<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]:
        # count after is n^2 time
        # can we reduce to nlogn
        # merge sort combins two sorted arrays 
        # let A = [3,5,9] B = [7,11,12]
        # only smaller is 7 to 9
        # must go right to left though, so one smaller in B is used on many biggers in A
        # add the bigger first -> added = processed 
        # normal merge sort with one modification
        # merge sort -> pass in unsorted nums -> return sorted nums
        # recursively sort left and rigth halves
        # post orderly combine both 
        # combine both changed so that we loop from back, then return the reversed (On)
        # if head at A is bigger than head at B put add len B to ind A then, A in and next
        # edge cases -> A empty? just keep adding B
        # B empty? just keep adding A
        # A = [2,4,5] B = []
        # 

        ans = [0] * len(nums)
        def merge(arr):
            if len(arr) == 1:
                return arr
            m = len(arr)//2
            A = merge(arr[:m])
            B = merge(arr[m:])
            res = []
            for i in range(len(arr)):
                #print(f"merging arr {arr}, A is {A}, B is {B}")
                if not A or (B and B[-1][1] >= A[-1][1]):
                    res.append((B[-1][0], B[-1][1]))
                    B.pop()
                else:
                    ans[A[-1][0]] += len(B)
                    #print(f"added {len(B)} to ans: {ans}")
                    res.append((A[-1][0], A[-1][1]))
                    A.pop()
            return res[::-1]

        enum = [(i,n) for i,n in enumerate(nums)]
        merge(enum)
        return ans
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