LeetCode Problem no. 50

Question no. 50. Pow(x, n)

class Solution {
    public double myPow(double x, int n) {
        long n1 = n;
        if(n1 == 1)
            return x;
        if(n1 == 0 || x == 1)
            return 1;
        if(x == 0)
            return 0;
        
        int b=0;
        if(n1 < 0){
            b = 1;
            n1 *= -1;
        }
        
        double a;
        a = 1;

        while(n1 > 0){
            if(n1%2 == 0){
                x = x*x;
                n1 /= 2;
            }
            else{
                a *= x;
                n1--;
            }
        }
        if(b == 1){
           a = 1/a; 
        }
        return a;
    }
}

class Solution:
    def myPow(self, x: float, n: int) -> float:
        if n == 1:
            return x
        if n == 0:
            return 1
        b = 0
        if n < 0:
            b = 1
            n -= n+n

        a = 1
        while(n):
            if n%2 == 0:
                x *= x
                n /= 2
            else:
                a *= x
                n -= 1
        
        if b == 1:
            a = 1/a
        
        return a

Normal explanation of the code:

Implement pow(x, n), which calculates x raised to the power n (i.e., xn).

In this task, we are instructed to create an efficient pow() function. The code must execute within a specified time frame. In my approach, I’ve employed the standard method, adding a condition: if the exponent becomes even, we can square the base and halve the exponent, thereby reducing the computation time.

Technical explanation of the code:

Special Cases:

• The code first checks for some special cases:
• If n is equal to 1, it directly returns x.
• If n is 0 or x is 1, it returns 1 (since any number raised to the power of 0 is 1).
• If x is 0, it returns 0 (since 0 raised to any power other than 0 is 0).

Negative Exponents:

• If n is negative, the code sets a flag b to 1 and makes n1 positive by multiplying it by -1.
• This step ensures that we handle negative exponents correctly.

Calculation Loop:

• The code initializes a variable a to 1. This will store the final result.
• It enters a loop while n1 is greater than 0.
• Within the loop:
• If n1 is even (i.e., divisible by 2), it squares the value of x and divides n1 by 2.
• Otherwise (when n1 is odd), it multiplies a by x and decrements n1 by 1.

For Negative Exponents:

• If the original exponent n was negative (indicated by b == 1), it computes the reciprocal of a (i.e., 1/a).

Return a.

Results for the Python and Java