<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]:
        def mergesort(numsF): 
            if len(numsF) <= 1: return numsF
            mid = len(numsF) // 2
            left = mergesort(numsF[:mid])
            right = mergesort(numsF[mid:])
            merge(left, right, numsF)
            return numsF

        def merge(a, b, arr): 
            count, i, j, k = 0, 0, 0, 0
            # arr = [None]*(len(a) + len(b))
            while i < len(a) and j < len(b): 
                if a[i][0] <= b[j][0]: 
                    arr[k] = [a[i][0], a[i][1] + count, a[i][2]]            
                    i += 1
                else:             
                    count += 1
                    cur = [b[j][0], b[j][1], b[j][2]]
                    arr[k] =  cur         
                    j += 1
                k += 1

            while i < len(a): 
                arr[k] = [a[i][0], a[i][1] + count, a[i][2]]       
                i += 1
                k += 1
            while j < len(b): 
                arr[k] = [b[j][0], b[j][1], b[j][2]]            
                j += 1
                k += 1

        numsF = []      # numsF[i] = [num, count, idx]
        for i, num in enumerate(nums):
            numsF.append([num, 0, i])    

        mergesort(numsF)
        numsF.sort(key = lambda x: x[2])
        return [x[1] for x in numsF]
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