<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:
    vector<int> countSmaller(vector<int>& nums) {
        int n = nums.size();
        vector<int> ans(n);
        
        array<int, 3> v[n];
        for(int i=0; i<n; ++i)
            v[i] = {nums[i], ans[i], i};
        mergeSort(0, n - 1, v);
        
        for(int i=0; i<n; ++i)
            ans[v[i][2]] = v[i][1];
        return ans;
    }
private:
    void mergeSort(int l, int r, array<int, 3>* v) {
        if(r <= l)
            return;
        
        int mid = l + ((r - l) >> 1);
        mergeSort(l, mid, v);
        mergeSort(mid + 1, r, v);
        for(int i=mid, j=r; i>=l && j>mid; --i){
            while(j > mid && v[i][0] <= v[j][0])
                --j;
            
            v[i][1] += j - mid;
        }
        
        inplace_merge(&v[0] + l, &v[0] + mid+1, &v[0] + r+1);
    }
};
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