<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,n):
        self.bits = [0]*(n+1)
        
    def update(self,bit):
        while(bit > 0):
            self.bits[bit] += 1
            bit -= (bit & -bit)
            
            
    def greaterCount(self,idx):
        count = 0
        while(idx < len(self.bits)):
            count += self.bits[idx]
            idx += (idx & -idx)
        return count
    
class Solution:
    def countSmaller(self, nums: List[int]) -> List[int]:
        N = len(nums)
        maxel = max(nums)
        obj = FenWicktree(N)
        ans = []
        temp = [(nums[i],i) for i in range(N)]
        temp.sort(key = lambda x:x[0])
        nums = [[i,0] for i in nums]
        for i in range(N):
            nums[temp[i][1]][1] = N-i
        ans.append(0)
        obj.update(nums[-1][1])
        for i in range(N-2,-1,-1):
            el,rank = nums[i][0],nums[i][1]
            c = obj.greaterCount(rank)
            ans.append(c)
            obj.update(rank)
        ans.reverse()
        return ans
            
53583