Classic sliding window with a zero-count tracker.
Use a sliding window (two-pointer) technique to find the longest subarray with at most k zeros. Expand the right pointer to include elements, and when the number of zeros exceeds k, shrink the window from the left until the condition is satisfied. Keep track of the maximum window length seen.
Pro tip: Clarify that the problem is equivalent to finding the longest subarray with at most k zeros, and mention that the sliding window approach runs in O(n) time and O(1) space, which is optimal. Also, discuss edge cases like k >= number of zeros or empty array.
Restate the problem: given a binary array and integer k, find the maximum length of a contiguous subarray that can be made all 1s by flipping at most k zeros. Confirm that flipping zeros to ones is equivalent to allowing at most k zeros in the subarray.
Recognize that this is a classic sliding window problem. Explain that a brute-force solution would be O(n^2) or O(n^3), but a two-pointer sliding window achieves O(n) time.
Initialize left and right pointers at 0, and a zero_count to track zeros in the current window. Expand right, increment zero_count if the new element is 0. While zero_count > k, move left forward and decrement zero_count if the element at left is 0. Update max_length at each step.
State that time complexity is O(n) because each element is visited at most twice, and space complexity is O(1). Discuss edge cases: k >= total zeros (return array length), empty array (return 0), all ones (return array length).
Walk through a small example, e.g., [1,0,1,0,1] with k=1, to demonstrate the algorithm and verify the output (expected 3).
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