1004. Max Consecutive Ones III

This requires counting the number of consecutive ones in the array if you are allowed to have up to k non-zeroes consecutively. We can solve this with a sliding window + counting the number of zeroes. If we have <= k zeroes, then we’re fine — we can count this. However, if there are more than k zeroes, then we increment our left side, incrementing until all of our left side is taken care of.

class Solution:
    def longestOnes(self, nums: List[int], k: int) -> int:
        l, zeroes, max_len = 0, 0, 0
        for r, val in enumerate(nums):
            if val == 0:
                zeroes += 1
                while zeroes > k:
                    if nums[l] == 0:
                        zeroes -= 1
                    l += 1
            max_len = max(max_len, r - l + 1)
        
        return max_len