<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) {
        unordered_multiset s(nums1.begin(), nums1.end());
        sort(nums1.begin(), nums1.end());
        vector<pair<int, int>> nums2idx;
        for (int i = 0; i <nums2.size(); ++i) nums2idx.push_back(make_pair(nums2[i], i));
        sort(nums2idx.begin(), nums2idx.end());
        int p = 0, q = 0;
        vector<int> ans(nums1.size(), -1);
        while (q < nums2idx.size() && p < nums1.size()) {
            while (p < nums1.size() && nums1[p] <= nums2idx[q].first) p++;
            if (p < nums1.size()) {ans[nums2idx[q].second] = nums1[p];
            auto const it = s.find(nums1[p]);
            s.erase(it);q++;p++;}
        }
        auto it = s.begin();
        for (int i = 0; i < ans.size(); ++i) {
            if (ans[i] == -1) {
                ans[i] = *it;
                ++it;
            }
        }
        return ans;
    }
};
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