<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) {
        sort(nums1.begin(),nums1.end());
        vector<int>temp=nums2;
        sort(temp.begin(),temp.end());
        int in=0;
        unordered_map<int,vector<int>>m;

        for(int i=0;i<nums1.size();i++)
        {
            if(in<temp.size() && nums1[i]>temp[in])m[temp[in]].push_back(nums1[i]),nums1[i]=-1,in++;
        }

        in=0;

        for(int i=0;i<nums2.size();i++)
        {
            if(m[nums2[i]].size())temp[i]=m[nums2[i]].back(),m[nums2[i]].pop_back();
            else {
                while(nums1[in]==-1)in++;
                temp[i]=nums1[in++];
            }
        }
        return temp;
    }
};
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