<p>Given an integer array <code>nums</code>, return<em> an integer array </em><code>counts</code><em> where </em><code>counts[i]</code><em> is the number of smaller elements to the right of </em><code>nums[i]</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [5,2,6,1]
<strong>Output:</strong> [2,1,1,0]
<strong>Explanation:</strong>
To the right of 5 there are <b>2</b> smaller elements (2 and 1).
To the right of 2 there is only <b>1</b> smaller element (1).
To the right of 6 there is <b>1</b> smaller element (1).
To the right of 1 there is <b>0</b> smaller element.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [-1]
<strong>Output:</strong> [0]
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [-1,-1]
<strong>Output:</strong> [0,0]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution {
public:
      #define iterator vector<vector<int>>::iterator
    void sort_count(iterator l, iterator r, vector<int>& count) {
        if (r - l <= 1) return;
        iterator m = l + (r - l) / 2;
        sort_count(l, m, count);
        sort_count(m, r, count);
        for (iterator i = l, j = m; i < m; i++) {
            while (j < r && (*i)[0] > (*j)[0]) j++;
            count[(*i)[1]] += j - m; // add the number of valid "j"s to the indices of *i
        }
        inplace_merge(l, m, r);
    }
    vector<int> countSmaller(vector<int>& nums) {
        vector<vector<int>> hold;
        int n = nums.size();
        for (int i = 0; i < n; ++i) hold.push_back(vector<int>({nums[i], i})); // "zip" the nums with their indices
        vector<int> count(n, 0);
        sort_count(hold.begin(), hold.end(), count);
        return count;
    }
};
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