← Two Sigma Interview Insights
The greedy angle clicked pretty fast: always pick the highest-profit project you can currently afford.
Use a min-heap to track feasible projects by required capital and a max-heap to select the most profitable among them. Iterate up to k times, each time moving all projects with capital requirement ≤ current capital into the max-heap, then adding the top profit to capital. This greedy strategy maximizes capital at each step.
Pro tip: Emphasize that the greedy choice is optimal because selecting the highest-profit feasible project never reduces future feasibility, and mention that the two-heap approach is a classic pattern for this problem (LeetCode 502).
Restate the problem: given k, initial capital W, and arrays Capital and Profits, return the maximum capital after at most k projects. Clarify that each project can be done at most once and that capital is cumulative.
Use a min-heap to store projects by required capital (to efficiently find feasible ones) and a max-heap to store feasible projects by profit (to pick the most profitable). Justify: min-heap gives O(log n) insertion and O(1) peek for feasibility; max-heap gives O(log n) insertion and O(1) peek for max profit.
Sort projects by capital requirement or push all into min-heap. For up to k iterations: move all projects from min-heap with capital ≤ current capital into max-heap; if max-heap is empty, break; else pop max profit, add to capital.
Time: O(n log n + k log n) where n is number of projects. Space: O(n) for the heaps. Explain that each project is inserted and removed at most once from each heap.
Argue that at each step, choosing the feasible project with maximum profit is optimal because it maximizes capital, which can only increase the set of feasible projects for future steps. This is a standard exchange argument.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Model the problem as a shortest-path search on a time-expanded graph where each state is (row, col, time), and use BFS to find the earliest arrival at the exit. At each step, check that the destination cell's flood time is strictly greater than the arrival time, and that the start cell is not already flooded at time 0.
Pro tip: Clarify edge cases upfront: if the start or exit is flooded at time 0, return -1 immediately; also mention that BFS is optimal because each move costs exactly one minute, and discuss how to handle large grids with a visited set keyed by (row, col, time) or by pruning dominated states.
Confirm grid dimensions, movement rules (4-directional?), and that flood time is the minute a cell becomes impassable. Check if start or exit floods at time 0 and handle immediately.
State is (row, col, time). From a state, move to adjacent open cells if the arrival time (time+1) is strictly less than the destination's flood time. Also ensure the current cell is not flooded at the current time.
Use BFS because each move takes exactly one minute and we want the minimum time. BFS explores states in increasing time order, guaranteeing the first time we reach the exit is optimal.
Maintain a visited set of (row, col, time) or a 3D boolean array to avoid revisiting states. Return the time when the exit is dequeued or reached. If BFS exhausts, return -1.
Time complexity is O(R*C*T) where T is the maximum possible time (bounded by R*C). Space is similar. Mention potential optimizations like bidirectional BFS or A* with a heuristic, and trade-offs.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.