<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 update(index, value, tree, size):
            while index < size:
                tree[index] += value
                index += index & -index

        def query(index, tree):
            sum = 0
            while index > 0:
                sum += tree[index]
                index -= index & -index
            return sum

        # Coordinate compression
        rank = {val: i + 1 for i, val in enumerate(sorted(set(nums)))}
        
        # Initialize BIT
        size = len(rank) + 1
        tree = [0] * size
        
        # Result list
        result = []
        
        # Traverse the nums array in reverse order
        for num in reversed(nums):
            pos = rank[num]
            result.append(query(pos - 1, tree))
            update(pos, 1, tree, size)
        
        return result[::-1]

# Example usage
solution = Solution()
print(solution.countSmaller([5, 2, 6, 1]))  # Output: [2, 1, 1, 0]
print(solution.countSmaller([-1]))          # Output: [0]
print(solution.countSmaller([-1, -1]))      # Output: [0, 0]

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