<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>

import copy
class Solution:
    def countSmaller(self, nums: List[int]) -> List[int]:
        """  
        return array of ints of how many are less than it on the right side


        merge sort
            l1
            l2
            if l1 is bigger than l2 or l2 iempty
                increment its counter
        
        """
        arr, ans = sorted(nums), []           #  <-- 1)
        
        for num in nums:
            i = bisect_left(arr,num)          #  <-- 2a)
            ans.append(i)                     #  <-- 2b)
            del arr[i]                        #  <-- 2c)
            
        return ans  

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