Leetcode: calculate the number of elements less than the current element on the right (sortedlist+bisect\u left)

Ideas ：
Look in the opposite direction
use sortedlist Add , then bisectleft Find something smaller than yourself

src：

from sortedcontainers import SortedList

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

        #  Reverse traversal 
        for i in range(n - 1, -1, -1):
            #  Find something smaller than your right 
            less = s.bisect_left(nums[i])
            #  Put it in the answer 
            ans[i] = less
            #  Add yourself 
            s.add(nums[i])
        
        return ans

summary ：
sortedlist + bisectleft The combination of