i + 1)
            j += 1
        return bestSize