Check if one integer is an integer power of another

Question

This is an interview question: \"Given 2 integers x and y, check if x is an integer power of y\" (e.g. for x = 8 and y = 2 the answer is \"true\", and for x = 10 and y = 2 \

Answer 1

On second thoughts, don't do this. It does not work for negative x and/or y. Note that all other log-based answers presented right now are also broken in exactly the same manner.

The following is a fast general solution (in Java):

static boolean isPow(int x, int y) {
    int logyx = (int)(Math.log(x) / Math.log(y));
    return pow(y, logyx) == x || pow(y, logyx + 1) == x;
}

Where pow() is an integer exponentiation function such as the following in Java:

static int pow(int a, int b) {
    return (int)Math.pow(a, b);
}

(This works due to the following guarantee provided by Math.pow: "If both arguments are integers, then the result is exactly equal to the mathematical result of raising the first argument to the power of the second argument...")

The reason to go with logarithms instead of repeated division is performance: while log is slower than division, it is slower by a small fixed multiple. At the same time it does remove the need for a loop and therefore gives you a constant-time algorithm.