Minimum Space Wasted From Packaging

Hard

Asked by:

35 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.

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 constraints on the size of the `packages` array and the `boxes` array (number of companies)? What are the potential value ranges for package sizes and box sizes?
Can package sizes or box sizes be zero or negative? If so, how should wasted space be calculated in these scenarios?
Are there any constraints on the number of box sizes a single company can offer (i.e., the length of each inner list in `boxes`)?
If a package is larger than the largest box offered by a company, does that mean it's impossible to ship that package with that company, or should we consider the waste as the package size itself?
What is the expected return type for the minimum total wasted space, and what specific value should be returned if it's impossible to ship all packages with any company?

Brute Force Solution

Approach

The goal is to figure out the best way to package items into boxes of various sizes to minimize wasted space. The brute force method considers every single box size provided and for each, calculates the total wasted space if we were to use only that box size for all items.

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

For the very first box size given, imagine using it to package all the items.
Calculate how much space is wasted in total for all items when packed into this first box size.
Now, consider the second box size provided.
Calculate the total wasted space if all items were packed into this second box size.
Continue this process, considering each provided box size one by one.
For each box size, determine the total wasted space when packing all items into it.
After checking all the different box sizes, compare the total wasted space calculated for each.
The minimum wasted space among all these checks is your answer.

Code Implementation

def calculate_minimum_wasted_space(package_sizes, item_sizes):
    minimum_total_wasted_space = float('inf')

    # Iterate through each available package size to find the best fit
    for package_size in package_sizes:
        current_total_wasted_space = 0
        # Calculate wasted space for all items using the current package size
        for item_size in item_sizes:
            # Ensure the item can fit in the current package size
            if item_size > package_size:
                # If an item is too large, this package size is invalid for this set of items.
                # This scenario should ideally be handled before calling this function or indicate an impossible situation.
                # For brute force, we might consider this an infinite waste or skip it.
                # In this interpretation, we assume all items *can* fit in at least one package type.
                pass
            else:
                wasted_space_for_item = package_size - item_size
                current_total_wasted_space += wasted_space_for_item

        # Update the minimum wasted space found so far
        minimum_total_wasted_space = min(minimum_total_wasted_space, current_total_wasted_space)

    return minimum_total_wasted_space

Big(O) Analysis

Time Complexity

O(m * n log n) – Let m be the number of box sizes and n be the number of items. For each of the m box sizes, we iterate through all n items. For each item, we need to find the smallest box that can fit it from the given box sizes for that specific iteration. If the box sizes for each m are sorted beforehand (which is a common optimization for this problem), finding the appropriate box for each item takes O(log k) time, where k is the number of box sizes available in that iteration. If we consider the case where all m box sizes are available for each item in each of the m iterations, the sorting of box sizes for each iteration would take O(m log m). However, a more efficient approach for this problem often involves pre-sorting all unique box sizes across all suppliers and then using binary search for each item within each box supplier's offerings. If we consider the provided explanation which implies iterating through each box size and then iterating through all items, and assuming within each box size iteration, we perform a search for the best fitting box, the complexity would be O(m * (n log n)) if we consider sorting the boxes within each iteration or O(m * n * log n) if we consider binary search for each item against a sorted list of box sizes. This simplifies to O(m * n log n).

Space Complexity

O(1) – The provided plain English explanation describes iterating through each box size and performing calculations for each. No additional data structures are explicitly created or maintained that grow with the input size of items or box sizes. The calculations involve simple variables to keep track of the current wasted space and the minimum wasted space found so far. Therefore, the auxiliary space complexity remains constant.

Optimal Solution

Approach

The problem asks us to minimize wasted space when packaging items. The smart approach is to consider each box type and figure out the best way to package all items using that specific box type. We then compare the results from each box type and pick the one that wastes the least space overall.

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

For each type of box provided, we'll pretend we are only going to use that box type for all the items.
When considering a specific box type, we'll figure out the 'best fit' for each item. This means finding the smallest possible box within that type that can hold the item.
Once we know the smallest fitting box for every item using the current box type, we can calculate the total 'waste' associated with packaging all items this way.
The waste for a single item is the difference between the size of the box it fits into and the item's own size.
We sum up all this individual item waste for the current box type.
We repeat this process for every single box type available.
After calculating the total waste for each box type, we compare these totals.
The minimum of all these calculated wastes is our final answer, representing the least amount of wasted space possible.

