I got the basic O(n) scan pretty fast, track running min and check profit at each step.
Use a single-pass greedy algorithm that tracks the minimum price seen so far and computes the potential profit at each step. Update the maximum profit and record the buy/sell indices when a new maximum is found. Handle edge cases by initializing with no profit and returning (0, -1, -1) if no profitable trade exists.
Pro tip: Clarify tie-breaking rules upfront (e.g., earliest buy day, then earliest sell day) and mention that the algorithm naturally handles ties by only updating on strictly greater profit. This shows attention to detail and avoids ambiguity.
Confirm the expected output format, tie-breaking rules, and edge cases like strictly decreasing arrays or multiple equal profits. This ensures alignment with the interviewer.
Explain that you will iterate through the array once, keeping track of the minimum price and its index, and compute profit at each step. This achieves O(n) time and O(1) space.
Describe the initialization and update logic: initialize min_price to infinity, max_profit to 0, and indices to -1. For each price, if it's lower than min_price, update min_price and its index; else compute profit and update max_profit and indices if profit is greater.
Explain how the algorithm handles strictly decreasing arrays (returns (0, -1, -1)) and ties (by only updating on strictly greater profit, preserving the earliest occurrence).
State that the algorithm runs in O(n) time and O(1) space, and summarize why it's optimal for this problem.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
First, clarify the problem constraints and confirm that BFS is appropriate for unweighted graphs. Then, describe a modified BFS that processes neighbors in sorted order without sorting at each step, using pre-sorted adjacency lists or a priority queue. Finally, explain how to reconstruct the lexicographically smallest shortest path by tracking predecessors and making greedy choices.
Pro tip: Mention that lexicographic order is determined by the sequence of node labels, so at each BFS level, you should explore neighbors in increasing order to ensure the first time you reach a node, it's via the lexicographically smallest path. This avoids the need for post-processing.
Restate the problem: find the shortest path from source to target avoiding forbidden nodes, and among all shortest paths, choose the lexicographically smallest. Confirm that the graph is unweighted and directed, with up to 100,000 nodes.
Remove forbidden nodes from consideration. Pre-sort the adjacency list of each node in increasing order of neighbor labels. This takes O(M log M) total, but since M can be up to 100,000, it's acceptable. Alternatively, if the graph is given with unsorted adjacency lists, you can sort each list once.
Perform BFS from the source. When expanding a node, iterate through its pre-sorted neighbors. For each unvisited neighbor, set its distance and predecessor. Because neighbors are processed in sorted order, the first time a node is visited, it is via the lexicographically smallest path among shortest paths.
Once BFS completes, if the target was reached, backtrack from target to source using the predecessor pointers to obtain the path. This path is guaranteed to be the lexicographically smallest shortest path.
The BFS itself runs in O(N+M) time. Pre-sorting adjacency lists takes O(M log M) time, which is dominated by O(N+M) if M is O(N). Space is O(N+M) for storing the graph and BFS data structures.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Sort by start, then sweep and merge when the current interval's start is less than or equal to the running end.
Start by clarifying the half-open interval semantics and edge cases, then propose sorting intervals by start and merging in a single pass. Discuss overflow-safe comparisons and prove the measure property using a sweep-line argument, and finally outline a comprehensive test suite including boundary and negative values.
Pro tip: Emphasize that half-open intervals eliminate ambiguity at boundaries and that using 64-bit arithmetic for length calculations prevents overflow even with 32-bit inputs. Mention that the measure property is a direct consequence of disjointness after merging, which is a key invariant to verify in tests.
Confirm that intervals are half-open [l, r), that touching intervals (e.g., [1,2) and [2,3)) should merge, and that inputs may include negative values and 32-bit boundaries. Discuss potential overflow when computing lengths or comparing endpoints.
Sort intervals by start (and then by end). Iterate through sorted intervals, merging with the last merged interval if they overlap or touch (i.e., if current.start <= last.end). Use 64-bit integers for safe comparisons and length calculations.
After merging, intervals are disjoint and sorted. The total covered measure is the sum of lengths of merged intervals because the union is a disjoint union. Use induction or a sweep-line argument to show that no measure is lost or double-counted.
Write code with careful handling of empty input, single interval, all overlapping, none overlapping, touching intervals, negative values, and intervals at 32-bit boundaries (e.g., [-2^31, 2^31-1)). Include tests that verify the measure property by comparing the sum of merged lengths to a brute-force union measure.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.