Use a sliding window (two-pointer) technique to efficiently find the maximum number of timestamps within any inclusive interval of the given window size. Initialize two pointers at the start, expand the right pointer to include timestamps within the window, and shrink from the left when the window is exceeded, keeping track of the maximum count. This yields O(n) time and O(1) extra space.
Pro tip: Clarify the inclusivity of the interval and whether the window size is inclusive or exclusive; explicitly state your assumption to avoid off-by-one errors. Also, mention that a binary search approach (O(n log n)) is possible but the sliding window is optimal for sorted arrays.
Confirm that the array is sorted, the window size is inclusive, and that timestamps are integers or comparable. Ask if the window size is given in the same units as timestamps.
Select the sliding window (two-pointer) approach for O(n) time. Explain why it works: since the array is sorted, any valid interval is contiguous, and the window can be adjusted monotonically.
Trace the algorithm on a small example (e.g., timestamps [1,2,3,5,8], window=3) to demonstrate how the pointers move and how the maximum is updated.
State that the time complexity is O(n) because each element is visited at most twice, and space complexity is O(1) beyond the input.
Discuss edge cases: empty array, window size larger than the range, duplicate timestamps, and negative timestamps. Explain how the algorithm handles them.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Model the process as each '1' moving right past '0's, where the time for a '1' to reach its final position depends on the number of '0's to its left and the time for the previous '1'. Compute the maximum time over all '1's using a single pass with a counter for zeros and a variable for the previous time.
Pro tip: Explain that the answer is the maximum over all '1's of (zeros seen so far + 1) but capped by the previous '1's time plus 1, which elegantly handles blocking. This shows you understand the dynamics beyond brute-force simulation.
Restate the problem: each second, all '01' pairs swap to '10'. We need the number of seconds until no '01' remains. Clarify that swaps happen simultaneously.
Recognize that each '1' moves right past '0's, and its movement is constrained by the '1' ahead of it. The time for a '1' to settle is the number of '0's to its left, but it cannot exceed the previous '1's time plus 1.
For each '1' at position i, let zeros be the count of '0's seen so far. The time for this '1' is max(zeros, prev_time + 1). Update prev_time and continue.
Write a single-pass algorithm: initialize zeros=0, prev_time=0, max_time=0. For each character: if '0', zeros++; if '1', prev_time = max(zeros, prev_time+1); max_time = max(max_time, prev_time). Return max_time.
Time complexity O(n), space O(1). Mention that brute-force simulation would be O(n^2) in worst case, so this is optimal.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.