<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 OldSegmentTree:

    def __init__(self, n):
        self.n = n
        self.tree = [0 for _ in range(2 * self.n)]
    
    def update(self, idx, val):
        idx += self.n
        self.tree[idx] += val
        while idx > 1:
            left = right = idx
            if idx % 2 == 1:
                # idx is the right child of the parent
                left -= 1
            else:
                # idx is the left child of the parent
                right += 1
            self.tree[idx // 2] = self.tree[left] + self.tree[right]
            idx = idx // 2
    
    def query(self, left, right):
        left += self.n
        right += self.n
        sum_val = 0
        while left < right:
            if left % 2 == 1:
                # right child of the parent, add the value here
                sum_val += self.tree[left]
                left += 1
            if right % 2 == 1:
                # right child of the parent, add the value of the left child here
                # NOTE: in such a way, we have sum over range [left, right) (exclusive of the right)
                right -= 1
                sum_val += self.tree[right]
            left = left // 2
            right = right // 2
        return sum_val


class SegmentTree(OldSegmentTree):
    
    def query(self, left, right):
        left += self.n
        right += self.n
        sum_val = 0
        while left <= right:
            if left % 2 == 1:
                # right child of the parent, add the value here
                sum_val += self.tree[left]
                left += 1
            if right % 2 == 0:
                # left child of the parent, add the value here
                sum_val += self.tree[right]
                right -= 1
            left = left // 2
            right = right // 2
        return sum_val


class Solution:
    def countSmaller(self, nums: List[int]) -> List[int]:
        max_val = 10**4
        offset = max_val
        n = 2 * max_val + 1
        tree = SegmentTree(n)
        res = []
        for num in nums[::-1]:
            # NOTE: we implement the SegTree query as [left, right], exclusive of the right side
            cnt = tree.query(0, num + offset - 1)
            res.append(cnt)
            tree.update(num + offset, 1)
        return res[::-1]
    
    def old_countSmaller(self, nums: List[int]) -> List[int]:
        max_val = 10**4
        offset = max_val
        n = 2 * max_val + 1
        tree = OldSegmentTree(n)
        res = []
        for num in nums[::-1]:
            # NOTE: we implement the SegTree query as [left, right], exclusive of the right side
            cnt = tree.query(0, num + offset)
            res.append(cnt)
            tree.update(num + offset, 1)
        return res[::-1]

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