Put Boxes Into the Warehouse I

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ArraysGreedy AlgorithmsTwo Pointers

You have n boxes with different colors labeled from 0 to n - 1. You are also given m warehouses.

Each warehouse i has some free space warehouse[i]. Boxes can be placed in the warehouse one by one only if there is enough space.

You are given two integer arrays boxes and warehouse of positive integers. boxes[i] is the size of the i^th box, and warehouse[j] is the available space of the j^th warehouse.

A box can only be put into a warehouse if the size of the box is less than or equal to the available space of the warehouse.

Your task is to put as many boxes as you can into the warehouses. Return the maximum number of boxes that can be placed into the warehouses.

Example 1:

Input: boxes = [1,2,2,3,4], warehouse = [3,4,1,2]
Output: 4
Explanation:
We can place the boxes of sizes 1, 2, 2, and 3 into the warehouses.
We cannot place the box of size 4 into any warehouse.

Example 2:

Input: boxes = [4,5,6], warehouse = [3,3,3]
Output: 0
Explanation:
We cannot place any of the boxes into the warehouses.

Constraints:

n == boxes.length
m == warehouse.length
1 <= n, m <= 10⁵
1 <= boxes[i], warehouse[j] <= 10⁵

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the size constraints for the boxes and warehouse arrays? Specifically, what is the maximum number of boxes and warehouses?
Can the dimensions (length) of the boxes or the warehouse compartments be zero or negative?
If a box cannot fit into any warehouse compartment, what should the function return?
Are the boxes and warehouse compartments already sorted in any way, or do I need to sort them myself?
If multiple placements of boxes into warehouses are possible, is any valid placement acceptable, or is there a specific optimization criteria, such as minimizing wasted space?

Brute Force Solution

Approach

The brute force method for putting boxes into the warehouse involves trying every possible combination of assigning boxes to warehouse sections. It's like testing out every single way you could potentially arrange the boxes and seeing if it works.

Here's how the algorithm would work step-by-step:

Start by trying to put all the boxes into the first warehouse section.
If that doesn't work because the boxes are too big, try putting all but the last box in the first section and the last box in the second section.
Keep shifting the boundary, putting more and more boxes into the second, third, etc. sections, and leaving less in the first.
When you've exhausted all combinations starting with boxes allocated to the first section, try combinations starting with no boxes in the first section.
Repeat this process, going through absolutely every single way to assign boxes to warehouse sections.
For each arrangement, check if all the boxes fit within the sizes of the warehouse sections assigned to them.
Finally, once you've tested all possible arrangements, choose the one that allows you to put the most boxes into the warehouse.

Code Implementation

def put_boxes_into_warehouse_brute_force(
    boxes, warehouse_sections
):
    max_boxes_placed = 0

    # Iterate through all possible combinations of box assignments
    for i in range(1 << (len(boxes) + len(warehouse_sections) - 1)):
        box_placements = []
        warehouse_section_index = 0
        current_placement = []
        
        for box_index in range(len(boxes)):
            current_placement.append(boxes[box_index])

            # Determine when to move to the next warehouse section
            if box_index < len(boxes) - 1 and (
                (i >> (box_index + warehouse_section_index)) & 1
            ):
                box_placements.append(
                    (warehouse_sections[warehouse_section_index], current_placement)
                )
                warehouse_section_index += 1
                current_placement = []
        
        # Add the last section and the remaining boxes
        box_placements.append(
            (warehouse_sections[warehouse_section_index], current_placement)
        )

        if len(box_placements) > len(warehouse_sections):
            continue

        boxes_placed = 0
        valid_placement = True

        # Validate if boxes can fit in allocated warehouse sections
        for warehouse_index, (warehouse_section_size, allocated_boxes) in enumerate(box_placements):
            if len(allocated_boxes) > 0:
                if sum(allocated_boxes) > warehouse_section_size:
                    valid_placement = False
                    break

        # Count boxes if placement is valid
        if valid_placement:
            boxes_placed = len(boxes)

        # Track maximum number of boxes placed
        if valid_placement:
            max_boxes_placed = max(max_boxes_placed, boxes_placed)

    return max_boxes_placed

Big(O) Analysis

Time Complexity

O(2^(n+m)) – The brute force approach explores all possible combinations of assigning n boxes to m warehouse sections. For each box, we have m+1 choices: assign it to one of the m warehouse sections or leave it unassigned. This leads to (m+1)^n possible combinations. In the worst case, m is comparable to n, so (n+1)^n grows exponentially. Further, for each arrangement, checking if the boxes fit requires iterating through all n boxes and potentially m warehouse sections, adding another O(n*m) factor which doesn't change the exponential dominance. Hence, the time complexity is approximately O(2^(n+m)).

