<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):
        self.size = size
        self.tree = [0] * (size + 1)
    
    def update(self, index, delta):
        while index <= self.size:
            self.tree[index] += delta
            index += index & -index
    
    def query(self, index):
        tsum = 0
        while index > 0:
            tsum += self.tree[index]
            index -= index & -index
        return tsum

class Solution:
    def countSmaller(self, nums: List[int]) -> List[int]:
        # O(Nlog(N)) | O(N)
        sorted_nums = sorted(set(nums))
        rank_map = {num: i + 1 for i, num in enumerate(sorted_nums)}

        f = FenwickTree(len(nums))
        res = []

        for num in reversed(nums):
            rank = rank_map[num]
            res.append(f.query(rank - 1))
            f.update(rank, 1)
        
        return res[::-1]
1376