50. Pow(x, n) - LeetcodeSolve

Explanation:

To calculate x raised to the power of n, we can use the concept of exponentiation by squaring. The idea is to break down the problem into smaller subproblems by dividing the power n by 2 in each step. By recursively solving these subproblems, we can efficiently compute the result. If n is even, we can square the result of x raised to the power of n/2. If n is odd, we need to multiply the result of x raised to the power of n/2 with itself and x. This approach optimizes the number of multiplications required.

Algorithm:

If n is 0, return 1.
If n is negative, update x to 1/x and n to -n to handle negative exponents.
Initialize a variable result to 1.
Loop while n is greater than 0:
- If n is odd, multiply result by x.
- Update x to x squared.
- Divide n by 2.
Return result.

Time Complexity:

The time complexity of this algorithm is O(log n) as we are reducing the problem size by half in each step.

Space Complexity:

The space complexity is O(1) as we are using a constant amount of extra space.

: :

View Leetcode Problem

class Solution {
    public double myPow(double x, int n) {
        if (n == 0) return 1;
        if (n < 0) {
            x = 1 / x;
            n = -n;
        }
        
        double result = 1;
        while (n > 0) {
            if (n % 2 == 1) {
                result *= x;
            }
            x *= x;
            n /= 2;
        }
        
        return result;
    }
}

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