<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]:
        lowest = min(nums)
        for i in range(len(nums)):
            nums[i] = nums[i]-lowest+1
        right = [0]*(max(nums)+1)
        for value in nums:
            right[value] += 1
        node = [0]*(len(right)*4+1)
        def build(arr, node, curr, l, r):
            if l == r:
                node[curr] = arr[l]
                return
            mid = l + (r-l)//2
            build(arr, node, curr*2+1, l, mid)
            build(arr, node, curr*2+2, mid+1, r)
            node[curr] = node[curr*2+1] + node[curr*2+2]
        def query(arr, node, curr, l, r, ll, rr):
            if l > rr or ll > r:
                return 0
            if l >= ll and rr >= r:
                return node[curr]
            mid = l + (r-l)//2
            left = query(arr, node, curr*2+1, l, mid, ll, rr)
            right = query(arr, node, curr*2+2, mid+1, r, ll, rr)
            return left + right
        def update(arr, node, curr, l, r, index, val):
            if l == r:
                node[curr] = val
                arr[index] = val
                return
            mid = l + (r-l)//2
            if index > mid:
                update(arr, node, curr*2+2, mid+1, r, index, val)
            else:
                update(arr, node, curr*2+1, l, mid, index, val)
            node[curr] = node[curr*2+1] + node[curr*2+2]
        ans = []
        build(right, node, 0, 0, len(right)-1)
        for i, value in enumerate(nums):
            ans.append(query(right, node, 0, 0, len(right)-1, 0, value-1))
            update(right, node, 0, 0, len(right)-1, value, right[value]-1)
            


        return ans


3604