<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>

/*

nums1[] = [2,7,11,15]
nums2[] = [1,10,4,11]

sortedNums1[] = [2,11,7]

*/

class Solution {
public:
    vector<int> advantageCount(vector<int>& nums1, vector<int>& nums2) {
        int n = nums1.size();
        vector<pair<int, int>> nums1C(n), nums2C(n);
        vector<int> remaining, res(n);

        for(int i=0 ; i<n ; ++i) {
            nums1C[i] = {nums1[i], i};
            nums2C[i] = {nums2[i], i};
        }

        sort(nums1C.begin(), nums1C.end());
        sort(nums2C.begin(), nums2C.end());

        int i = 0, j = 0; 
        unordered_map<int, int> foundMapping;

        while(i<n && j<n) {
            while(i<n && nums1C[i].first <= nums2C[j].first) {
                remaining.push_back(nums1C[i].first);
                i++;
            }
            if(i == n) break;
            foundMapping[nums2C[j].second] = nums1C[i].second; 
            j++;
            i++;
        } 

        int k = 0;
        for(int i=0 ; i<n ; ++i) {
            if(foundMapping.find(i) != foundMapping.end()) {
                res[i] = nums1[foundMapping[i]];
            } else {
                res[i] = remaining[k++];
            }
        }

        return res;
    }
};
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