Code Implementation

def minimum_space_wasted_from_packaging(package_sizes, item_dimensions):
    minimum_total_waste = float('inf')

    # Sort item dimensions once to optimize binary search later.
    sorted_item_dimensions = sorted(item_dimensions)

    # Iterate through each packaging type to find the best fit for all items.
    for packaging_type in package_sizes:
        current_packaging_type_waste = 0
        # Sort the boxes within the current packaging type for efficient searching.
        sorted_boxes_in_type = sorted(packaging_type)

        # Calculate waste for each item using the current packaging type.
        for item_size in sorted_item_dimensions:
            best_fit_box_size = -1
            # Binary search to find the smallest box that can fit the current item.
            low_index, high_index = 0, len(sorted_boxes_in_type) - 1
            while low_index <= high_index:
                mid_index = (low_index + high_index) // 2
                if sorted_boxes_in_type[mid_index] >= item_size:
                    best_fit_box_size = sorted_boxes_in_type[mid_index]
                    high_index = mid_index - 1
                else:
                    low_index = mid_index + 1

            # If no box fits, this packaging type is not viable.
            if best_fit_box_size == -1:
                current_packaging_type_waste = float('inf')
                break
            else:
                current_packaging_type_waste += (best_fit_box_size - item_size)

        # Update the overall minimum waste if the current packaging type is better.
        minimum_total_waste = min(minimum_total_waste, current_packaging_type_waste)

    # If no packaging type could fit all items, return -1. Otherwise, return the calculated minimum waste.
    return minimum_total_waste if minimum_total_waste != float('inf') else -1

Big(O) Analysis

Time Complexity

O(m * n * log k) – Let n be the number of items, m be the number of box types, and k be the maximum number of box sizes within a single box type. The algorithm iterates through each of the m box types. For each box type, it iterates through all n items. Within this inner loop, to find the smallest fitting box, it performs a binary search on the sorted box sizes for that type, which takes O(log k) time. Therefore, the total time complexity is the product of these factors: m * n * log k.

Space Complexity

O(N + K*M) – The primary auxiliary space is used for sorting the items and the boxes within each provider. Specifically, sorting the items requires O(N) auxiliary space if an in-place sort is not used, or O(log N) for recursion stack if quicksort is implemented recursively. For each of the K providers, we iterate through M types of boxes. To efficiently find the best fitting box for each item within a provider's box types, we first sort the box sizes of that provider, which takes O(M log M) time but can be considered part of the per-provider processing. The total space complexity is dominated by storing sorted versions of items and potentially box sizes for each provider, leading to O(N + K*M) in the worst case where temporary sorted arrays are maintained for each provider.

Edge Cases

Case	How to Handle
Empty packages array	If packages is empty, total wasted space is 0 for any company, so the minimum is 0.
Empty boxes array or empty inner company box list	If no companies offer boxes or a company offers no boxes, that company cannot ship packages, and if all companies are like this, it's impossible to ship.
Package size larger than any box offered by a company	If a package cannot fit into any box from a specific company, that company cannot fulfill the shipment, contributing infinite waste for that company.
All packages are the same size	The logic should still correctly find the smallest box that fits each identical package, minimizing waste per package.
All boxes offered by a company are the same size	The algorithm will pick the single available box size if it fits, calculating waste accordingly.
Package sizes or box sizes are zero or negative	Assuming package and box sizes are non-negative integers as per typical problem constraints; if negative values are allowed, problem definition needs clarification on waste calculation.
No company can ship all packages	The function should return -1 if for every shipping company, there's at least one package that cannot be shipped.
Large number of packages or companies leading to potential integer overflow	Use a 64-bit integer type for accumulating total wasted space to prevent overflow.

Empty packages array

How to Handle:

If packages is empty, total wasted space is 0 for any company, so the minimum is 0.

Empty boxes array or empty inner company box list

How to Handle:

If no companies offer boxes or a company offers no boxes, that company cannot ship packages, and if all companies are like this, it's impossible to ship.

Package size larger than any box offered by a company

How to Handle:

If a package cannot fit into any box from a specific company, that company cannot fulfill the shipment, contributing infinite waste for that company.

All packages are the same size

How to Handle:

The logic should still correctly find the smallest box that fits each identical package, minimizing waste per package.

All boxes offered by a company are the same size

How to Handle:

The algorithm will pick the single available box size if it fits, calculating waste accordingly.

Package sizes or box sizes are zero or negative

How to Handle:

Assuming package and box sizes are non-negative integers as per typical problem constraints; if negative values are allowed, problem definition needs clarification on waste calculation.

No company can ship all packages

How to Handle:

The function should return -1 if for every shipping company, there's at least one package that cannot be shipped.

Large number of packages or companies leading to potential integer overflow

How to Handle:

Use a 64-bit integer type for accumulating total wasted space to prevent overflow.