Solution Explanation

This problem involves optimally placing boxes of varying heights into a warehouse with rooms of varying heights. The key constraint is that boxes can only be placed from left to right, and a box cannot be placed in a room if its height exceeds the room's height. This immediately suggests a greedy approach focusing on efficiently using the available space.

The solution employs a clever preprocessing step to determine the minimum height encountered so far from the left end of the warehouse. This is efficiently calculated in a single pass. Then, the boxes are sorted in ascending order of height. This allows the algorithm to greedily attempt to fit the smallest boxes first.

Algorithm:

1. Preprocessing: Create an array left of the same size as warehouse. left[i] will store the minimum height encountered from warehouse[0] to warehouse[i]. This can be computed in O(n) time.
2. Sort Boxes: Sort the boxes array in ascending order. This ensures that we attempt to place smaller boxes first, maximizing our chances of filling the warehouse. The time complexity of this step is O(m log m), where m is the number of boxes.
3. Greedy Placement: Iterate through the sorted boxes and the left array simultaneously. For each box, find the rightmost room in left that can accommodate it (i.e., where left[j] >= boxes[i]). If such a room is found, place the box and move to the next box. If no suitable room is found, the process stops.

Time Complexity Analysis:

• Preprocessing left array: O(n)
• Sorting boxes: O(m log m)
• Greedy placement: O(min(n,m)) in the worst case because at most min(n, m) iterations are performed.

Therefore, the overall time complexity is O(n + m log m), dominated by the sorting step if m is significantly larger than n. The space complexity is O(n) due to the left array.

Code Explanation (Python):

class Solution:
    def maxBoxesInWarehouse(self, boxes: List[int], warehouse: List[int]) -> int:
        n = len(warehouse)
        left = [warehouse[0]] * n  # Initialize left array with the first warehouse height
        for i in range(1, n):
            left[i] = min(left[i - 1], warehouse[i]) #Calculate minimum height so far from left
        boxes.sort() #Sort boxes for greedy approach
        i, j = 0, n - 1  #Initialize pointers for boxes and warehouse
        while i < len(boxes):
            while j >= 0 and left[j] < boxes[i]: #Find a suitable room
                j -= 1
            if j < 0: #No suitable room found
                break
            i, j = i + 1, j - 1 # Move to next box and previous room
        return i #Return the number of boxes placed

The code in other languages follows the same logic with minor syntactic differences. The core idea remains the same: efficient preprocessing, sorting, and a greedy iterative placement strategy.