The sliding window click took me a second longer than I'd like to admit.
Use a sliding window (two-pointer) technique to maintain a window that contains at most k zeros. Expand the right pointer to include new elements, and when the number of zeros exceeds k, shrink the window from the left until the zero count is within k. Track the maximum window length throughout.
Pro tip: Clarify that the array is binary and that flipping zeros to ones is equivalent to finding the longest subarray with at most k zeros. Mention that the solution runs in O(n) time and O(1) space, which is optimal.
Restate the problem: find the longest contiguous subarray that contains at most k zeros, since flipping those zeros yields consecutive 1s. Confirm that the array is binary and k is non-negative.
Explain that a sliding window with two pointers (left and right) efficiently tracks a window with at most k zeros. Initialize left = 0, zero_count = 0, and max_length = 0.
Iterate right from 0 to n-1: if the current element is 0, increment zero_count. While zero_count > k, if the element at left is 0, decrement zero_count, then increment left. Update max_length with the current window size (right - left + 1).
After the loop, max_length holds the maximum number of consecutive 1s achievable by flipping at most k zeros. Return max_length.
State that the time complexity is O(n) because each element is visited at most twice (once by right, once by left), and space complexity is O(1) as only a few variables are used.
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