Minimum Space Wasted From Packaging

Hard

Two Sigma

2 views

Topics:

ArraysBinary SearchGreedy Algorithms

You have n packages that you are trying to place in boxes, one package in each box. There are m suppliers that each produce boxes of different sizes (with infinite supply). A package can be placed in a box if the size of the package is less than or equal to the size of the box.

The package sizes are given as an integer array packages, where packages[i] is the size of the i^th package. The suppliers are given as a 2D integer array boxes, where boxes[j] is an array of box sizes that the j^th supplier produces.

You want to choose a single supplier and use boxes from them such that the total wasted space is minimized. For each package in a box, we define the space wasted to be size of the box - size of the package. The total wasted space is the sum of the space wasted in all the boxes.

For example, if you have to fit packages with sizes [2,3,5] and the supplier offers boxes of sizes [4,8], you can fit the packages of size-2 and size-3 into two boxes of size-4 and the package with size-5 into a box of size-8. This would result in a waste of (4-2) + (4-3) + (8-5) = 6.

Return the minimum total wasted space by choosing the box supplier optimally, or -1 if it is impossible to fit all the packages inside boxes. Since the answer may be large, return it modulo 10⁹ + 7.

Example 1:

Input: packages = [2,3,5], boxes = [[4,8],[2,8]]
Output: 6
Explanation: It is optimal to choose the first supplier, using two size-4 boxes and one size-8 box.
The total waste is (4-2) + (4-3) + (8-5) = 6.

Example 2:

Input: packages = [2,3,5], boxes = [[1,4],[2,3],[3,4]]
Output: -1
Explanation: There is no box that the package of size 5 can fit in.

Example 3:

Input: packages = [3,5,8,10,11,12], boxes = [[12],[11,9],[10,5,14]]
Output: 9
Explanation: It is optimal to choose the third supplier, using two size-5 boxes, two size-10 boxes, and two size-14 boxes.
The total waste is (5-3) + (5-5) + (10-8) + (10-10) + (14-11) + (14-12) = 9.

Constraints:

n == packages.length
m == boxes.length
1 <= n <= 10⁵
1 <= m <= 10⁵
1 <= packages[i] <= 10⁵
1 <= boxes[j].length <= 10⁵
1 <= boxes[j][k] <= 10⁵
sum(boxes[j].length) <= 10⁵
The elements in boxes[j] are distinct.

Minimum Space Wasted From Packaging

Hard

Two Sigma

2 views

Topics:

ArraysBinary SearchGreedy Algorithms

You have n packages that you are trying to place in boxes, one package in each box. There are m suppliers that each produce boxes of different sizes (with infinite supply). A package can be placed in a box if the size of the package is less than or equal to the size of the box.

The package sizes are given as an integer array packages, where packages[i] is the size of the i^th package. The suppliers are given as a 2D integer array boxes, where boxes[j] is an array of box sizes that the j^th supplier produces.

You want to choose a single supplier and use boxes from them such that the total wasted space is minimized. For each package in a box, we define the space wasted to be size of the box - size of the package. The total wasted space is the sum of the space wasted in all the boxes.

For example, if you have to fit packages with sizes [2,3,5] and the supplier offers boxes of sizes [4,8], you can fit the packages of size-2 and size-3 into two boxes of size-4 and the package with size-5 into a box of size-8. This would result in a waste of (4-2) + (4-3) + (8-5) = 6.

Return the minimum total wasted space by choosing the box supplier optimally, or -1 if it is impossible to fit all the packages inside boxes. Since the answer may be large, return it modulo 10⁹ + 7.

Example 1:

Input: packages = [2,3,5], boxes = [[4,8],[2,8]]
Output: 6
Explanation: It is optimal to choose the first supplier, using two size-4 boxes and one size-8 box.
The total waste is (4-2) + (4-3) + (8-5) = 6.

Example 2:

Input: packages = [2,3,5], boxes = [[1,4],[2,3],[3,4]]
Output: -1
Explanation: There is no box that the package of size 5 can fit in.

Example 3:

Input: packages = [3,5,8,10,11,12], boxes = [[12],[11,9],[10,5,14]]
Output: 9
Explanation: It is optimal to choose the third supplier, using two size-5 boxes, two size-10 boxes, and two size-14 boxes.
The total waste is (5-3) + (5-5) + (10-8) + (10-10) + (14-11) + (14-12) = 9.

Constraints:

n == packages.length
m == boxes.length
1 <= n <= 10⁵
1 <= m <= 10⁵
1 <= packages[i] <= 10⁵
1 <= boxes[j].length <= 10⁵
1 <= boxes[j][k] <= 10⁵
sum(boxes[j].length) <= 10⁵
The elements in boxes[j] are distinct.

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 upper and lower bounds for the sizes of the `packages` and `boxes` arrays? What is the maximum value within those arrays?
Can the dimensions in `packages` and `boxes` be zero or negative?
If no box can accommodate all packages, what value should I return?
Are there any duplicate dimensions within the `packages` array or within any individual box in the `boxes` array, and how should that be handled?
Is the goal to minimize the *total* wasted space across *all* packages for a given box, or is there a different definition of 'minimum space wasted'?

