<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) {
        priority_queue<pair<int,int>, vector<pair<int,int>>, greater<>> n1, n2;
        for(int i = 0; i < nums1.size(); ++i)
        {
            n1.push({nums1[i],i});
            n2.push({nums2[i],i});
        }

        priority_queue<pair<int,int>, vector<pair<int,int>>, greater<>> kalan;
        while(!n2.empty())
        {
            if(n1.empty()){
                nums1[n2.top().second] = kalan.top().first;
                n2.pop();
                kalan.pop();
            }
            else{
                if(n1.top().first > n2.top().first){
                    nums1[n2.top().second] = n1.top().first;
                    n2.pop();
                    n1.pop();
                } else {
                    kalan.push(n1.top());
                    n1.pop();
                }     
            }
        }
        return nums1;
    }
};
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