<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>

// // approach 1 : using N*N approach , it will give tle 

// class Solution {
// public:
//     vector<int> countSmaller(vector<int>& nums) {
        
//         vector<int> ans;

//         for(int i = 0 ; i + 1 < nums.size() ; i++){

//             int count = 0;

//             for(int j = i + 1 ; j < nums.size() ; j++){

//                 if(nums[j] < nums[i])       count++;
//             }

//             ans.push_back(count);
//         }

//         ans.push_back(0);

//         return ans;
//     }
// };










// appraoch 2 : using ordered_set 

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;

#define pbds tree<pair<int,int>, null_type,less<pair<int,int>>, rb_tree_tag,tree_order_statistics_node_update>
// order_of_key (k) : Number of items strictly smaller than k .
// find_by_order(k) : K-th element in a set (counting from zero).

// tc : O(NlogN) , sc : O(N)
class Solution {
public:
    vector<int> countSmaller(vector<int>& nums) {
        
        vector<int> ans;

        pbds st;

        for(int i = 0 ; i < nums.size() ; i++)      st.insert({nums[i] , i});     // bcoz 1 ele 1 se jyada baar aa sakta hn , isiliye pair me index karke daal rha

        for(int i = 0 ; i + 1 < nums.size() ; i++){

            st.erase({nums[i] , i});

            int count = st.order_of_key({nums[i] , i});

            ans.push_back(count);
        }

        ans.push_back(0);

        return ans;
    }
};
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