1889. Minimum Space Wasted From Packaging.py

""" 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. """


#package 排序， boxes内部排序
        #对于每个supplier， 如果最大box容纳不了最大package， 则跳过这个supplier
        #从小到大尝试每个box， 如果box太小则跳过， 用box去二分搜索packages， 知道合适位置的最右（这样相同size的package在左边）
        #记录下前次二分搜索的位置， 则前次和本次位置中间的package都能用当前盒子容纳。

import bisect
class Solution:
    def minWastedSpace(self, packages: List[int], boxes: List[List[int]]) -> int:
        #package 排序， boxes内部排序
        #对于每个supplier， 如果最大box容纳不了最大package， 则跳过这个supplier
        #从小到大尝试每个box， 如果box太小则跳过， 用box去二分搜索packages， 知道合适位置的最右（这样相同size的package在左边）
        #记录下前次二分搜索的位置， 则前次和本次位置中间的package都能用当前盒子容纳。
        ans = math.inf
        packages.sort()
        total = sum(packages)
        for supplier in boxes:
            supplier.sort()
            if len(supplier) == 0: continue
            if supplier[-1] < packages[-1]: continue
            prev_idx = 0
            used = 0
            #print('box: ', box)
            for b in supplier:
                if b < packages[0]: continue
                cur_idx = bisect.bisect_right(packages, b)
                #[prev_idx, cur_idx) package can be put in be

                used += b*(cur_idx-prev_idx)
                prev_idx = cur_idx

            wasted = used - total
            ans = min(wasted, ans)
        return ans % (10**9 + 7) if ans!= math.inf else -1