Program to return number of smaller elements at right of the given list in Python

In Python, a list is a ordered and mutable data structure where elements are enclosed in square braces []. In some scenarios, we may need to count how many elements to the right of each element in the list are smaller than it.

Using Binary Search with bisect_left() Function

The bisect_left() Function of the bisect method is used locate the insertion point for an element in a sorted list to maintain the list's sorted order. The insertion point returned by bisect_left() is the index where the element can be inserted without violating the sort order by placing it before any existing entries with the same value.

Example-1

Following is the example in which we go through the input list from right to left by using the bisect_left() function to insert elements into a sorted list by tracking the position of insertion, which represents the count of smaller elements -

import bisect

def count_smaller_elements(arr):
    result = []
    sorted_list = []
    
    for num in reversed(arr):
        index = bisect.bisect_left(sorted_list, num)
        result.append(index)
        bisect.insort(sorted_list, num)
    
    return result[::-1]

# Sample input list
numbers = [5, 2, 6, 1]
print("Count of smaller elements to the right:", count_smaller_elements(numbers))

Here is the output of the above example -

Count of smaller elements to the right: [2, 1, 1, 0]

Example-2

Here is another example to return number of smaller elements at right of the given list in python using the bisect_left() function of the bisect module -

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))

Below is the output of the above example -

[1, 1, 2, 1, 0]