Problem Description

In this problem, we are given n packages with different sizes, and we aim to place each package into a box from one of m suppliers. Each supplier supplies an unlimited number of boxes of various sizes. A package can fit in a box if its size doesn't exceed the size of the box. The goal is to minimize the total wasted space, which is the sum of the differences between the sizes of the boxes and the packages they contain.

The sizes of the packages are given in an integer array packages, and the suppliers' boxes' sizes are given in a 2D integer array boxes. Our task is to select boxes from one supplier such that the total wasted space is minimized. If it's not possible to fit all packages into boxes, we should return -1. Otherwise, we compute the total wasted space and return it modulo 10^9 + 7, which is a way of keeping the returned value within a reasonable range for very large numbers.

An example is provided where we have packages of sizes [2,3,5] and a supplier with box sizes [4,8]. The smallest total wasted space in this scenario is 6, achieved by placing the packages into the respective boxes such that the package of size 2 and the package of size 3 go into boxes of size 4, and the package of size 5 goes into a box of size 8.

The problem requires optimization and an efficient approach to checking all possible suppliers while avoiding unnecessary calculations to find the solution within a reasonable time.

Intuition

The solution to this problem is based on sorting and binary search. Firstly, we recognize that to fit all packages from a supplier's boxes, the largest box from the supplier must be equal to or greater than the largest package size. If that's the case, we can then proceed to calculate the total wasted space using the boxes from this supplier.

The intuition behind the sorting of both "packages" and the "boxes" array for each supplier is that it aligns the packages in ascending order, making it easier to find the smallest box in which each package will fit using binary search. We use the bisect_right function from the bisect module (in Python) for this purpose, which finds the insertion point in the array to the right of the target value, efficiently determining the number of packages a specific box can accommodate.

By iterating over each box size for the chosen supplier, we can keep track of how many packages have been placed (i represents the index within the package array), and how much wasted space has been incurred thus far (s is the cumulative waste). For each box size b, we find the index j up to which we can place the packages starting from index i. Then we simply multiply b with the number of packages that can fit in this box (j - i) and add it to the total waste accumulator s.

At each step, we compare the total waste s with the best answer found so far (ans). If s is smaller, we update ans.

If after checking all possible suppliers, ans still holds the value inf, it implies that it was not possible to fit all packages into any supplier's boxes, and we return -1. If we found a valid configuration, we return ans - sum(packages) (this is the total wasted space) modulo 10^9 + 7.

The solution approach efficiently narrows down suppliers and calculates waste without unnecessary checks or iterations, hence optimizing the process.

Learn more about Binary Search, Prefix Sum and Sorting patterns.

Solution Approach

The solution involves two major components: sorting and binary search. These algorithms are crucial in optimizing the process to minimize wasted space. The steps for the solution are implemented as follows:

1. Sort Package Sizes: We begin by sorting the array packages in non-decreasing order. This is crucial for the binary search step as binary search requires a sorted array to function correctly.

2. Iterate Over Suppliers: For each supplier in boxes, we proceed with the following steps:

   • Sort Box Sizes: Sort the array of box sizes for the current supplier. This not only assists with the binary search but also ensures that we can check from the smallest to the largest box when trying to place packages, which helps in finding the optimal fit.

   • Check Largest Box: Before calculating the wasted space for a supplier, check if the largest box of the supplier is at least as large as the largest package size. If not, we cannot use this supplier, as not all packages can fit. This is done by comparing packages[-1] (the largest package, since the array is sorted) to box[-1] (the largest box size for the current supplier).

   • Initialize Variables: Before traversing the box sizes, we initialize s to 0, which will hold the cumulative wasted space, and i to 0, which will track the current index in the packages array.

3. Binary Search for Package Placement: For each box size b of the current supplier:

   • Utilize bisect_right(packages, b, lo=i) to find the index j in the sorted packages array. This index j is such that all packages in packages[i:j] fit in the box size b. The lo=i argument ensures the binary search starts from the current index i in the packages array.

   • Calculate Wasted Space: Add (j - i) * b to s, which represents the space taken by the boxes that could have been filled with packages in packages[i:j].

   • Update Package Counter: Assign j to i to update the index in packages for the next iteration.

4. Update Answer: After iterating through all the box sizes for the current supplier, check if the accumulated wasted space s is less than the current answer ans. If so, update ans to s.

