<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) {
    //     multiset<int> s;
    //     int n=nums1.size();
    //     for(int i=0;i<n;i++){
    //         s.insert(nums1[i]);
    //         nums1[i]=-1;
    //     }
    //     for(int i=0;i<n;i++){
    //         auto it=s.upper_bound(nums2[i]);
    //         if(it!=s.end()){
    //             nums1[i]=*it;
    //             s.erase(it);
    //         }
    //     }
    //     for(int i=0;i<n;i++){
    //         if(nums1[i]==-1){
    //             auto it=s.begin();
    //             nums1[i]=*it;
    //             s.erase(it);
    //         }
    //     }
    //     return nums1;
    // }

    vector<int> advantageCount(vector<int>& nums1, vector<int>& nums2) {
        int n=nums1.size();
        vector<pair<int,int>> v;
        for(int i=0;i<n;i++){
            v.push_back({nums2[i],i});
        }
        sort(v.begin(),v.end());
        sort(nums1.begin(),nums1.end());
        vector<int> ans(n,-1);
        int i=0,j=0;
        vector<int> vis(n,0);
        while(i<n and j<n){
            if(v[j].first<nums1[i]){
                ans[v[j].second]=nums1[i];
                vis[i]=1;
                i++;
                j++;
            }else i++;
        }
        for(int k=0;k<n;k++){
            if(vis[k]==0){
                ans[v[j].second]=nums1[k];
                j++;
            }
        }
        return ans;
    }
};
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