<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]:
        def mapping(mapping_x, nums):
            for i, num in enumerate(nums):
                if num in mapping_x:
                    mapping_x[num].append(i)
                else:
                    mapping_x[num] = [i]
            return mapping_x
        mapping_2 = mapping({}, nums2)
        nums1.sort()
        nums2.sort()
        res = [0] * len(nums1)
        i = len(nums1) - 1
        m = i
        j = 0
        print(nums1)
        print(nums2)
        while i >= 0:
            if nums1[m] <= nums2[i]:
                res[mapping_2[nums2[i]].pop(0)] = nums1[j]
                j += 1
            else:
                res[(mapping_2[nums2[i]]).pop(0)] = nums1[m]
                m -= 1
            i -= 1
        return res        
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