<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(nums1)
        shuffled = [-1 for i in range(n)]
        
        for i in range(n):
            nums1[i] = (nums1[i], i)
            nums2[i] = (nums2[i], i)
        
        nums1.sort()
        nums2.sort()

        l, r = n - 1, n - 1
        least_ind = 0

        while l >= 0 and r >= 0:
            val1 = nums1[l][0]
            val2, ind2 = nums2[r]
            if val1 > val2:
                shuffled[ind2] = val1
                r -= 1
                l -= 1
            else:
                shuffled[ind2] = nums1[least_ind][0]
                least_ind += 1
                r -= 1
        
        # while l >= 0:
        #     val1 = nums1[l][0]
        #     shuffled[least_ind] = val1
        #     least_ind += 1
        #     l += 1
        
        return shuffled
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