Advantage Shuffle
<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>
Greedy; Array; Two Pointers; Sorting
func advantageCount(nums1 []int, nums2 []int) []int {
	n := len(nums1)
	t := make([][]int, n)
	for i, v := range nums2 {
		t[i] = []int{v, i}
	}
	sort.Slice(t, func(i, j int) bool {
		return t[i][0] < t[j][0]
	})
	sort.Ints(nums1)
	ans := make([]int, n)
	i, j := 0, n-1
	for _, v := range nums1 {
		if v <= t[i][0] {
			ans[t[j][1]] = v
			j--
		} else {
			ans[t[i][1]] = v
			i++
		}
	}
	return ans
}