<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 FenwickTree:
    def __init__(self, size, original_value):
        self.size = size
        self.tree = [original_value] * size

    def __lowbit(self, index):
        return index & -index

    def update(self, index, delta):
        while index < self.size:
            self.tree[index] += delta
            index += self.__lowbit(index)

    def query(self, index):
        ans = 0
        while index:
            ans += self.tree[index]
            index -= self.__lowbit(index)
        return ans


class Solution:
    def countSmaller(self, nums: List[int]) -> List[int]:
        transfer, ans = {num: i + 1 for i, num in enumerate(sorted(set(nums)))}, []
        ft = FenwickTree(len(transfer) + 1, 0)
        for num in reversed(nums):
            ans.append(ft.query(transfer[num] - 1))
            ft.update(transfer[num], 1)
        return ans[::-1]

        
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