The base sliding window solution came to me pretty quickly, two pointers, track zero count, shrink left when zeros exceed k.
Use a sliding window (two-pointer) technique to maintain a window with at most k zeros, expanding the right pointer and shrinking the left when zeros exceed k. Track the maximum window length seen. For streaming, adapt to a circular buffer or maintain a queue of zero indices to handle unbounded data.
Pro tip: Emphasize that the sliding window is optimal for this problem and that for streaming, you can maintain a deque of zero positions to efficiently shrink the window when needed, achieving O(n) time and O(k) space.
Confirm that the array is binary, k is non-negative, and we want the longest contiguous subarray after flipping at most k zeros. Discuss edge cases like k=0 or all ones.
Initialize left=0, zeros=0, max_len=0. Iterate right from 0 to n-1: if arr[right]==0, increment zeros. While zeros > k, if arr[left]==0 decrement zeros; increment left. Update max_len = max(max_len, right-left+1).
Time complexity is O(n) since each element is visited at most twice. Space complexity is O(1) for the basic approach, as we only use a few variables.
For streaming, we cannot store the entire array. Use a queue (or deque) to store indices of zeros. When zeros exceed k, remove the oldest zero index and set left to that index+1. Maintain the current window length and max length. Space is O(k) for the queue.
Mention that the streaming approach uses O(k) space, which is efficient if k is small. If k is large, consider alternative approaches like maintaining a count of zeros in a sliding window with a fixed-size buffer, but note that the deque method is optimal.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.