<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) {
        const int N = nums1.size();
        vector<int> index1(N);
        vector<int> index2(N);
        for (int i = 0; i < N; i++) {
            index1[i] = index2[i] = i;
        }

        sort(index1.begin(), index1.end(), [&nums1](auto lhs, auto rhs) {
            return nums1[lhs] < nums1[rhs];
        });
        sort(index2.begin(), index2.end(), [&nums2](auto lhs, auto rhs) {
            return nums2[lhs] < nums2[rhs];
        });

        vector<int> res(N);
        int left = 0;
        int right = N - 1;
        for (int i = 0; i < N; i++) {
            if (nums1[index1[i]] > nums2[index2[left]]) {
                res[index2[left++]] = nums1[index1[i]];
            } else {
                res[index2[right--]] = nums1[index1[i]];
            }
        }
        return res;
    }
};
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