<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:
    void update(vector<int> &t, int v, int tl, int tr, int pos){
        if(tl==tr)
        {
            t[v]++;
            return;
        }
        int tm = (tl + tr)/2;
        if(pos<=tm)
            update(t, 2*v, tl, tm, pos);
        else
            update(t, 2*v+1, tm+1, tr, pos);

        t[v]=t[2*v]+t[2*v+1];

    }

    int query(int v, int tl, int tr, int l, int r, vector<int> &t)
    {
        if(r<tl || l>tr) return 0;
        if(l<=tl &&  r>=tr) return t[v];

        int tm=(tl+tr)/2;

        return query(2*v, tl, tm, l, r, t) + query(2*v+1, tm+1, tr, l, r, t);
    }
    vector<int> countSmaller(vector<int>& nums) {
        int n=nums.size();
        vector<int> t(400005, 0);
        for(int i=0; i<n ; i++) nums[i]+=10000;
        vector<int> ans;
        for(int i=n-1; i>=0; i--){
            ans.push_back(query(1, 0, 20000, 0, nums[i]-1, t));
            update(t, 1,0,20000, nums[i]);
        }

        reverse(ans.begin(), ans.end());

        return ans;
    }
};
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