<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());
        unordered_map<int,vector<int>>mp;
        int n = nums1.size();
        for(int i=0;i<n;i++)
        {
            mp[nums2[i]].push_back(i);
        }
        sort(nums2.begin(),nums2.end());
        int i = 0,j = 0;
        vector<bool> v(n,false);
        vector<int> result(n);
        while(i<n && j<n)
        {
            if(nums1[i]>nums2[j])
            {
                result[mp[nums2[j]][0]] = nums1[i];
                mp[nums2[j]].erase(mp[nums2[j]].begin());
                j += 1;
            }
            else
            {
                v[i] = true;
            }
            i += 1;
        }
        for(int k=j;k<n;k++)
        {
            int idx = 0;
            for(int l=0;l<n;l++)
            {
                if(v[l]==true)
                {
                    idx = l;
                    v[l] = false;
                    break;
                }
            }
            result[mp[nums2[k]][0]] = nums1[idx];
            mp[nums2[k]].erase(mp[nums2[k]].begin());
        }
        return result;
    }
};
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