5. Final Check: If after examining all suppliers, ans remains math.inf, it's impossible to fit all packages into boxes, and we return -1. Otherwise:

   • Calculate Total Wasted Space: Subtract the sum of package sizes from ans to get the total wasted space.

   • Modulo Operation: Return the result of this subtraction modulo 10^9 + 7 to handle the case of large numbers, as required by the problem statement.

The patterns used in this solution, such as sorting and binary search, are well-known for their efficiency. Sorting the arrays allows for the use of binary search, which is an efficient algorithm with a time complexity of O(log n) for finding elements or insertion points in a sorted array. These algorithms, combined with a careful approach to examine each supplier and efficiently calculate the wasted space, allow the solution to minimize the total wasted space while fitting the packages into boxes optimally.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach using the problem's logic.

Suppose we have packages of sizes [3,5,8] and boxes from suppliers [[2,4,7],[3,6,9]]. The goal is to find the minimum total wasted space by optimally fitting these packages into the boxes provided by one of the suppliers.

Step 1: Sort Package Sizes

• We sort the packages array: [3, 5, 8]. It’s already sorted in this example.

Step 2: Iterate Over Suppliers

• For the first supplier boxes[0], the box sizes are [2,4,7].

  Step 2.1: Sort Box Sizes

  • Sort the box sizes: [2,4,7].

  Step 2.2: Check Largest Box

  • We check if the largest box size (7) is at least as large as the largest package size (8). It is not, so we cannot use this supplier since not all packages would fit.

• For the second supplier boxes[1], the box sizes are [3,6,9].

  Step 2.1: Sort Box Sizes

  • Sort the box sizes: [3,6,9] (also already sorted).

  Step 2.2: Check Largest Box

  • The largest box size (9) is larger than the largest package size (8), so we can continue with this supplier.

  Step 2.3: Initialize Variables

  • Set s = 0 and i = 0.

Step 3: Binary Search for Package Placement

• We iterate over the boxes [3,6,9] from the current supplier.

  For box size b = 3:

  • Using binary search with bisect_right(packages, 3, lo=0), we find index j = 1 in packages.
  • Wasted space: (1 - 0) * 3 = 3.
  • Update i to 1.

  For box size b = 6:

  • With bisect_right(packages, 6, lo=1), we get j = 2.
  • Wasted space: (2 - 1) * 6 = 6 and now s = 3 + 6 = 9.
  • Update i to 2.

  For box size b = 9:

  • With bisect_right(packages, 9, lo=2), we get j = 3.
  • Wasted space: (3 - 2) * 9 = 9 and now s = 9 + 9 = 18.
  • i becomes 3, indicating all packages are placed.

Step 4: Update Answer

• As this is the first supplier that can fit all packages, we set ans = 18.

Step 5: Final Check

• After checking all suppliers, the minimum ans found is 18.

  • Calculating the total wasted space: ans - sum(packages) = 18 - (3 + 5 + 8) = 18 - 16 = 2.
  • The result modulo 10^9 + 7 is still 2.

Therefore, for this example, the minimum total wasted space is 2, and that is achieved by using the boxes from the second supplier.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the given code can be analyzed as follows:

• Sorting the packages list takes O(NlogN) time, with N being the length of the packages.
• The for loop iterates through each box in boxes. Let's say there are M boxes.
• Each box is also sorted, taking O(BlogB) time, where B is the maximum number of items in a single box.
• The inner for loop iterates through each item b in the box. The bisect_right function is also called inside this loop, which works in O(logN) time complexity for each b because it uses a binary search algorithm.
  • In the worst case, every call to bisect_right can iterate over all elements of packages, thus the combined complexity of the loop with the bisect_right is O(BlogN). However, since i = j assigns the new starting index after every bisect, it ensures that each package is considered only once across all boxes. Hence, the complexity for all boxes together should be O(BlogN), not O(M*B*logN).
• The overall time complexity is O(NlogN + M*B*logN) since the sorting of the packages and the boxes are the dominating factors.

Space Complexity

The space complexity of the code can be determined as follows:

• The sorted packages list and box list require additional space, which contributes to the space complexity. This could be, in the worst case, O(N + B) respective space for sorted packages and the largest box.
• The variables ans, s, i, j, and mod use constant space, which does not depend on the input size, hence contributing O(1) space.
• Consequently, the total space complexity is O(N + B).

Note: inf and mod are constants defined in the global namespace, and their space is considered as O(1).

Learn more about how to find time and space complexity quickly using problem constraints.