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: 8Input: 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:
- Create a recursive function with parameters number N and power P.
- If P = 0 return 1.
 - Else return N times result of the recursive call for N and P-1.
 
 
Below is the implementation of the above approach.
Python3
# Python3 code to recursively find# the power of a number# Recursive function to find N^P.def power(N, P):    # If power is 0 then return 1    # if condition is true    # only then it will enter it,    # otherwise not    if P == 0:        return 1    # Recurrence relation    return (N*power(N, P-1))# Driver codeif __name__ == '__main__':    N = 5    P = 2    print(power(N, P)) | 
25
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
# Python3 code to recursively find# the power of a number# Recursive function to find N^P.def power(N, P):    # If power is 0 then return 1    if P == 0:        return 1    # Recurrence relation    if P%2 == 0:      result = power(N, P//2)      return result * result    else :      result = power(N, (P-1)//2)      return N * result * result # Driver codeif __name__ == '__main__':    N = 5    P = 2    print(power(N, P)) | 
25
Time Complexity: O(log P), For log2P recursive calls.
Auxiliary Space: O(log P), For recursion call stack.
