<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 findIndex(self, number, s):
        low = 0
        high = len(s)
        while low < high:
            mid = low + (high -low)//2
            if s[mid] < number:
                low = mid + 1
            else:
                high = mid
        
        s.insert(low, number)
        return low

    def countSmaller(self, nums: List[int]) -> List[int]:
        if len(nums) == 0:
            return None
        ans = [0 for i in range(len(nums))]
        s = []
        for i in range(len(nums)-1, -1, -1):
            if len(s) == 0:
                ans[i] = 0
                s.append(nums[i])
                continue
            
            index = self.findIndex(nums[i], s)
            ans[i] = index
        
        return ans
    



        
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