Put Boxes Into the Warehouse II

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ArraysGreedy Algorithms

You are given two arrays of integers, boxes and warehouse, representing the height of some boxes of unit width and the heights of the available spaces in a warehouse. The warehouse's width is the length of the warehouse array.

You are asked to put the boxes into the warehouse such that each box is placed vertically into a space of the warehouse. Boxes can be rearranged arbitrarily.

If the height of a box is greater than the height of a space, that box cannot be placed in the warehouse.
You can reorder the boxes and the warehouse.
You can only put one box in each available space of the warehouse.

Return the maximum number of boxes that can be placed in the warehouse.

Note: Since the warehouse may have more spaces than the number of boxes, for an optimal solution, you should rearrange the spaces such that the first spaces in the warehouse have the lowest height.

Example 1:

Input: boxes = [4,3,4,1], warehouse = [5,3,3,4,1]
Output: 3
Explanation: 
We can first reorder the warehouse to [1,3,3,4,5]. Then we can place the boxes as follows:
boxes[3] -> warehouse[0]
boxes[1] -> warehouse[1]
boxes[0] -> warehouse[3]

Example 2:

Input: boxes = [1,2,2,3,4], warehouse = [3,4,1,2]
Output: 4
Explanation: 
We can first reorder the warehouse to [1,2,3,4]. Then we can place the boxes as follows:
boxes[0] -> warehouse[0]
boxes[1] -> warehouse[1]
boxes[2] -> warehouse[2]
boxes[3] -> warehouse[3]

Example 3:

Input: boxes = [1,2,3], warehouse = [1,2,3,4]
Output: 3
Explanation: 
We can place all the boxes in the warehouse.

Constraints:

n == boxes.length
m == warehouse.length
1 <= n <= m <= 10⁵
1 <= boxes[i], warehouse[i] <= 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 `warehouses` arrays?
Can box or warehouse sizes be zero, negative, or non-integer?
If a box cannot fit into any warehouse, what should the function return?
Are the `boxes` and `warehouses` arrays guaranteed to be sorted, and if not, can I sort them in-place?
Is there a specific order or format expected for returning the result (e.g., number of boxes placed, list of warehouse assignments)?
If multiple possible assignments exist, is any valid assignment acceptable?

Brute Force Solution

Approach

The brute force method to fit boxes into warehouses involves testing every possible combination of boxes and warehouse pairings. We essentially try placing each box in every possible warehouse and see if it fits, repeating this process for all boxes until we find a valid arrangement or exhaust all possibilities. This method guarantees finding a solution if one exists, but it can be quite inefficient.

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

Consider all the ways to order the boxes.
For each box ordering, start by trying to fit the first box into the first warehouse.
If the box fits, try to fit the next box into the same warehouse or the next warehouse.
If the box does not fit, try the next warehouse.
Continue trying to fit boxes into warehouses in all possible combinations until all boxes are placed or no warehouse can fit the current box.
If all boxes are placed, record this successful arrangement.
Repeat the process for every possible order of boxes and every possible combination of warehouse assignments.
Finally, among all the successful arrangements, select the one that uses the fewest warehouses.

Code Implementation

def put_boxes_into_warehouse_brute_force(boxes, warehouses):
    from itertools import permutations

    best_arrangement = None

    for box_permutation in permutations(boxes):
        for i in range(2**(len(boxes) - 1) if len(boxes) > 0 else 1):
            warehouse_arrangement = []
            current_warehouse = []
            warehouse_index = 0
            binary_representation = bin(i)[2:].zfill(len(boxes) - 1)

            for box_index in range(len(box_permutation)):
                current_warehouse.append(box_permutation[box_index])

                if box_index < len(box_permutation) - 1 and binary_representation[box_index] == '1':
                    warehouse_arrangement.append(current_warehouse)
                    current_warehouse = []
                    warehouse_index += 1

            warehouse_arrangement.append(current_warehouse)

            # Validate if boxes fit the warehouses. 
            valid_arrangement = True

            for warehouse_index, boxes_in_warehouse in enumerate(warehouse_arrangement):
                warehouse_size = warehouses[warehouse_index] if warehouse_index < len(warehouses) else 0

                for box_size in boxes_in_warehouse:
                    if box_size > warehouse_size:
                        valid_arrangement = False
                        break

                if not valid_arrangement:
                    break

            if valid_arrangement:
                # Compare the number of used warehouses.
                num_warehouses_used = len(warehouse_arrangement)
                if best_arrangement is None or num_warehouses_used < len(best_arrangement):
                    best_arrangement = warehouse_arrangement

    if best_arrangement:
        return len(best_arrangement)
    else:
        return -1

Big(O) Analysis

Time Complexity

O(n! * m^n) – The provided brute force solution considers all possible orderings of the n boxes, which takes O(n!) time. For each ordering of boxes, we try to fit each box into the m warehouses. In the worst case, for each of the n boxes, we might try every warehouse before finding one that fits (or determining it doesn't fit). This results in m options for the placement of each box, leading to m^n possible warehouse assignments for a fixed box ordering. Therefore, the overall time complexity is approximately O(n! * m^n) as we explore all box permutations and warehouse placements for each permutation.

