<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 {
    private:
        void swap(int &a, int &b) {
            int temp = a;
            a = b;
            b = temp;
        }
        int patition(vector<int> &array, vector<int> &indexes, int left, int right) {
            int pivotVal = array[right], i = left;
            for (int j = left; j < right; j++) {
                if (array[j] < pivotVal) {
                    swap(array[i], array[j]);
                    swap(indexes[i], indexes[j]);
                    i++;
                }
            }
            swap(array[i], array[right]);
            swap(indexes[i], indexes[right]);
            return i;
        }
        void quickSort(vector<int> &array, vector<int> &indexes, int left, int right) {
            if (left < right) {
                int pivot = patition(array, indexes, left, right);
                quickSort(array, indexes, left, pivot - 1);
                quickSort(array, indexes, pivot + 1, right);
            }
        }
    public:
        vector<int> advantageCount(vector<int> &nums1, vector<int> &nums2) {
            int len = nums1.size();
            vector<int> indexes(len, 0);
            for (int i = 0; i < len; i++) indexes[i] = i;
            sort(nums1.begin(), nums1.end());
            quickSort(nums2, indexes, 0, len - 1);

            vector<int> result(len, 0);
            int minPtr1 = 0, maxPtr1 = len - 1;
            int maxPtr2 = len - 1;
            while (maxPtr2 >= 0) {
                if (nums2[maxPtr2] >= nums1[maxPtr1]) {  // disadvantage
                    result[indexes[maxPtr2]] = nums1[minPtr1];
                    minPtr1++;
                } else if (nums2[maxPtr2] < nums1[maxPtr1]) {   // advantage
                    result[indexes[maxPtr2]] = nums1[maxPtr1];
                    maxPtr1--;
                }
                maxPtr2--;
            }

            return result;
        }
};
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