<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:
std::vector<int> advantageCount(std::vector<int>& nums1, std::vector<int>& nums2) {
    int n = nums1.size();
    std::priority_queue<std::pair<int, int>, std::vector<std::pair<int,int>>, std::less<>> maxpq;
    for (int i = 0; i < nums2.size(); i++) {
        maxpq.emplace(nums2[i], i);
    }
    std::sort(nums1.begin(), nums1.end());
    int left = 0, right = nums1.size() - 1;
    std::vector<int> res(nums1.size());
    while (!maxpq.empty()) {
        auto pair = maxpq.top(); maxpq.pop();
        int max_val = pair.first, i = pair.second;
        if (max_val < nums1[right]) {
            res[i] = nums1[right];
            right--;
        } else {
            res[i] = nums1[left];
            left++;
        }
    }
    return res;
}
};
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