<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]:        
        res  = [0] * len(nums)                                    #note1
        enum = list(enumerate(nums))                              #note2
        
        self.mergeSort(enum, 0, len(nums) - 1, res)
        return res
    
    def mergeSort(self, enum, start, end, res):
        if start >= end: return
        
        mid = start + (end - start) // 2
        self.mergeSort(enum, start, mid, res)
        self.mergeSort(enum, mid + 1, end, res)
        self.merge(enum, start, mid, end, res)
    
    def merge(self, enum, start, mid, end, res):
        p, q = start, mid + 1
        inversion_count = 0                                      #note3
        temp = []
        
        while p <= mid and q <= end:
            if enum[p][1] <= enum[q][1]:
                temp.append(enum[p])
                res[enum[p][0]] += inversion_count               #note5
                p += 1
            else:
                temp.append(enum[q])
                inversion_count += 1                             #note4
                q += 1
        
        while p <= mid:
            temp.append(enum[p])
            res[enum[p][0]] += inversion_count                  #note6
            p += 1
        
        while q <= end:         
            temp.append(enum[q])
            q += 1
        
        enum[start:end+1] = temp                                #note7        
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