Check Even or Odd Numbers using recursion in Python
We use the concept of getting the remainder without using the modulus operator by subtracting the number by 2. In the below code, we are calling the recursion function by subtracting the number by 2. In the base condition, we are checking whether the n becomes equal to 1 or 0. If n==1 it is odd and if n==0 number is even.
Python3
# defining the function having the one parameter as inputdef evenOdd(n): # if remainder is 0 then num is even if(n==0): return True # if remainder is 1 then num is odd elif(n==1): return False else: return evenOdd(n-2) num=33if(evenOdd(num)): print(num,"num is even")else: print(num,"num is odd") |
Output
33 num is odd
Time Complexity: O(n/2)
Auxiliary Space: O(1)
Python Program to Check if a Number is Odd or Even
Given a number and our task is to check number is Even or Odd using Python. Those numbers which are completely divisible by 2 or give 0 as the remainder are known as even number and those that are not or gives a remainder other than 0 are known as odd numbers.
Example:
Input: 2
Output: Even number
Input: 41
Output: Odd Number