<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):
        def merge_sort(indexes):
            half = len(indexes) // 2
            if half:
                left, right = merge_sort(indexes[:half]), merge_sort(indexes[half:])
                for i in range(len(indexes))[::-1]:
                    if not right or left and nums[left[-1]] > nums[right[-1]]:
                        smaller[left[-1]] += len(right)
                        indexes[i] = left.pop()
                    else:
                        indexes[i] = right.pop()
            return indexes
        
        smaller = [0] * len(nums)
        merge_sort(list(range(len(nums))))
        return smaller

# Example usage:
solution = Solution()
nums = [5, 2, 6, 1]
print(solution.countSmaller(nums))  # Output: [2, 1, 1, 0]
58952