Introduction

In this blog post, we will explore how to solve the LeetCode problem "1889" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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.

Explanation

Here's the breakdown of the solution:

• Sort: Sort the packages array to enable efficient searching for suitable boxes. Also, sort each supplier's box sizes to enable binary search.
• Iterate and Calculate Waste: Iterate through each supplier. For each supplier, check if all packages can be accommodated. If so, calculate the total wasted space for that supplier. Use binary search to find the smallest box size from the supplier that can contain each package.

• Minimize: Maintain a running minimum of the total wasted space across all suppliers. If no supplier can accommodate all packages, return -1.

• Runtime Complexity: O(m (n log(n) + n * log(k))) where n is the number of packages, m is the number of suppliers, and k is the maximum number of box sizes offered by a supplier. Sorting the packages takes O(n log n). For each supplier, we iterate through the packages (O(n)) and perform a binary search on the supplier's box sizes (O(log k)).

• Storage Complexity: O(n + k), where n is the number of packages and k is the maximum number of boxes from a supplier (for sorting the boxes array)

Code

    def minWastedSpace(packages, boxes):
    packages.sort()
    n = len(packages)
    min_waste = float('inf')
    modulo = 10**9 + 7

    for supplier_boxes in boxes:
        supplier_boxes.sort()
        if supplier_boxes[-1] < packages[-1]:
            continue

        total_waste = 0
        possible = True
        for package_size in packages:
            l, r = 0, len(supplier_boxes) - 1
            best_box = -1

            while l <= r:
                mid = (l + r) // 2
                if supplier_boxes[mid] >= package_size:
                    best_box = supplier_boxes[mid]
                    r = mid - 1
                else:
                    l = mid + 1

            if best_box == -1:
                possible = False
                break
            total_waste = (total_waste + (best_box - package_size)) % modulo

        if possible:
            min_waste = min(min_waste, total_waste)

    if min_waste == float('inf'):
        return -1
    else:
        return min_waste