<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>ans;
    void merge(vector<pair<int,int>>&arr,int s,int mid,int e){
        vector<pair<int,int>>temp(e-s+1);
        int i=s,j=mid+1,idx=0;
        while(i<=mid && j<=e){
            if(arr[i].first >arr[j].first){
                temp[idx++]=arr[j];
                // ans[arr[i].first]++;
                  j++;
            }else{
              // now right side is equal or more
              ans[arr[i].second]+=j-mid-1;
              temp[idx++]=arr[i];
              i++;

            }
        }
        while(i<=mid){
              temp[idx++]=arr[i];
              ans[arr[i].second]+=j-mid-1;
              i++;
        }
        while(j<=e){
              temp[idx++]=arr[j];
              j++;

        }
        for(int i=0;i<temp.size();i++){
            arr[s+i]=temp[i];
        }
    }
    void ms(vector<pair<int,int>>&arr,int s,int e){
        if(s>=e)
        return;
        int mid=(e-s)/2 +s;
        ms(arr,s,mid);
        ms(arr,mid+1,e);

        merge(arr,s,mid,e);
    }
    vector<int> countSmaller(vector<int>& nums) {
        vector<pair<int,int>>arr;
        for(int i=0;i<nums.size();i++){
            arr.push_back({nums[i],i});
        }
        int n=nums.size();
        ans.resize(n,0);
        ms(arr,0,n-1);
        return ans;
        
    }
};
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