<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: List[int], nums2: List[int]) -> List[int]:
        n = len(nums2)
        
        nums1_std = sorted(nums1, reverse=True)
        
        cnt1 = 0
        for i in range(n):
            if nums1[i] > nums2[i]:
                cnt1 += 1

        indx_nums2 = []
        for i, n in enumerate(nums2):
            indx_nums2.append((i, n))
        indx_nums2 = sorted(indx_nums2, key=lambda tup: tup[1], reverse=True)

        pairs = {}

        while nums1_std and indx_nums2:
            if nums1_std[0] > indx_nums2[0][1]:
                pairs.update({indx_nums2.pop(0)[0]: nums1_std.pop(0)})
            else:
                pairs.update({indx_nums2.pop(0)[0]: nums1_std.pop(-1)})
       
        for i in pairs.keys():
            nums1[i] = pairs[i]

        return nums1
45517