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Had actually drilled this one before the call so it went fine.
First, clarify the problem constraints and edge cases, then propose a greedy strategy: sort boxes in descending order and preprocess the warehouse to compute the effective height at each position (the minimum height from the left up to that position). Iterate through the warehouse from right to left, placing the largest possible box at each step, and count how many boxes can be placed.
Pro tip: Mention that the greedy approach works because placing larger boxes first maximizes the remaining space, and the preprocessing step ensures we respect the non-increasing height constraint. Also, discuss time complexity: O(n log n + m) where n is number of boxes and m is warehouse length.
Restate the problem: given boxes and warehouse heights, determine the maximum number of boxes that can be placed from left to right without exceeding the height at any position. Clarify that boxes can only be placed if their height is ≤ the effective height at that position, and the effective height is the minimum of all heights to the left.
Compute an array of effective heights by taking the running minimum from left to right. This ensures that at any position i, the height is the minimum of warehouse[0..i], reflecting the constraint that boxes cannot be placed if they exceed any previous height.
Sort the boxes in descending order. Then iterate through the warehouse from right to left, placing the largest box that fits at each position. If a box fits, increment the count and move to the next box; otherwise, move to the next position.
Discuss time complexity: sorting boxes takes O(n log n), preprocessing warehouse takes O(m), and the greedy placement takes O(n + m). Space complexity is O(m) for the effective heights array. Consider edge cases: empty boxes, empty warehouse, boxes taller than any warehouse position.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Recognize that maximizing total height is a weighted interval scheduling problem where each box has a value equal to its height. Adapt the DP recurrence to maximize sum of heights instead of count, and discuss trade-offs like time/space complexity and potential optimizations.
Pro tip: Explicitly state that the problem is a variation of the classic 'box stacking' or 'weighted interval scheduling' problem, and show how a small change in the DP state transition yields the solution. This demonstrates pattern recognition and adaptability.
Confirm that boxes have dimensions (length, width, height) and can be rotated, and that stacking requires strictly smaller base dimensions. Also confirm that 'total height' means the sum of heights of all boxes in the stack.
Define dp[i] as the maximum total height of a stack with box i at the bottom (or top). The recurrence: dp[i] = height[i] + max(dp[j]) for all j that can be placed on i.
Sort boxes by base area (or one dimension) to ensure we only consider valid stacking orders. Iterate through boxes, computing dp[i] using previously computed dp[j] for smaller boxes.
Generate all possible orientations of each box (up to 3 unique ones) and treat each as a separate item. Ensure that when comparing, the base dimensions are strictly smaller.
The answer is the maximum value in dp. Discuss time complexity O(n^2) and space O(n), and mention possible optimizations like using a segment tree for O(n log n) if needed.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.