Python program to find the power of a number using recursion (original) (raw)

Last Updated : 02 May, 2023

Given a number N and power P, the task is to find the power of a number ( i.e. NP ) using recursion.

Examples:

Input: N = 2 , P = 3
Output: 8

Input: N = 5 , P = 2
Output: 25

Approach: Below is the idea to solve the above problem:

The idea is to calculate power of a number ‘N’ is to multiply that number ‘P’ times.

Follow the below steps to Implement the idea:

Below is the implementation of the above approach.

Python3

def power(N, P):

`` if P = = 0 :

`` return 1

`` return (N * power(N, P - 1 ))

if __name__ = = '__main__' :

`` N = 5

`` P = 2

`` print (power(N, P))

Time Complexity: O(P), For P recursive calls.
Auxiliary Space: O(P), For recursion call stack.

Optimized Approach :

Calling the recursive function for (n, p) -> (n, p-1) -> (n, p-2) -> … -> (n, 0) taking P recursive calls. In the optimized approach the idea is to
decrease the number of functions from p to log p.

Let’s see how.

we know that

if p is even we can write N p = N p/2 * N p/2 = (N p/2) 2 and

if p is odd we can wrte N p = N * (N (p-1)/2 * N (p-1)/2) = N * (N (p-1)/2) 2

for example : 24 = 22 * 22

also, 25 = 2 * (22 * 22)

From this definaion we can derive this recurrance relation as

if p is even

result = ( func(N, p/2) ) 2

else

result = N * ( func(N, (p-1)/2) ) 2

Below is the implementation of the above approach in python3

Python3

def power(N, P):

`` if P = = 0 :

`` return 1

`` if P % 2 = = 0 :

`` result = power(N, P / / 2 )

`` return result * result

`` else :

`` result = power(N, (P - 1 ) / / 2 )

`` return N * result * result

if __name__ = = '__main__' :

`` N = 5

`` P = 2

`` print (power(N, P))

Time Complexity: O(log P), For log2P recursive calls.
Auxiliary Space: O(log P), For recursion call stack.

Similar Reads

Python Matrix Exercises

Python Functions Exercises

Python Lambda Exercises

Python Pattern printing Exercises