README.md

315. Count of Smaller Numbers After Self

Given an integer array nums, return an integer arraycountswherecounts[i]is the number of smaller elements to the right ofnums[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

Solutions (Python)

1. Solution

fromsortedcontainersimportSortedListclassSolution: defcountSmaller(self, nums: List[int]) ->List[int]: sortednums=SortedList() counts= [0] *len(nums) foriinrange(len(nums) -1, -1, -1): counts[i] =sortednums.bisect_left(nums[i]) sortednums.add(nums[i]) returncounts