<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]:
        nums1.sort()
        nums2_unsorted = nums2.copy()

        nums2.sort()

        length = len(nums1)

        max_advantage = -1
        max_offset = 0

        for offset in range(length):
            advantage = 0
            for j in range(length):
                if nums2[j % length] < nums1[(j + offset) % length]:
                    advantage += 1

            if advantage >= max_advantage:
                max_advantage = advantage
                max_offset = offset
            else:
                break
        
        nums2_to_nums1_permutation = {}

        # not a bijection a priori, so use array and pick and choose from it
        # order is redundant, so non-uniqueness is fine
        for j in range(length):
            if not nums2[j] in nums2_to_nums1_permutation:
                nums2_to_nums1_permutation[nums2[j]] = nums1[(j + max_offset) % length]
            else:
                curr_value = nums2_to_nums1_permutation[nums2[j]]

                if isinstance(curr_value, int):
                    nums2_to_nums1_permutation[nums2[j]] = [curr_value, nums1[(j + max_offset) % length]]
                else:
                    curr_value.append(nums1[(j + max_offset) % length])

        # applying the quasi-permutation to recover nums1 from the nums2 permutation
        nums2_magically_sorted = [0] * length
        for j in range(length):
            curr_value = nums2_to_nums1_permutation[nums2_unsorted[j]]

            if isinstance(curr_value, int):
                nums2_magically_sorted[j] = curr_value
            else:
                nums2_magically_sorted[j] = curr_value.pop()

        return nums2_magically_sorted
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