Solution Explanation: Put Boxes Into the Warehouse II

This problem involves optimally placing boxes of varying heights into a warehouse with rooms of varying heights, aiming to maximize the number of boxes placed. The key constraint is that boxes cannot be stacked, and they can be inserted from either the left or right side of the warehouse.

Approach:

The solution employs a greedy approach combined with preprocessing and sorting. Here's a breakdown:

1. Preprocessing: The warehouse's height array is preprocessed to determine the minimum height constraint for each room. For each room, we consider the minimum height available by considering both left and right constraints. This means for each room, we determine the minimum of:

   • Its own height
   • The minimum height of all rooms to its left
   • The minimum height of all rooms to its right

2. Sorting: Both the boxes array (box heights) and the modified warehouse array (minimum room heights) are sorted in ascending order. This allows us to efficiently find the best placement for boxes in a greedy manner.

3. Greedy Placement: We iterate through the sorted boxes array. For each box, we iterate through the sorted warehouse array, searching for the first room with a minimum height greater than or equal to the box's height. If a suitable room is found, the box is placed, and we increment the count of placed boxes. If no suitable room is found, the box cannot be placed.

Time Complexity Analysis:

• Preprocessing: The preprocessing step takes O(n) time, where n is the number of rooms in the warehouse.
• Sorting: Both sorting operations (for boxes and warehouse) take O(n log n) time.
• Greedy Placement: The nested loops in the greedy placement step have a combined time complexity of O(n) in the worst-case scenario.

Therefore, the overall time complexity is dominated by the sorting step, resulting in O(n log n).

Space Complexity Analysis:

The solution uses auxiliary arrays (left, right) for preprocessing, which have a size of O(n). The space complexity is therefore O(n).

Code Implementation (Python):

class Solution:
    def maxBoxesInWarehouse(self, boxes: List[int], warehouse: List[int]) -> int:
        n = len(warehouse)
        left = [0] * n
        right = [0] * n
        left[0] = float('inf')  # Initialize with infinity
        right[-1] = float('inf') # Initialize with infinity
        for i in range(1, n):
            left[i] = min(left[i - 1], warehouse[i - 1])
        for i in range(n - 2, -1, -1):
            right[i] = min(right[i + 1], warehouse[i + 1])
        for i in range(n):
            warehouse[i] = min(warehouse[i], max(left[i], right[i]))
        boxes.sort()
        warehouse.sort()
        ans = i = 0
        for x in boxes:
            while i < n and warehouse[i] < x:
                i += 1
            if i == n:
                break
            ans, i = ans + 1, i + 1
        return ans
 

The code in other languages (Java, C++, Go) follows a very similar structure, reflecting the same algorithmic approach and complexity analysis. The key differences are in the syntax and library functions used for sorting and handling infinity values.