<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
        enum = list(enumerate(nums))

        self.mergeSort(enum, 0, n-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
        inv_count = 0

        while p<=mid and q<=end:
            if enum[p][1] <= enum[q][1]:
                res[enum[p][0]]+=inv_count
                p+=1
            else:
                inv_count += 1
                q += 1

        while p <= mid:
            res[enum[p][0]] += end-mid
            p+=1
        enum[start:end+1] = sorted(enum[start:end+1], key=lambda e:e[1])
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