<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();
        sort(nums1.begin(),nums1.end());
        priority_queue<pair<int,int>,vector<pair<int,int>>,greater<pair<int,int>>>pq;
        int m=nums2.size();
        for(int i=0;i<m;i++){
            pq.push({nums2[i],i});
        }
        int i=0;
        stack<int>sta;
        vector<int>ans(n,-1);
        while(!pq.empty()&&(i<n)){
            pair<int,int>a=pq.top();
            pq.pop();
            while((i<n)&&nums1[i]<=a.first){
                      sta.push(nums1[i]);
                i++;
          
            }
            if(i<n){
            ans[a.second]=nums1[i];
            i++;
            }
        }
        for(int i=0;i<n;i++){
            if(ans[i]==-1){
                ans[i]=sta.top();
                sta.pop();
            }
        }
        return ans;
        
    }
};
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