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Abhishek Chaudhary

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# Count of Smaller Numbers After Self

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`

SOLUTION:

``````import bisect

class Solution:
def countSmaller(self, nums: List[int]) -> List[int]:
n = len(nums)
smallers = []
for i in range(n - 1, -1, -1):
currctr = bisect.bisect_left(smallers, nums[i])
bisect.insort(smallers, nums[i])
nums[i] = currctr
return nums
``````