This is the classic LIS problem but they wanted the actual indices too, not just the length, which tripped me up.
Use the patience sorting algorithm with binary search to compute the LIS length in O(n log n). Maintain an array of tails and parent pointers to reconstruct one valid subsequence, then convert the indices to 1-based.
Pro tip: Clarify upfront whether the subsequence must be strictly increasing and whether any valid subsequence is acceptable; this shows attention to detail and avoids wasted effort.
Confirm that the subsequence must be strictly increasing, indices are 1-based, and any valid subsequence is acceptable. Discuss handling of empty arrays or duplicates.
Describe patience sorting: maintain an array 'tails' where tails[i] is the smallest tail of an increasing subsequence of length i+1. Use binary search to update tails.
While updating tails, store the index of the previous element in the subsequence for each element. Keep track of the index of the last element of the longest subsequence.
Starting from the last index, follow parent pointers backwards to build the subsequence, then reverse it to get the correct order. Convert indices to 1-based.
State that time complexity is O(n log n) due to binary search, and space complexity is O(n). Walk through a small example to verify correctness.
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Fenwick tree question, pretty standard if you've seen it before.
Start by clarifying the problem constraints and requirements, then propose a Fenwick tree (Binary Indexed Tree) as an efficient solution with O(log n) per operation. Explain the implementation details, including 1-based indexing and update/query logic, and discuss complexity and potential alternatives.
Pro tip: Mention that a Fenwick tree is simpler and more memory-efficient than a segment tree for this specific problem, and briefly note that both achieve O(log n) per operation. This shows you understand trade-offs and can choose the right tool.
Confirm the array size (up to 200,000), operation types (point update, range sum), and expected performance. Ask if updates and queries are interleaved and if there are any memory constraints.
Select a Fenwick tree (BIT) for its simplicity and efficiency, or a segment tree if more complex operations are anticipated. Explain why O(log n) per operation is optimal for this scale.
Describe how the tree is built using 1-based indexing, how point updates propagate by adding to indices i += i & -i, and how prefix sums are computed by subtracting i -= i & -i.
Show that range sum [l, r] = prefix_sum(r) - prefix_sum(l-1), and explain how each prefix sum is computed in O(log n) time.
State time complexity O(log n) per operation and space O(n). Discuss handling of large values (use 64-bit integers), and mention alternative approaches like segment trees or sqrt decomposition.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
I liked this one more than the others, felt more like something you'd actually encounter.
Sort both the release timestamps and the deployment windows, then process the windows in chronological order while maintaining a max-heap of available releases. For each window, push all releases with timestamps ≤ window timestamp into the heap, then pop the largest (most recent) release; if the heap is empty, return -1. This yields O((n+m) log(n+m)) time due to sorting and heap operations.
Pro tip: Clarify whether the releases and windows are already sorted or if sorting is required, and discuss the trade-offs between sorting and using a balanced BST for dynamic insertion. Also, mention that using a max-heap ensures we always pick the most recent available release efficiently.
Restate the problem: for each window, assign the most recent unused release ≤ window timestamp, else -1. Note the required time complexity O((n+m) log(n+m)).
Use sorting for releases and windows, and a max-heap to efficiently retrieve the most recent available release. Alternatively, consider a balanced BST if releases are added dynamically.
Sort windows by timestamp. Iterate through windows, adding all releases with timestamp ≤ current window to the heap. Then pop the max from the heap if available.
If heap is empty, assign -1. Ensure each release is used at most once. Return the list of assignments in the original window order.
Sorting takes O(n log n + m log m), heap operations O((n+m) log n). Overall O((n+m) log(n+m)). Space O(n+m).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.