Minimum Space Wasted From Packaging

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 **10^9 + 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 <= 10^5
1 <= m <= 10^5
1 <= packages[i] <= 10^5
1 <= boxes[j].length <= 10^5
1 <= boxes[j][k] <= 10^5
sum(boxes[j].length) <= 10^5
The elements in boxes[j] are distinct.

Solution

class Solution {
    private static final int MOD = 1_000_000_007;

    public int minWastedSpace(int[] packages, int[][] boxes) {
        int numPackages = packages.length;
        Arrays.sort(packages);
        long[] preSum = new long[numPackages];
        preSum[0] = packages[0];
        for (int i = 1; i < packages.length; i++) {
            preSum[i] = packages[i] + preSum[i - 1];
        }
        long ans = Long.MAX_VALUE;
        for (int[] box : boxes) {
            Arrays.sort(box);
            // Box of required size not present
            if (packages[numPackages - 1] > box[box.length - 1]) {
                continue;
            }
            // Find the total space wasted
            long totalWastedSpace = 0;
            int prev = -1;
            for (int j : box) {
                if (prev == packages.length - 1) {
                    break;
                }
                if (j < packages[0] || j < packages[prev + 1]) {
                    continue;
                }
                // Find up to which package the current box can fit
                int upper = findUpperBound(packages, j);
                if (upper == -1) {
                    continue;
                }
                // The current box will be able to handle the packages from
                // prev + 1 to the upper index
                long totalSpace = ((long) upper - (long) prev) * j;
                long packageSum = preSum[upper] - (prev >= 0 ? preSum[prev] : 0);
                long spaceWastedCurr = totalSpace - packageSum;
                totalWastedSpace += spaceWastedCurr;
                prev = upper;
            }
            ans = Math.min(ans, totalWastedSpace);
        }
        return ans == Long.MAX_VALUE ? -1 : (int) (ans % MOD);
    }

    private int findUpperBound(int[] packages, int key) {
        int l = 0;
        int h = packages.length;
        while (l < h) {
            int m = l + (h - l) / 2;
            if (packages[m] <= key) {
                l = m + 1;
            } else {
                h = m;
            }
        }
        return h - 1;
    }
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).