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
class Solution {
public:
    vector<int> advantageCount(vector<int>& nums1, vector<int>& nums2) {
        int n = nums1.size();
        vector<pair<int, int>> t;
        for (int i = 0; i < n; ++i) t.push_back({nums2[i], i});
        sort(t.begin(), t.end());
        sort(nums1.begin(), nums1.end());
        int i = 0, j = n - 1;
        vector<int> ans(n);
        for (int v : nums1) {
            if (v <= t[i].first)
                ans[t[j--].second] = v;
            else
                ans[t[i++].second] = v;
        }
        return ans;
    }
};