<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]:
        n = len(nums)
        res = [0] * n
        
        def merge_sort(arr):
            if len(arr) <= 1:
                return arr
            m = len(arr)//2
            left = merge_sort(arr[:m])
            right = merge_sort(arr[m:])
            return merge(left, right)
        
        def merge(A, B):
            merged = []
            ps = 0
            i = j = 0
            while i < len(A) and j < len(B):
                if nums[A[i]] > nums[B[j]]:
                    ps += 1
                    merged.append(B[j])
                    j += 1
                else:
                    res[A[i]] += ps
                    merged.append(A[i])
                    i += 1
            while i < len(A):
                res[A[i]] += ps
                merged.append(A[i])
                i += 1
            while j < len(B):
                merged.append(B[j])
                j += 1
            return merged
        merge_sort(list(range(n)))
        return res
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