<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) {
        vector<pair<int,int>>v;
        for(int i=0;i<nums2.size();i++){
            v.push_back({nums2[i],i});
        }
        // sort(v.begin(),v.end());
      multiset<int>s(nums1.begin(),nums1.end());vector<int>ans(nums1.size());
      
       for(int i=0;i<v.size();i++){
        int ii=v[i].second;
       auto it=s.upper_bound(v[i].first);
       if(it!=s.end()){
        ans[ii]=*it;
        s.erase(it);
       }
       else{
         
        ans[ii]=*s.begin();
        s.erase(s.begin());
       }
       }
       return ans;
    }
};
81200