Program to return number of smaller elements at right of the given list in Python
Suppose we have a list of numbers called nums, we will create a new list where each element in
the new list is the number of smaller elements to the right hand side of that element in the
original input list.
So, if the input is like nums = [4, 5, 9, 7, 2], then the output will be [1, 1, 2, 1, 0], as there is 1
smaller element to the right of 4, there is 1 smaller element to the right of 5, there are 2
smaller elements to the right of 9, there is 1 smaller element to the right of 7, there are no
smaller elements to the right of 2.
To solve this, we will follow these steps −
res := a new list, inc := a new list
while nums is not empty, do
num := delete last element from nums
insert left most index to insert num in inc at the end of res
sorted list after inserting num in inc
return a list res[from index 0 to end]
Let us see the following implementation to get better understanding−
Example
Live Demo
import bisect
class Solution:
def solve(self, nums):
res, inc = [], []
while nums:
num = nums.pop()
res.append(bisect.bisect_left(inc, num))
bisect.insort(inc, num)
return res[::-1]
ob = Solution()
nums = [4, 5, 9, 7, 2]
print(ob.solve(nums))
Input
[4, 5, 9, 7, 2]
Output
[1, 1, 2, 1, 0]
Published on 10-Oct-2020 15:13:43
