<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 {
public:
    vector<int> advantageCount(vector<int>& nums1, vector<int>& nums2) {
        int n = nums1.size();
        vector<int> ans(n);

        vector<pair<int, int>> nums2order(n);
        for(int i = 0; i < n; i++){
            nums2order[i] = {nums2[i], i};
        }
        sort(nums2order.begin(), nums2order.end());

        multiset<int> s;
        for(int i = 0; i < n; i++){
            s.insert(nums1[i]);
        }
        
        for(int i = n - 1; i >= 0; i--) {
            auto [element, index] = nums2order[i];
            // smallest number from nums1 > element
            auto it = s.upper_bound(element);
            if(it == s.end()){
                it = s.begin();
            }
            ans[index] = *it;
            s.erase(it);
        }
        return ans;
    }
};
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