<p>You are given two integer arrays <code>nums1</code> and <code>nums2</code> both of the same length. The <strong>advantage</strong> of <code>nums1</code> with respect to <code>nums2</code> is the number of indices <code>i</code> for which <code>nums1[i] &gt; nums2[i]</code>.</p>

<p>Return <em>any permutation of </em><code>nums1</code><em> that maximizes its <strong>advantage</strong> with respect to </em><code>nums2</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> nums1 = [2,7,11,15], nums2 = [1,10,4,11]
<strong>Output:</strong> [2,11,7,15]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> nums1 = [12,24,8,32], nums2 = [13,25,32,11]
<strong>Output:</strong> [24,32,8,12]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums1.length &lt;= 10<sup>5</sup></code></li>
	<li><code>nums2.length == nums1.length</code></li>
	<li><code>0 &lt;= nums1[i], nums2[i] &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def advantageCount(self, nums1, nums2):
        n = len(nums1)
  
        # Step 1: Create a max-heap for nums2, paired with indices.
        # Negate the values to simulate a max-heap using heapq.
        maxpq = [(-num, i) for i, num in enumerate(nums2)]
        heapq.heapify(maxpq)
        
        # Step 2: Sort nums1 in ascending order.
        nums1.sort()

        # Step 3: Use two pointers to assign values from nums1.
        left, right = 0, n - 1
        res = [0] * n

        while maxpq:
            # Step 4: Pop the largest element from nums2 (with its index).
            maxval, i = heapq.heappop(maxpq)
            maxval *= -1  # Revert the negation to get the original value.

            # Step 5: Check if nums1[right] can beat the largest value in nums2.
            if nums1[right] > maxval:
                # If yes, assign the largest available value from nums1.
                res[i] = nums1[right]
                right -= 1  # Decrease the right pointer (use the next largest).
            else:
                # Otherwise, assign the smallest available value from nums1.
                res[i] = nums1[left]
                left += 1  # Increase the left pointer (use the next smallest).
        
        # Step 6: Return the resulting array.
        return res
41432