<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]:
        counts = [0] * len(nums) # to save counts
        indexed_nums = [(nums[key_index], key_index) for key_index in range(len(nums))] # change the nums into (number, index) pairs
        desc_nums, counts = self.mergeSort(indexed_nums, counts) # the helper function returns the descendingly sorted indexed_nums and the counts
        return counts
	
    def mergeSort(self, nums, counts):
        if len(nums) == 1:
            return nums, counts

        left, counts = self.mergeSort(nums[: len(nums) // 2], counts)
        right, counts = self.mergeSort(nums[len(nums) // 2 :], counts)
        cur_stack, counts = self.merge(left, right, counts)
			
        return cur_stack, counts

    def merge(self, left, right, counts):
        cur_stack = []

        while left and right:    
            if left[0][0] > right[0][0]:
                counts[left[0][1]] += len(right) # left[0] gives a (number, index) pair as we defined at the beginning, and left[0][1] returns the original index of the number.
                cur_stack.append(left.pop(0))
            else:
                cur_stack.append(right.pop(0))
                
        if left:
            cur_stack.extend(left)
        elif right: #else:
            cur_stack.extend(right)
			
        return cur_stack, counts
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