<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>

from bisect import bisect_left
class Solution:
    def countSmaller(self, nums: List[int]) -> List[int]:
        def merge(l1, l2):
            res = []
            n = len(l1)
            i = j = 0
            while i < n:
                if j == n or l1[i] < l2[j]:
                    res.append(l1[i])
                    i += 1
                else:
                    res.append(l2[j])
                    j += 1
            return res + l2[j:]
        stk = []
        res = []
        for i in nums[::-1]:
            temp = 0
            for l in stk:
                temp += bisect_left(l, i)
            res.append(temp)
            l = [i]
            while stk and len(stk[-1]) == len(l):
                l = merge(l, stk.pop())
            stk.append(l)
        return res[::-1]
1785