<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>

#include <iostream>
#include <queue>
#include <vector>
#include <algorithm>

using namespace std;

class Solution {
public:
    vector<int> advantageCount(vector<int>& nums1, vector<int>& nums2) {
        // Step 1: Sort nums1
        sort(nums1.begin(), nums1.end());

        // Step 2: Insert elements into priority queue (max heap)
        priority_queue<vector<int>> q;
        for(int i = 0; i < nums2.size(); i++){
            q.push({nums2[i], i});
        }

        // Step 3: Compare each value in nums1 to each value in nums2
        int l = 0;
        int r = nums1.size() - 1;
        vector<int> ans(nums1.size(), 0);
        while(!q.empty()){
            int val = q.top()[0];
            int index = q.top()[1];
            q.pop();

            if(nums1[r] > val){
                ans[index] = nums1[r];
                r--;
            } else {
                ans[index] = nums1[l];
                l++;
            }
        }
        return ans;
    }
};


84200