<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, A: List[int], B: List[int]) -> List[int]:

        # keep A sorted in ascending order
        A.sort()
        
        
        # array size
        size = len(A)
        
        # best advantage shuffle of A
        best_shuffle = [ 0 for _ in range(size) ]
        
        
        # two pointers for A and best shuffle
        left, right = 0, size-1
        
        
        # B_tuple
        # first item: negative value of b
        # second item: index of b
        B_tuple = [ (-b, idx) for idx, b in enumerate(B) ]
        
        
        # build a min heap from B_tuple
        # larger elements will be on the top naturally (because what we record in tuple is -b, negative value )
        heapq.heapify(B_tuple)
        
        
        # find best shuffle of A against B from largest value
        
        while B_tuple:
            
            
            largest_b_value, largest_b_idx = heapq.heappop(B_tuple)
        
            if A[right] > abs(largest_b_value):
                
                # largest of A is bigger than largest of B
                # pick A's largest value as best shuffle
                
                best_shuffle[largest_b_idx] = A[right]
                right -= 1
                
            else:
                
                # largest of A is smaller than or equal to B
                # no chance to win, pick A's smallest value as best shuffle
                
                best_shuffle[largest_b_idx] = A[left]
                left += 1
                
                
        return best_shuffle
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