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g1801_1900.s1889_minimum_space_wasted_from_packaging.readme.md Maven / Gradle / Ivy

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 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**.