<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:
    def countSmaller(self, nums: List[int]) -> List[int]:
        d=[0]*len(nums)
        def merge(left,right):
            l=r=0
            n1,n2=len(left),len(right)
            a=[]
            while l<n1 and r<n2:
                if left[l][0]>right[r][0]:
                    a.append(left[l])
                    d[left[l][1]]+=n2-r
                    l+=1
                else:
                    a.append(right[r])
                    r+=1
                if r==n2:
                    a+=left[l:]
                if l==n1:
                    a+=right[r:]
            return a
        def divide(l,h):
            if l==h:
                return [[nums[l],l]]
            m=(l+h)//2
            return merge(divide(l,m),divide(m+1,h))
        a=divide(0,len(nums)-1)
        print(a)
        print(d)
        return d


        
47765