<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>

import heapq
class Solution:
    def advantageCount(self, nums1: List[int], nums2: List[int]) -> List[int]:
        #create 2 min heaps for storing nums1 and nums2
        #basic idea is finding smallest value still greater than
        #curr element in nums2
        retArr = [0]*len(nums2)
        
        nums1Heap = []
        nums2Heap = []
        
        for i in range(0,len(nums2)):
            heapq.heappush(nums1Heap,(nums1[i],i))
            heapq.heappush(nums2Heap,(nums2[i],i))
        
        remainNums =[]
        while nums2Heap:
            val,index = heapq.heappop(nums2Heap)
            while nums1Heap and nums1Heap[0][0] <= val:
                currVal,currIndex = heapq.heappop(nums1Heap)
                remainNums.append(currVal)
            if nums1Heap:
                currVal2,currIndex2 = heapq.heappop(nums1Heap)
                retArr[index] = currVal2
            
        for i in range(0,len(retArr)):
            if retArr[i] == 0:
                retArr[i] = remainNums.pop(len(remainNums)-1)
        return retArr
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