1889 - Minimum Space Wasted From Packaging.

Hard

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 <code>i^th</code> package. The suppliers are given as a 2D integer array boxes, where boxes[j] is an array of box sizes that the <code>j^th</code> 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 <code>10⁹ + 7</code>.

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

• <code>1 <= n <= 10⁵</code>

• <code>1 <= m <= 10⁵</code>

• <code>1 <= packagesi<= 10⁵</code>

• <code>1 <= boxesj.length <= 10⁵</code>

• <code>1 <= boxesk<= 10⁵</code>

• <code>sum(boxesj.length) <= 10⁵</code>

• The elements in boxes[j] are distinct.