Skip to content

315. Count of Smaller Numbers After Self

Difficulty Topics

Description

Given an integer array nums, return an integer array counts 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

count-of-smaller-numbers-after-self.py
from sortedcontainers import SortedList

class Solution:
    def countSmaller(self, nums: List[int]) -> List[int]:
        n = len(nums)
        res = [0] * n
        sl = SortedList([nums[-1]])

        for i in range(n - 2, -1, -1):
            index = sl.bisect(nums[i] - 1)

            res[i] = index

            sl.add(nums[i])

        return res