<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) {
           unordered_map<int,int> mp;
            vector<pair<int,int>> v1;
            vector<pair<int,int>> v2;
            for(int i=0;i<nums2.size();i++)  {
                v2.push_back({nums2[i],i});
                v1.push_back({nums1[i],i});
            }
            sort(v1.begin(),v1.end());

            sort(v2.begin(),v2.end());
           
            int i=nums1.size()-1;
            int j=i;
            while(i>=0&&j>=0){
                if(v1[i].first>v2[j].first){
                    mp[v2[j].second]=v1[i].second;
                    i--;
                }
                j--;

            }
            j=nums1.size()-1;
            while(i>=0&&j>=0){

                if(mp.find(v2[j].second)==mp.end()){
                mp[v2[j].second]=v1[i].second;
                i--;}
                j--;

            }
            vector<int> ans;
            for(int i=0;i<nums2.size();i++){
                 ans.push_back(nums1[mp[i]]);

            }
            return ans;
    }
};
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