<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]:
        smaller_rights, sorted_nums = [], []
        count = 0  # Count of sorted integers.

        for num in nums[::-1]:  # Iteration: from rightmost to leftmost.
            if sorted_nums and num > sorted_nums[-1]:
                smaller_rights.append(count)
                sorted_nums.append(num)
            
            elif sorted_nums and num <= sorted_nums[0]:
                smaller_rights.append(0)
                sorted_nums.insert(0, num)
            
            else:
                smaller_right = self._count_smaller_right(num, sorted_nums, count)
                smaller_rights.append(smaller_right)
                sorted_nums.insert(smaller_right, num)
            
            count += 1
        
        return smaller_rights[::-1]  # Reverse to make order from leftmost to rightmost.
    
    @staticmethod
    def _count_smaller_right(target: int, sorted_integers: list[int], size: int) -> int:
        if size == 0:
            return 0

        back_idx, front_idx = 0, size - 1
        while back_idx <= front_idx:
            mid_idx = (back_idx + front_idx) // 2
            if sorted_integers[mid_idx] < target:
                back_idx = mid_idx + 1
                continue
            front_idx = mid_idx - 1

        return back_idx  # Number of ints < target, implying insertion idx.

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