My first instinct was to just sort the timestamps and do binary search on both ends, which works fine.
Clarify that timestamps are fixed-format and can be converted to seconds since midnight, then preprocess by sorting and building a prefix-count array or using binary search. For each query, use binary search to find the first and last indices within the range and return the difference in counts.
Pro tip: Mention that if the number of events is huge and queries are frequent, you can bucket by second (86400 buckets) to achieve O(1) query time with O(86400) space, but binary search on sorted timestamps is more space-efficient.
Ask about the number of events, number of queries, whether timestamps are unique, and if the range is inclusive. Confirm that timestamps are in HH:MM:SS format and can be converted to seconds.
Convert each timestamp to an integer (seconds since midnight). Decide between sorting the array and using binary search, or using a prefix sum array over the 86400 possible seconds.
If using sorted array: sort the integer timestamps. If using prefix sum: create an array of size 86401 where each index i stores the count of events up to second i.
For sorted array: use binary search to find the leftmost index >= start and rightmost index <= end, then return the difference. For prefix sum: return prefix[end] - prefix[start-1].
Discuss time and space complexity: sorting O(n log n) preprocessing, O(log n) per query; prefix sum O(n + 86400) preprocessing, O(1) per query. Mention memory vs speed trade-off.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Model the problem as a bipartite matching between target characters and cards, where each card can match at most one character and only if the character appears on either side. Use a greedy approach with frequency counts or a max-flow formulation to determine if a perfect matching exists for all target characters.
Pro tip: Clarify constraints upfront (e.g., target length vs. number of cards, character set) and discuss trade-offs between greedy and flow-based solutions; this shows you consider scalability and edge cases.
Confirm that each card can be used at most once and only one side can be chosen per card. Ask about constraints like target length, number of cards, and character set.
Represent each target character as a demand and each card as a supply that can satisfy one demand if the character is on either side. This is a bipartite matching problem.
For small constraints, use backtracking or max-flow. For larger constraints, use a greedy approach with frequency counts: count available cards per character and ensure counts meet target needs, considering cards with two different characters.
Consider cards with identical letters on both sides (only one character available) and cards with two different letters (can satisfy either). Ensure the greedy assignment doesn't starve other characters.
Discuss time and space complexity. For greedy, O(n + m) where n is target length and m is number of cards; for flow, O(V^2 E). Walk through examples and edge cases.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Use a min-heap to efficiently merge the N sorted lists by repeatedly extracting the smallest element from the heap and inserting the next element from the same list. This achieves O(M log N) time where M is the total number of elements, which is optimal for comparison-based merging.
Pro tip: Mention that if N is very large, you can use a tournament tree or divide-and-conquer pairwise merging to reduce heap overhead, and discuss the trade-offs between different approaches.
Ask about the number of lists (N), total elements (M), memory limits, and whether lists are sorted in ascending order. Confirm if the output should be a new list or in-place.
Explain that you will use a min-heap of size N, initially containing the first element of each list along with its list index. Repeatedly extract the minimum, append to result, and insert the next element from the same list.
State that each insertion and extraction takes O(log N), and there are M total elements, so overall time is O(M log N). Space is O(N) for the heap plus O(M) for the output.
Mention divide-and-conquer pairwise merging (O(M log N) time, O(M) space) and tournament trees. Compare with naive concatenation and sorting (O(M log M)).
Address empty lists, N=1, very large N, and memory constraints. Suggest streaming if memory is limited.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.