<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]:
        # This list will hold the count of smaller numbers for each index
        count = [0] * len(nums)
        
        # We define a list of tuples (num, index) to keep track of the original indices
        enumerated_nums = list(enumerate(nums))
        
        # We implement the merge sort logic with a modified merging step
        def merge_sort(enum):
            if len(enum) <= 1:
                return enum
            
            mid = len(enum) // 2
            left = merge_sort(enum[:mid])
            right = merge_sort(enum[mid:])
            
            # Merge the two halves and count the smaller elements on the right
            merged = []
            i = j = 0
            while i < len(left) and j < len(right):
                if left[i][1] > right[j][1]:
                    # Since right[j] is smaller than left[i], increment the count
                    count[left[i][0]] += len(right) - j
                    merged.append(left[i])
                    i += 1
                else:
                    merged.append(right[j])
                    j += 1
            
            # Append the remaining elements
            merged.extend(left[i:])
            merged.extend(right[j:])
            
            return merged
        
        # Call merge_sort on the enumerated list
        merge_sort(enumerated_nums)
        
        return count

# 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]

1603