Power Of Numbers

Problem Statement - link #

Given a number and its reverse. Find that number raised to the power of its own reverse. Note: As answers can be very large, print the result modulo 10^9 + 7.

Your Task: You don’t need to read input or print anything. You just need to complete the function pow() that takes two parameters N and R denoting the input number and its reverse and returns power of (N to R)mod(109 + 7).

Expected Time Complexity: O(LogN).
Expected Auxiliary Space: O(LogN).

Examples #

Example 1:

Input:
N = 2
Output: 4
Explanation: The reverse of 2 is 2
and after raising power of 2 by 2 
we get 4 which gives remainder as 
4 by dividing 1000000007.

Example 2:

Input:
N = 12
Output: 864354781
Explanation: The reverse of 12 is 21
and 1221 , when divided by 1000000007 
gives remainder as 864354781.

Constraints #

• 1 <= N <= 10^9

Solutions #

class Solution
{
    public:
    //You need to complete this fucntion
    int mod = 1000000007;
    long long power(int N,int R)
    {
       //Your code here
       if(R == 1) return N;
        if(R == 0) return 1;
        long long int ans= power(N,R/2)%mod;
        ans = (ans*ans)%mod;
        return (R%2 == 0)?ans:(N*ans)%mod;
    }
};