We use a sliding window of which the zeroes count is at most k. We expand the right pointer and shrink the left pointer as needed.

1. l = 0, zeros = 0, res = 0.
2. For each r:
   • If nums[r] == 0, zeros += 1.
   • While zeros > k:
     • If nums[l] == 0, zeros -= 1.
     • l += 1.
   • res = max(res, r - l + 1).
3. Return res.

• Time Complexity: O(N).
• Space Complexity: O(1).

def longest_ones(nums, k):
    l = 0
    zeros = 0
    res = 0
    for r in range(len(nums)):
        if nums[r] == 0:
            zeros += 1
        while zeros > k:
            if nums[l] == 0:
                zeros -= 1
            l += 1
        res = max(res, r - l + 1)
    return res