<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>& a, vector<int>& b) {
        
        sort(a.begin(),a.end());
        map<int,int> m;
        for(auto x:a)m[x]++;
        vector<int> ans(a.size());
         
         priority_queue<pair<int,int>, vector< pair<int,int> >> pq;

         for(int i=0;i<b.size();i++) pq.push({b[i],i});
          int i=0;
         int j=a.size()-1;
        while(pq.size()>0)
        {    
               int val=pq.top().first;
               int idx=pq.top().second;
               if(val<a[j])
               {
                 ans[idx]=a[j];
                 j--;
               }
               else
               {
                 ans[idx]=a[i];
                 i++;
               }
               
               pq.pop();
        }

        return ans;
    }
};
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