# 315. Count of Smaller Numbers After Self

#### Hard

***

You are given an integer array `nums` and you have to return a new `counts` array. The `counts` array has the property where `counts[i]` is the number of smaller elements to the right of `nums[i]`.

 

**Example 1:**

```
Input: nums = [5,2,6,1]
Output: [2,1,1,0]
Explanation:
To the right of 5 there are 2 smaller elements (2 and 1).
To the right of 2 there is only 1 smaller element (1).
To the right of 6 there is 1 smaller element (1).
To the right of 1 there is 0 smaller element.
```

**Example 2:**

```
Input: nums = [-1]
Output: [0]
```

**Example 3:**

```
Input: nums = [-1,-1]
Output: [0,0]
```

 

**Constraints:**

* `1 <= nums.length <= 105`
* `-104 <= nums[i] <= 104`

```python
class Solution:
    def countSmaller(self, nums: List[int]) -> List[int]:
        # Complexity : O(n^2)
        # Here we are using a simple idea that when we insert element in sorted array index at which position element gets inserted will tell then number of elements which are greater or less than current number.
        sorted_arr = []
        result = []
        for num in nums[::-1]:
            index = bisect_left(sorted_arr, num)
            result.append(index)
            sorted_arr.insert(index, num)
        return result[::-1]
        
```