Question

Given two integers x and n, describe and prove an algorithm that computes xⁿ.

Is there any way to go faster than O(n)?

Answer 1

function power(x, n)
    if n is 0
        return 1
    else if n is odd
        temp = power(n / 2)
        return x * temp * temp
    else
        temp = power(n / 2)
        return temp * temp
    end if
end

time complexity of this algorithm is O(log(n))