<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]:
        res = [-1] * len(A)
        sA = [(a,i) for i,a in enumerate(A)]
        sB = [(b,i) for i,b in enumerate(B)]
        heapq.heapify(sA)
        heapq.heapify(sB)
        remaining  = []
        while sA:
            a,i = heapq.heappop(sA)
            b,j = heapq.heappop(sB)
            if a>b:
                res[j] = a
            else:
                remaining.append(a)
                heapq.heappush(sB,(b,j))
        for i in range(len(res)):
            if res[i]==-1:
                res[i]=remaining.pop()
        return res
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