Power of Two - LeetCode

Given an integer, write a function to determine if it is a power of two.

Algorithm:

The intuition here is compare all the 2 ^ i values with n. If they are equal return true else if 2 ^ i is greater than n then return false

Code:

/**
 * @param {number} n
 * @return {boolean}
 */
var isPowerOfTwo = function(n) {
    let cur = 0
    let temp = 1
    while(temp <= n) {
        if(temp == n) {
            return true
        }
        temp = Math.pow(2, cur);
        cur++
    }

    return false
};

Alternative ways:

/**
 * @param {number} n
 * @return {boolean}
 */
var isPowerOfTwo = function(n) {
    return Math.log2(n)%1 === 0
};

/**
 * @param {number} n
 * @return {boolean}
 */
var isPowerOfTwo = function(n) {
    return n < 1 ? false : Number.MAX_VALUE % n == 0
};

Comments

[Ravi Mishra]

There's an interesting bit property that we can use to solve this. If you AND a number with number - 1, and it's 0 then it is a power of 2. Binary representation of any number which is a power of 2 will have only one set bit and the number less than that will have all bits set except the left most bit. Take 32 for example
32 - 100000
31 - 011111
32 & 31 - 000000

This property won't hold true for numbers which aren't power of 2.