Power Of Numbers

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Sun Feb 06 2022 23:19:49 GMT+0000 (Coordinated Universal Time)

Saved by @Uttam #java #gfg #geeksforgeeks #recursion #power #powerof numbers

class Solution
{
    public long modfun(long n, long  R)
    {
        // Base cases 
        if (n == 0) 
            return 0; 
        // power zero mean answer is 1
        if (R == 0)  
            return 1; 
        // If R is even 
        long y; 
        if (R % 2 == 0) { 
            // finding r/2 power as power is even then storing answer in y and if power is even like 2^4 we can simply do (2^2)*(2^2)
            y = modfun(n, R/2);  
            y = (y * y) % 1000000007; 
        } 
      
        // If R is odd 
        else { 
            y = n % 1000000007; 
            // reduce the power by 1 to make it even. The reducing power by one can be done if we take one n out. Like 2^3 can be written as 2*(2^2)
            y = (y * modfun(n, R - 1) % 1000000007) % 1000000007; 
        } 
        // finally return the answer (y+mod)%mod = (y%mod+mod%mod)%mod = (y%mod)%mod
        return ((y + 1000000007) % 1000000007);  
        }
        
        
    long power(int N,int R)
    {
        return modfun(N,R)%1000000007;
        
    }

}
content_copyCOPY

Power Of Numbers 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. 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. 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(10^9 + 7). Expected Time Complexity: O(LogN). Expected Auxiliary Space: O(LogN). Constraints: 1 <= N <= 10^9

https://practice.geeksforgeeks.org/problems/power-of-numbers-1587115620/1/?track=DSASP-Recursion&batchId=190