#1889

Hard

auto_awesomeBinary search over the valid answer space

LeetCode Problem Workspace

Minimum Space Wasted From Packaging

Minimize the wasted space when packaging items into boxes, considering package and box size constraints.

Problem Statement

You are given a set of packages and suppliers that each provide different box sizes. Each package needs to be placed in a box where the box size must be greater than or equal to the size of the package. The goal is to minimize the wasted space when placing the packages in the boxes by choosing one supplier and using their box sizes.

For each supplier, you calculate the wasted space for each box by subtracting the package size from the box size. The total wasted space is the sum of all individual wasted spaces. Your task is to determine which supplier results in the minimum total wasted space. If no box can fit a package, return -1.

Examples

Example 1

Input: packages = [2,3,5], boxes = [[4,8],[2,8]]

Output: 6

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

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

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 Approach

Binary Search over Valid Box Sizes

Perform a binary search to find the optimal box size that minimizes the wasted space. This allows you to efficiently check possible box sizes while ensuring the minimum wasted space is achieved for each package.

Sorting Packages and Boxes

Sort the packages and box sizes for each supplier. Sorting allows for easier allocation of packages to boxes using binary search, ensuring that you can quickly determine the best-fitting box for each package.

Prefix Sum for Efficient Wasted Space Calculation

Using prefix sum techniques, compute the total wasted space for each valid configuration efficiently. This enables quick accumulation of wasted space when multiple boxes are used, avoiding the need for repetitive calculations.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the sorting and binary search steps. Sorting the boxes takes O(m log m), and binary searching for each package can take O(log m). Thus, the overall complexity is O(n log m), where n is the number of packages and m is the number of suppliers. The space complexity is dominated by the space required for storing the packages and boxes, which is O(n + m).

What Interviewers Usually Probe

Ability to optimize search within the box sizes using binary search.
Proficiency in sorting algorithms and their applications to packing problems.
Understanding the use of prefix sums in cumulative calculations.

Common Pitfalls or Variants

Common pitfalls

Failing to handle cases where no box fits a package, leading to incorrect results.
Not optimizing the solution with binary search, resulting in inefficient brute-force solutions.
Overcomplicating the solution with unnecessary data structures when a simple sorting and binary search approach is sufficient.

Follow-up variants

Allowing multiple suppliers to provide the same box sizes and exploring different packaging strategies.
Extending the problem to support 3D box sizes and packages.
Adding constraints that limit the number of boxes that can be used per supplier.

FAQ

What is the optimal way to minimize wasted space in the Minimum Space Wasted From Packaging problem?

The optimal approach involves using binary search to find the smallest possible box size that fits the packages, combined with sorting and prefix sums for efficient calculations.

How does binary search help in the Minimum Space Wasted From Packaging problem?

Binary search allows you to efficiently explore the valid space of box sizes, narrowing down the best fitting box for each package while minimizing the total wasted space.

Can there be cases where no box fits the package in the Minimum Space Wasted From Packaging problem?

Yes, if no box is large enough to fit a package, the answer should be -1, indicating that the package cannot be packed.

What is the time complexity of the solution for the Minimum Space Wasted From Packaging problem?

The time complexity is O(n log m), where n is the number of packages and m is the number of suppliers, primarily due to the sorting and binary search steps.

Are there any edge cases to consider in the Minimum Space Wasted From Packaging problem?

Yes, edge cases include scenarios where no box fits a package or when multiple suppliers have similar box sizes. It's important to handle these cases correctly to avoid incorrect results.

terminal

Solution

Solution 1

#### Python3

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class Solution:
    def minWastedSpace(self, packages: List[int], boxes: List[List[int]]) -> int:
        mod = 10**9 + 7
        ans = inf
        packages.sort()
        for box in boxes:
            box.sort()
            if packages[-1] > box[-1]:
                continue
            s = i = 0
            for b in box:
                j = bisect_right(packages, b, lo=i)
                s += (j - i) * b
                i = j
            ans = min(ans, s)
        if ans == inf:
            return -1
        return (ans - sum(packages)) % mod

Continue Practicing

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