<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> <i>if it is <strong>impossible</strong> to fit all the packages inside boxes. </i>Since the answer may be <strong>large</strong>, return it <strong>modulo </strong><code>10<sup>9</sup> + 7</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> packages = [2,3,5], boxes = [[4,8],[2,8]]
<strong>Output:</strong> 6
<strong>Explanation</strong>: 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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> packages = [2,3,5], boxes = [[1,4],[2,3],[3,4]]
<strong>Output:</strong> -1
<strong>Explanation:</strong> There is no box that the package of size 5 can fit in.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> packages = [3,5,8,10,11,12], boxes = [[12],[11,9],[10,5,14]]
<strong>Output:</strong> 9
<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.
The total waste is (5-3) + (5-5) + (10-8) + (10-10) + (14-11) + (14-12) = 9.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == packages.length</code></li>
	<li><code>m == boxes.length</code></li>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= m &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= packages[i] &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= boxes[j].length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= boxes[j][k] &lt;= 10<sup>5</sup></code></li>
	<li><code>sum(boxes[j].length) &lt;= 10<sup>5</sup></code></li>
	<li>The elements in <code>boxes[j]</code> are <strong>distinct</strong>.</li>
</ul>

from bisect import bisect_right
from typing import List

class Solution:
    def minWastedSpace(self, packages: List[int], boxes: List[List[int]]) -> int:
        MOD = 10**9 + 7
        n = len(packages)
        packages.sort()
        
        # Compute prefix sums of package sizes
        prefix_sums = [0] * n
        prefix_sums[0] = packages[0]
        for i in range(1, n):
            prefix_sums[i] = prefix_sums[i - 1] + packages[i]
        
        # Collect all unique box sizes across all suppliers
        all_box_sizes = set()
        for supplier_boxes in boxes:
            all_box_sizes.update(supplier_boxes)
        all_box_sizes = sorted(all_box_sizes)
        
        # Compute total_packages and total_size for each box size
        total_packages = {}
        total_size = {}
        idx = 0  # Pointer over packages
        count = 0
        size_sum = 0
        for b in all_box_sizes:
            while idx < n and packages[idx] <= b:
                count += 1
                size_sum += packages[idx]
                idx += 1
            total_packages[b] = count
            total_size[b] = size_sum
        
        min_total_waste = float('inf')
        total_packages_in_packages = n  # Total number of packages
        
        # Process each supplier
        for supplier_boxes in boxes:
            supplier_boxes = sorted(supplier_boxes)
            if supplier_boxes[-1] < packages[-1]:
                continue  # Cannot fit all packages with this supplier
            
            last_total_packages = 0
            last_total_size = 0
            total_waste = 0
            for b in supplier_boxes:
                if b not in total_packages:
                    continue  # No packages can fit into this box size
                curr_total_packages = total_packages[b]
                curr_total_size = total_size[b]
                num_packages = curr_total_packages - last_total_packages
                size_packages = curr_total_size - last_total_size
                total_waste += num_packages * b - size_packages
                last_total_packages = curr_total_packages
                last_total_size = curr_total_size
                if last_total_packages == n:
                    break  # All packages have been assigned
            if last_total_packages == n:
                min_total_waste = min(min_total_waste, total_waste)
        
        return min_total_waste % MOD if min_total_waste != float('inf') else -1

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