Space Complexity

O(1) – The brute force approach, as described, does not create any significant auxiliary data structures. It iterates through possible arrangements potentially using index variables to track box and warehouse section assignments. These variables occupy constant space, irrespective of the number of boxes or warehouse sections. Thus, the auxiliary space complexity is O(1).

Optimal Solution

Approach

The best way to pack boxes into a warehouse with size restrictions is to prioritize using the smallest boxes first and placing them in the smallest warehouses that can fit them. This prevents larger boxes from blocking access to smaller spaces. In essence, efficiently match the smallest box with the smallest available warehouse.

Here's how the algorithm would work step-by-step:

First, arrange the boxes from smallest to largest and the warehouses from smallest to largest.
Then, go through each box and try to fit it into the smallest warehouse available that's big enough for the box.
If a box fits, mark that warehouse as used and move to the next box.
If a box doesn't fit in any of the remaining warehouses, it means it cannot be stored.
Repeat until all boxes have been considered or there are no more warehouses available.
The goal is to maximize the number of boxes that can be stored by using the smallest possible space for each box.

Code Implementation

def put_boxes_into_the_warehouse_i(
    boxes_array, warehouse_array
):
    boxes_array.sort()
    warehouse_array.sort()
    box_index = 0
    warehouse_index = 0
    boxes_stored = 0

    # Iterate through boxes and warehouses, matching smallest to smallest.
    while box_index < len(boxes_array) and warehouse_index < len(
        warehouse_array
    ):
        if boxes_array[box_index] <= warehouse_array[warehouse_index]:
            boxes_stored += 1

            # Increment box index since it fit.
            box_index += 1

        # Always increment warehouse index to find the next available.
        warehouse_index += 1

    return boxes_stored

Big(O) Analysis

Time Complexity

O(n log n) – Sorting both the boxes and warehouses initially takes O(n log n) time, where n is the number of boxes or warehouses (assuming they are of similar magnitude). After sorting, we iterate through each box trying to find a suitable warehouse. The nested loop checks each warehouse for each box in the worst case. However, because the warehouses are also sorted and we mark warehouses as used, each warehouse is only checked a maximum of once across all boxes. Thus, the overall time complexity is dominated by the initial sorting step, resulting in O(n log n).

Space Complexity

O(1) – The algorithm sorts the boxes and warehouses in place and uses only a few integer variables for indexing during the matching process. No additional data structures that scale with the input size are created, such as new arrays or hash maps to store intermediate results. Therefore, the auxiliary space required is constant, independent of the number of boxes or warehouses.

Edge Cases

Case	How to Handle
Empty boxes or warehouse arrays	Return 0 because no boxes can be placed if either the boxes or warehouse is empty.
boxes array is larger than warehouse array	Iterate only through the min(boxes.length, warehouse.length) to avoid index out of bounds.
Warehouse array with all identical values	The algorithm should still work correctly as it iterates and compares; no special handling is needed.
Boxes array with all identical values	The algorithm should still work correctly; the number of placements depends on if the warehouse array has large enough elements.
Warehouse array is non-increasing.	Sort warehouse array and boxes array to find optimal number of fits.
Boxes array has larger values than any value in warehouse array	No box can fit, so the algorithm should return 0 after iterating through both arrays.
Integer overflow if using sum of sizes of boxes.	Avoid calculating the sum of sizes of boxes and instead focus on comparisons.
Very large arrays for both boxes and warehouse exceeding memory constraints	Consider using a streaming approach or external sorting if the data cannot fit into memory.

Empty boxes or warehouse arrays

How to Handle:

Return 0 because no boxes can be placed if either the boxes or warehouse is empty.

boxes array is larger than warehouse array

How to Handle:

Iterate only through the min(boxes.length, warehouse.length) to avoid index out of bounds.

Warehouse array with all identical values

How to Handle:

The algorithm should still work correctly as it iterates and compares; no special handling is needed.

Boxes array with all identical values

How to Handle:

The algorithm should still work correctly; the number of placements depends on if the warehouse array has large enough elements.

Warehouse array is non-increasing.

How to Handle:

Sort warehouse array and boxes array to find optimal number of fits.

Boxes array has larger values than any value in warehouse array

How to Handle:

No box can fit, so the algorithm should return 0 after iterating through both arrays.

Integer overflow if using sum of sizes of boxes.

How to Handle:

Avoid calculating the sum of sizes of boxes and instead focus on comparisons.

Very large arrays for both boxes and warehouse exceeding memory constraints

How to Handle:

Consider using a streaming approach or external sorting if the data cannot fit into memory.