1889 Minimum Space Wasted From Packaging

🔀 Pick One

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 ith package. The suppliers are given as a 2D integer array boxes, where boxes[j] is an array of box sizes that the jth 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.
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class Solution:
    def minWastedSpace(self, packages: List[int], boxes: List[List[int]]) -> int:
        MOD = 10 ** 9 + 7
        packages.sort()
        res = math.inf

        for supp in boxes:
            supp.sort()
            if supp[-1] < packages[-1]:
                continue
            i = 0
            space = 0
            for box in supp:
                j = bisect.bisect_right(packages, box)
                space += box * (j - i)
                i = j
            res = min(res, space)
        
        if res < math.inf:
            return (res - sum(packages)) % MOD

        return -1