<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, nums2):
        sorted_1 = sorted(nums1.copy())
        sorted_2 = sorted(enumerate(nums2.copy()), key=lambda x: x[1])
        
        ret = []
        
        min1 = 0
        max1 = len(sorted_1) - 1

        current = len(sorted_2) - 1

        while current >= 0:
            og_index, number = sorted_2[current]
        
            if sorted_1[max1] > number:
                ret += [(og_index, sorted_1[max1])]
                max1 -= 1
            else:
                ret += [(og_index, sorted_1[min1])]
                min1 += 1

            current -= 1

        # print(sorted(ret, key=lambda x: x[0]))
        return list(map(lambda x: x[1], sorted(ret, key=lambda x: x[0])))
        

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