<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]:
        idx_nums2 = defaultdict(list)
        for idx, num in enumerate(nums2):
            idx_nums2[num].append(idx)
        print("idx_nums2 -- tells us original index before sort", idx_nums2)
        nums1.sort()
        nums2.sort()
        unused = []
        perm = [None] * len(nums1)
        n1_idx = 0
        n2_idx = 0
        while n1_idx < len(nums2): #if we get to end, pair the rest w/whatever is unused
            if nums2[n2_idx] >= nums1[n1_idx]: #advance 1 ptr, 2 ptr still needs to be paired
                unused.append(nums1[n1_idx])
                n1_idx += 1
            else:
                perm[n2_idx] = nums1[n1_idx]
                n1_idx += 1
                n2_idx += 1
        #print("perm before pairing w nones", perm)
        unused_idx = 0
        #print("unused", unused)
        for i in range(len(perm)):
            if perm[i] is None:
                #print("i is none")
                perm[i] = unused[unused_idx]
                unused_idx += 1
        #print("perm after getting all nones", perm)
        perm_2 = [0] * len(perm)
        for i in range(len(perm)): #unshuffle
            new_list = idx_nums2[nums2[i]]
            new_idx = new_list[-1]
            new_list.pop()
            if len(new_list) == 0:
                idx_nums2.pop(nums2[i])
            perm_2[new_idx] = perm[i]
        return perm_2


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