<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>& a1, vector<int>& a2) {
        int n=a1.size();
        vector<pair<int,int>> v(n);
        for(int i=0;i<n;i++)
        {
            v[i]={a2[i],i};
        }
        sort(v.begin(),v.end());
        vector<int> ans(n,-1);
        priority_queue<int,vector<int>,greater<int>> q;
        for(auto &i:a1)
        {
            q.push(i);
        }
        int c=0;
        vector<int> waste;
        while(!q.empty())
        {
            int x=q.top();
            q.pop();
            if(x>v[c].first)
            {
                ans[v[c].second]=x;
                c++;
            }
            else
            {
                waste.push_back(x);
            }
        }
        c=0;
        for(int i=0;i<n;i++)
        {
            if(ans[i]==-1)
            {
                ans[i]=waste[c++];
            }
        }
        return ans;
    }
};
66200