50. Pow(x, n)

//🎯DAY 25 PROBLEM 1
//EDGE CASES TO BE CONSIDERED: the range of int is -2^(31) to 2^(31)-1 so abs(-2 ^(31)) goes out of range of int so we use long long
//The idea is to use log(n) time complexity for rate of decrease of n as that is what constraints allow. 
//If we have to calculate 2^(10), 2^(10)=> (2*2)^5 => 4^(5)=>4 * 4^(4)=> 4* (4*4)^2=> 4* (16)^2 => 4* (16*16)^1, we reached base case and we keep on returning 256 upwards
//and once multiply it by 4.
class Solution {
    double pow_calc(double x,long long n)
    {
    //to be dealt explicitly
        if(n==0)
            return 1;
           if(n<=1)
               return x;
        if(n%2==0)
            return pow_calc(x*x,n/2);
        else
            return x*pow_calc(x,n-1);
    }
public:
    double myPow(double x, int n) {
        long org_n=n;
        n=abs(n);
        double ans=pow_calc(x,n);
        if(org_n>0) 
            return ans;
        else
            return (double)1/(ans);
    }
};