<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:
int n =200005;
vector<int>tree;

 int sum(int ind){
          int res=0;

          while(ind>=1){
                res+=tree[ind];
                ind-=(ind&-ind);
          }
          return res;
 }

  void add(int i,int x){
                   for(int  k=i;k<=n;k+=(k&-k)){
                        tree[k]+=x;
                   }
  }


    vector<int> countSmaller(vector<int>& nums) {
vector<int>res(nums.size()) ;
          tree.resize(n+5,0);

          for(int i=nums.size()-1;i>=0;i--){
            nums[i]=nums[i]+10001;
            res[i]=sum(nums[i]-1);
               add(nums[i],1);
          }
         
          return res;
    }
};
142106