<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> mergeSort(vector<int>&cnt,vector<int>&nums,int l,int r){
    if(l==r)return {l};
    vector<int>a=mergeSort(cnt,nums,l,(l+r)/2);
    vector<int>b=mergeSort(cnt,nums,(l+r)/2+1,r);
    int i=0,j=0;
    for(int i=0;i<a.size();i++){
        while(j<b.size()&&nums[a[i]]>nums[b[j]])j++;
        cnt[a[i]]+=j;
    }
    vector<int>res;
    i=0,j=0;
    while(i<a.size()||j<b.size()){
        if(i==a.size()){
            while(j<b.size())res.push_back(b[j++]);
        }else if(j==b.size()){
            while(i<a.size())res.push_back(a[i++]);
        }else{
            if(nums[a[i]]<nums[b[j]]){
                res.push_back(a[i++]);
            }else{
                res.push_back(b[j++]);
            }
        }
    }
    return res;
}
    vector<int> countSmaller(vector<int>& nums) {
        int n=nums.size();
        vector<int>cnt(n,0);
        mergeSort(cnt,nums,0,n-1);
        return cnt;
    }
};
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