<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:
    static bool compare(pair<int,int> A, pair<int,int> B)
    {
        return A.first < B.first;
    }

    vector<int> advantageCount(vector<int>& nums1, vector<int>& nums2) 
    {
        int n = nums1.size();

        //sort both the arrays, but keep index of nums 2
        vector<pair<int,int>> sorted_nums2(n);
        for(int i=0;i < n;i++)
            sorted_nums2[i] = {nums2[i], i};
        
        sort(nums1.begin(),nums1.end());
        sort(sorted_nums2.begin(), sorted_nums2.end(), compare);

        int l = 0, r = n-1;
        vector<int> ans(n);

        for(int i=n-1;i >=0 ;i--)
        {
            if(nums1[r] > sorted_nums2[i].first)
            {
                ans[sorted_nums2[i].second] = nums1[r];
                r--;
            }
            else
            {
                ans[sorted_nums2[i].second] = nums1[l];
                l++;
            }
        } 

        return ans;
    }
};
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