# Python program to find the power of a number using recursion

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:

• 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 code``if` `__name__ ``=``=` `'__main__'``:``    ``N ``=` `5``    ``P ``=` `2` `    ``print``(power(N, P))`

Output
`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 = 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 code``if` `__name__ ``=``=` `'__main__'``:``    ``N ``=` `5``    ``P ``=` `2` `    ``print``(power(N, P))`

Output
```25
```

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

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