<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]:

        counts=[0]*len(nums) # to save counts
        idx_nums=[(nums[i],i) for i in range(len(nums))]

        desc_nums,counts=self.merge(idx_nums,counts)

        return counts

    def merge(self,nums,counts):

        if len(nums)==1:
            return nums,counts

        left,counts=self.merge(nums[:len(nums)//2],counts)
        right,counts=self.merge(nums[len(nums)//2:],counts)

        cur_stack=[]

        while left and right:
            if left[0]>right[0]:
                counts[left[0][1]]+=len(right)
                cur_stack.append(left.pop(0))
            else:
                cur_stack.append(right.pop(0)) 

        if left:
            cur_stack.extend(left)

        if right:
            cur_stack.extend(right) 

        return cur_stack,counts                          
        
        
3195