Space Complexity

O(N!) – The brute force approach considers all permutations of boxes, which requires storing the current permutation of boxes being tested. Generating all permutations of N boxes can take O(N!) space in the worst case if we explicitly store each permutation, though more efficient algorithms may generate them iteratively with lower space requirements. Furthermore, storing the successful arrangement that uses the fewest warehouses requires space proportional to the number of boxes, at most N. Dominating the space complexity is the generation and potentially storing of the box permutations, leading to O(N!).

Optimal Solution

Approach

The problem is about fitting boxes of different sizes into warehouses of different sizes to maximize the number of boxes you can fit. The core idea is to sort both boxes and warehouses and then try to match them in a clever way using two pointers to avoid unnecessary comparisons.

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

First, sort the box sizes in ascending order (smallest to largest) and the warehouse sizes in ascending order too.
Use two pointers, one to track the current box and the other to track the current warehouse. Start from the largest warehouse since smaller boxes are more likely to fit into a larger warehouse.
For each warehouse, iterate through the boxes from the smallest to largest and see if the current box fits in the current warehouse. If it does, put the box in the warehouse and move to the next smallest box and next largest warehouse.
If the current box does not fit into the current warehouse, skip to the next smaller warehouse to see if it fits there.
Continue this process until you've considered all the boxes or warehouses. The number of boxes you've placed into warehouses is your answer.

Code Implementation

def put_boxes_into_warehouse_ii(boxes, warehouses):
    boxes.sort()
    warehouses.sort()

    box_index = 0
    warehouse_index = len(warehouses) - 1
    boxes_placed = 0

    # Iterate from the largest warehouse to smallest.
    while warehouse_index >= 0 and box_index < len(boxes):
        # Find if the current smallest box fits.
        if boxes[box_index] <= warehouses[warehouse_index]:
            boxes_placed += 1
            box_index += 1
            warehouse_index -= 1
        else:
            # Go to the next smallest warehouse.
            warehouse_index -= 1

    return boxes_placed

Big(O) Analysis

Time Complexity

O(n log n) – The dominant operation in this solution is the sorting of both the boxes and the warehouses. Assuming we use an efficient sorting algorithm like merge sort or quicksort, this takes O(n log n) time for each array, where n is the number of boxes and m is the number of warehouses. The subsequent two-pointer iteration comparing boxes to warehouses takes O(n+m) in the worst case, where n is the number of boxes and m is the number of warehouses. Since sorting dominates the complexity, and assuming that n and m are roughly the same size so that n+m is O(n) and m log m is O(n log n), the overall time complexity is O(n log n) + O(m log m) + O(n+m), which simplifies to O(n log n).

Space Complexity

O(1) – The algorithm primarily uses two pointers to iterate through the sorted box and warehouse lists. It sorts the input lists in place, which means no auxiliary data structures of significant size are created. The space used for the pointers is constant and does not depend on the number of boxes or warehouses. Therefore, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
Empty boxes or warehouses array	Return 0 immediately as no boxes can be placed in an empty warehouse.
Boxes array contains very large box sizes	Ensure the data type used for box and warehouse sizes (e.g., int, long) can accommodate the maximum possible size to prevent overflow or incorrect comparisons.
Warehouses array contains very small warehouse sizes	The solution should correctly handle warehouses that are too small to fit any box, likely resulting in those warehouses being skipped during the matching process.
No warehouse is big enough for any box	The algorithm should return 0 since no boxes can be placed.
All boxes are identical in size	Sorting can be skipped if all warehouse sizes are the same to avoid unnecessary operations.
All warehouses are identical in size	Sorting of boxes and assigning to each warehouse sequentially can address this case.
Boxes are already sorted (best case scenario)	The algorithm should be efficient even if the boxes or warehouses are already sorted, avoiding unnecessary sorting operations if possible, or taking advantage of it.
Integer overflow during calculations of space remaining or comparison of large sizes	Use appropriate data types (e.g., long) to avoid integer overflow during calculations when dealing with large box or warehouse sizes.

Empty boxes or warehouses array

How to Handle:

Return 0 immediately as no boxes can be placed in an empty warehouse.

Boxes array contains very large box sizes

How to Handle:

Ensure the data type used for box and warehouse sizes (e.g., int, long) can accommodate the maximum possible size to prevent overflow or incorrect comparisons.

Warehouses array contains very small warehouse sizes

How to Handle:

The solution should correctly handle warehouses that are too small to fit any box, likely resulting in those warehouses being skipped during the matching process.

No warehouse is big enough for any box

How to Handle:

The algorithm should return 0 since no boxes can be placed.

All boxes are identical in size

How to Handle:

Sorting can be skipped if all warehouse sizes are the same to avoid unnecessary operations.

All warehouses are identical in size

How to Handle:

Sorting of boxes and assigning to each warehouse sequentially can address this case.

Boxes are already sorted (best case scenario)

How to Handle:

The algorithm should be efficient even if the boxes or warehouses are already sorted, avoiding unnecessary sorting operations if possible, or taking advantage of it.

Integer overflow during calculations of space remaining or comparison of large sizes

How to Handle:

Use appropriate data types (e.g., long) to avoid integer overflow during calculations when dealing with large box or warehouse sizes.