231. Power of Two

Description of Problem

Given an integer n, return true if it is a power of two. Otherwise, return false.

An integer n is a power of two, if there exists an integer x such that n == 2^x.

Example 1:

Input: n = 1
Output: true
Explanation: 20 = 1

Example 2:

Input: n = 16
Output: true
Explanation: 24 = 16

Example 3:

Input: n = 3
Output: false

Constraints:

-2^31 <= n <= 2^31 - 1

Follow up: Could you solve it without loops/recursion?

Solution 1 - Bitwise operations

Explanation

Consider binary number representation, each position is either 0 or 1 and it represents \(2^i\).

For any power of two, the binary representation should be \(1000......000000_2\).

Code (Java)

class Solution {
    public boolean isPowerOfTwo(int n) {

        if ( n <= 0 ){
            return false;
        }
        
        while((n & 1) == 0){
            n = n >> 1;
        }

        return n == 1 ;
    }
}

Complexity

n is the input number

Time complexity:

\( T(n) = O(\log_2(n) + 1) \)
- Assume division takes constant time.

Auxiliary Space:

\( S(n) = O(1) \)

Solution 2 - Meet the requirement of the follow-up

Explanation

The maximum power of two in i32 integer is \( 2^{30}=1073741824 \). This number is divsible by any power of two.

Code (Java)

class Solution {
    public boolean isPowerOfTwo(int n) {
        if (n <= 0) {
            return false;
        }else{
            return (1073741824 % n) == 0;
        } 
    }
}

Complexity

n is the input number

Time complexity:

\( T(n) = O(1) \)
- Assume division takes constant time.

Auxiliary Space:

\( S(n) = O(1) \)

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