- Difficulty: Hard
- Tags: LeetCode, Hard, Array, Binary Search, Prefix Sum, Sorting, leetcode-1889, O(mlogm + nlogn + mlogn), O(1)
Problem
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.