<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 helper(L, l, r):
            if l >= r:
                return None
            mid = l + (r-l)//2
            helper(L, l, mid)
            helper(L, mid+1, r)

            i, j, p = l, mid+1, mid+1
            # temp will be used for merged sorted list
            temp = []
            cnt = 0
            while i <= mid:
                while p <= r and L[i][0] > L[p][0]:
                    cnt += 1
                    p += 1
                res[L[i][1]] += cnt
                while j <= r and L[i][0] > L[j][0]:
                    temp.append(L[j])
                    j += 1
                temp.append(L[i])
                i += 1
            while j <= r:
                temp.append(L[j])
                j += 1
            L[l:r+1] = temp[:]

        newnums = []
        for i, n in enumerate(nums):
            newnums.append((n, i))

        res = [0] * len(nums)
        helper(newnums, 0, len(newnums)-1)

        return res

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