"Power of Three" LeetCode problem

[chadxz]

Given an integer n, return true if it is a power of three. Otherwise, return false. An integer n is a power of three, if there exists an integer x such that n == 3^x.

https://leetcode.com/explore/interview/card/top-interview-questions-easy/102/math/745/

[chadxz]

First answer was a recursive solution, but javascript does not support tail call recursion so it overflows the stack:

function isPowerOfThree(y) {
    return y === 3 || y % 3 === 0 && isPowerOfThree(y / 3);
}

Tail call optimization was added to Node.js v6 behind a flag, but was removed in Node.js v7.

Second answer without recursion:

function isPowerOfThree(y) {
    if (y === 3) {
        return true;
    }

    let current = 1;
    while (current < y) {
        current = current * 3;
    }

    return current === y;
};

To do it without looping, you'd need to use logarithms

3^x = y ==> log3(y) = x Math.log() ==> ln() ==> log base e can use logarithm change of base formula to calculate this based on natural log: log3(y) = ln(y) / ln(3) Number.isSafeInteger to check if the number is fractional Math.log() is not precise Math.round() rounds to the nearest integer

So Math.log(y) / Math.log(3) should do it, but an example of the precision issue is Math.log(27) / Math.log(3) = 3.0000000000000004

Rounding to 10 decimal places looks like: Math.round(3.9999999999999 * 10000000000) / 10000000000 = 4

Combining these ideas you get: Math.round(Math.log(y) / Math.log(3) * 10000000000) / 10000000000;

• Math.round(Math.log(3) / Math.log(3) * 10000000000) / 10000000000 = 1
• Math.round(Math.log(9) / Math.log(3) * 10000000000) / 10000000000 = 2
• Math.round(Math.log(27) / Math.log(3) * 10000000000) / 10000000000 = 3
• Math.round(Math.log(6) / Math.log(3) * 10000000000) / 10000000000 = 1.63093

Full logarithm-based answer:

function isPowerOfThree(y) {
    return Number.isSafeInteger(Math.round(Math.log(y) / Math.log(3) * 10000000000) / 10000000000);
};