<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>

#define it  vector<vector<int>>::iterator
class Solution {
    vector<vector<int>> vec;
    void helper(it l, it r, vector<int>& res) {
        if(r-l <= 1)
            return;
        it m = l + (r-l) / 2;
        helper(l, m, res);
        helper(m, r, res);
        for(it i = l, j = m; i < m; i++) {
            while(j < r && (*i)[0] > (*j)[0])
                j++;
            res[(*i)[1]] += j-m;
        }
        inplace_merge(l, m, r);
    }
public:
    vector<int> countSmaller(vector<int>& nums) {
        for(int i = 0; i < nums.size(); i++) {
            vec.push_back({nums[i], i});
        }
        vector<int> res(nums.size(), 0);
        helper(vec.begin(), vec.end(), res);
        return res;
    }
};
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