<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]:
        # we wanna want a solution faster than O(n^2)
        # use binary search, by creating a arr (sorted), when going through the nums list
        # use the sorted array to find smaller elements.
        # delete the n in each iteration, so we can find only smaller elements in the right
        # For example, suppose nums = [5,2,6,1] Then arr = [1,2,5,6].
        # num = 5 => binsearch: arr = [1,2,/\5,6], i = 2 => ans = [2,_,_,_], del 5
        # num = 2 => binsearch: arr = [1,/\2,6],   i = 1 => ans = [2,1,_,_], del 2
        # num = 6 => binsearch: arr = [1,/\6],     i = 1 => ans = [2,1,1,_], del 6
        # num = 1 => binsearch: arr = [/\1],       i = 0 => ans = [2,
        
        arr, ans = sorted(nums), []           #  <-- 1)
        
        for n in nums:
            i = bisect_left(arr,n)          #  <-- 2a)
            ans.append(i)                     #  <-- 2b)
            del arr[i]                        #  <-- 2c)
            
        return ans                            #  <-- 3)
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