Package g1801_1900.s1889_minimum_space_wasted_from_packaging

Class Solution

java.lang.Object
g1801_1900.s1889_minimum_space_wasted_from_packaging.Solution
public class Solution extends Object
1889 - Minimum Space Wasted From Packaging.<p>Hard</p> <p>You have <code>n</code> packages that you are trying to place in boxes, <strong>one package in each box</strong>. There are <code>m</code> suppliers that each produce boxes of <strong>different sizes</strong> (with infinite supply). A package can be placed in a box if the size of the package is <strong>less than or equal to</strong> the size of the box.</p> <p>The package sizes are given as an integer array <code>packages</code>, where <code>packages[i]</code> is the <strong>size</strong> of the <code>i<sup>th</sup></code> package. The suppliers are given as a 2D integer array <code>boxes</code>, where <code>boxes[j]</code> is an array of <strong>box sizes</strong> that the <code>j<sup>th</sup></code> supplier produces.</p> <p>You want to choose a <strong>single supplier</strong> and use boxes from them such that the <strong>total wasted space</strong> is <strong>minimized</strong>. For each package in a box, we define the space <strong>wasted</strong> to be <code>size of the box - size of the package</code>. The <strong>total wasted space</strong> is the sum of the space wasted in <strong>all</strong> the boxes.</p> <ul> <li>For example, if you have to fit packages with sizes <code>[2,3,5]</code> and the supplier offers boxes of sizes <code>[4,8]</code>, you can fit the packages of size-<code>2</code> and size-<code>3</code> into two boxes of size-<code>4</code> and the package with size-<code>5</code> into a box of size-<code>8</code>. This would result in a waste of <code>(4-2) + (4-3) + (8-5) = 6</code>.</li> </ul> <p>Return <em>the <strong>minimum total wasted space</strong> by choosing the box supplier <strong>optimally</strong> , or</em> <code>-1</code> <em>if it is <strong>impossible</strong> to fit all the packages inside boxes.</em> Since the answer may be <strong>large</strong> , return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p> <p><strong>Example 1:</strong></p> <p><strong>Input:</strong> packages = [2,3,5], boxes = [[4,8],[2,8]]</p> <p><strong>Output:</strong> 6</p> <p><strong>Explanation:</strong> It is optimal to choose the first supplier, using two size-4 boxes and one size-8 box.</p> <p>The total waste is (4-2) + (4-3) + (8-5) = 6.</p> <p><strong>Example 2:</strong></p> <p><strong>Input:</strong> packages = [2,3,5], boxes = [[1,4],[2,3],[3,4]]</p> <p><strong>Output:</strong> -1</p> <p><strong>Explanation:</strong> There is no box that the package of size 5 can fit in.</p> <p><strong>Example 3:</strong></p> <p><strong>Input:</strong> packages = [3,5,8,10,11,12], boxes = [[12],[11,9],[10,5,14]]</p> <p><strong>Output:</strong> 9</p> <p><strong>Explanation:</strong> It is optimal to choose the third supplier, using two size-5 boxes, two size-10 boxes, and two size-14 boxes.</p> <p>The total waste is (5-3) + (5-5) + (10-8) + (10-10) + (14-11) + (14-12) = 9.</p> <p><strong>Constraints:</strong></p> <ul> <li><code>n == packages.length</code></li> <li><code>m == boxes.length</code></li> <li><code>1 <= n <= 10<sup>5</sup></code></li> <li><code>1 <= m <= 10<sup>5</sup></code></li> <li><code>1 <= packages[i] <= 10<sup>5</sup></code></li> <li><code>1 <= boxes[j].length <= 10<sup>5</sup></code></li> <li><code>1 <= boxes[j][k] <= 10<sup>5</sup></code></li> <li><code>sum(boxes[j].length) <= 10<sup>5</sup></code></li> <li>The elements in <code>boxes[j]</code> are <strong>distinct</strong>.</li> </ul>

  • Constructor Details

    • Solution

      public Solution()

  • Method Details

    • minWastedSpace

      public int minWastedSpace(int[] packages, int[][] boxes)