<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>

using qpair = pair<int, int>;

class Solution {
public:
    vector<int> advantageCount(vector<int>& a, vector<int>& b) {
        int len = a.size(), curr;
        priority_queue<int, vector<int>, greater<int>> qa, leftovers;
        priority_queue<qpair, vector<qpair>, greater<qpair>> qb;
        for (int i = 0; i < len; i ++) {
            qa.push(a[i]);
            qb.push({b[i], i});
        }
        while (qa.size()) {
            curr = qa.top();
            if (curr > qb.top().first) {
                a[qb.top().second] = curr;
                qa.pop(), qb.pop();
            }
            else {
                leftovers.push(curr);
                qa.pop();
            }
        }
        while (leftovers.size()) {
            curr = leftovers.top();
            a[qb.top().second] = curr;
            leftovers.pop(), qb.pop();
        }
        return a;
    }
};
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