Brute Force Solution

Approach

We want to find the best way to pack items into boxes of various sizes, minimizing wasted space. The brute force approach involves trying every single possible combination of boxes to pack all the items. It's like trying every possibility until you find the one that wastes the least space.

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

Consider each box size individually.
For each box size, figure out all the ways to pack the items into boxes of that size, allowing for some items to potentially be packed into other boxes.
Calculate the amount of wasted space for each possible packing arrangement for that box size.
Repeat the process for every other box size.
Compare the wasted space from all the box sizes and their packing arrangements.
Choose the box size and its packing arrangement that results in the least wasted space.

Code Implementation

def min_space_wasted_brute_force(packages_sizes, boxes):    minimum_wasted_space = float('inf')
    for box_size in boxes:
        # Iterate through each box size to find the optimal waste.
        total_packages_size = sum(packages_sizes)
        if total_packages_size <= box_size * len(packages_sizes):
            wasted_space = (box_size * len(packages_sizes)) - total_packages_size
            minimum_wasted_space = min(minimum_wasted_space, wasted_space)
        else:
            # Box size inadequate to hold all packages, wasted space not valid
            continue
    if minimum_wasted_space == float('inf'):
        return -1
    else:
        return minimum_wasted_space

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(m * (2^n)) – The algorithm iterates through each of the 'm' box sizes. For each box size, it considers all possible subsets of items (power set) to determine how to pack them into boxes of that size. Generating the power set of 'n' items takes O(2^n) time. Therefore, the overall time complexity is O(m * (2^n)), where 'm' is the number of box sizes and 'n' is the number of items to pack. This is because for each box size we must generate all combinations of items.

Space Complexity

O(N!) – The algorithm explores all possible packing arrangements for each box size. To represent these arrangements, temporary data structures are required to store which items are packed into which boxes. In the worst case, this could involve generating all permutations of items, leading to auxiliary data structures that can grow factorially with the number of items N. Specifically, we need to store all possible combinations of items being packed or not packed into a box and given we are checking *all* combinations this can result in a factorial space complexity. Where N represents the number of items needing packing.

Optimal Solution

Approach

To minimize wasted space when packaging items, we need to find the best box size for each provider. We aim to identify the box size from each provider that is just big enough to hold all items, and then compare the wasted space of these optimal boxes across all providers.

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

First, add up the total size of all the items you need to pack.
For each packaging provider, go through their available box sizes.
For each box size, check if the box can fit all the items. If it's smaller than the total item size, it can't fit.
If a box can fit all the items, calculate how much space would be wasted by subtracting the total item size from the box size.
Keep track of the smallest wasted space for each provider. Ignore box sizes that are too small to fit everything.
Once you've checked all providers, find the provider with the overall smallest wasted space. That's your answer.

Code Implementation

def minimum_space_wasted(packages, boxes):
    total_item_size = sum(packages)
    minimum_wasted_space = float('inf')

    for provider_boxes in boxes:
        provider_boxes.sort()
        provider_min_waste = float('inf')

        # Iterate through each box size of the provider.
        for box_size in provider_boxes:
            if box_size >= total_item_size:

                wasted_space = box_size - total_item_size

                # Find min wasted space for the provider.
                provider_min_waste = min(provider_min_waste, wasted_space)

        # Choose best provider
        minimum_wasted_space = min(minimum_wasted_space, provider_min_waste)

    if minimum_wasted_space == float('inf'):
        return -1
    else:
        return minimum_wasted_space

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(m*n) – Let n be the number of providers and m be the maximum number of box sizes any provider offers. We iterate through each of the n providers. For each provider, we iterate through at most m box sizes. Inside this inner loop, we perform a constant-time calculation (checking if the box fits and calculating wasted space). Therefore, the dominant operations are dictated by the nested loops, resulting in a time complexity proportional to m*n. Hence, the overall time complexity is O(m*n).

Space Complexity

O(1) – The algorithm primarily uses a few variables to store the total item size, the minimum wasted space for each provider, and potentially a few loop counters. The space required for these variables does not depend on the number of items or providers. Therefore, the auxiliary space used by the algorithm is constant, resulting in a space complexity of O(1).

Edge Cases

Case	How to Handle
Boxes or packages array is null or empty.	Return 0 if either boxes or packages is null or empty, as no waste can be calculated.
Only one box is present.	The smallest possible waste can be calculated by using the largest package dimension scaled to fill the largest package count.
Very large box or package dimensions resulting in integer overflow.	Use long data type for calculations involving the product of dimensions to avoid integer overflow.
All packages fit perfectly into a single box.	The waste will be equal to the difference between the box volume and the total package volume.
No box can accommodate any of the packages (all boxes are too small).	Return -1 to indicate that it is impossible to contain packages in boxes.
Boxes or packages array contains zero or negative values	Throw an exception if boxes or packages contains zero or negative dimensions as those are invalid.
Extremely large number of boxes and packages exceeding memory limits.	Consider using an iterative approach rather than recursion to reduce memory usage, and ensure that the number of elements is within reasonable bounds for the system's memory capacity.
Multiple boxes result in the same minimal waste.	Return the calculated minimal waste as only one value is needed to represent the minimum wastage.