342. Power of Four

LeetCode 342. Power of Four

Description

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

An integer n is a power of four, if there exists an integer x such that n == 4^x.

Example 1:

Input: n = 16
Output: true

Example 2:

Input: n = 5
Output: false

Example 3:

Input: n = 1
Output: true

Constraints:

• -2^31 <= n <= 2^31 - 1

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

Tags

Bit Manipulation

Solution

From LeetCode 231. Power of Two we know that a positive number n is a power of 2 when n&(n-1) == 0. We also know if a positive number n is a power of 4, it must a power of 2. Therefore, we can start off the characteristics of n = 4^x.

1. Such numbers represented in binary are like 1, 100, 10000, ... . We can find that 1 appears at odd-bits. Thus, we can use a number whose even-bits are all filled with 1 and use it to test with n;

2. 4^x % 3 == 1.

We can derive 2 statements from both characteristics, and append either of them to the solution of Power of Two.

1. n&0xaaaaaaaa == 0;

2. n%3 == 1.

Complexity

• Time complexity: $O(1)$

• Space complexity: $O(1)$

Code

Solution 1:

func isPowerOfFour(n int) bool {
   return n > 0 && n&(n-1) == 0 && n%3 == 1
}

Solution 2:

func isPowerOfFour(n int) bool {
	return n > 0 && n&(n-1) == 0 && n&0xaaaaaaaa == 0
}

Reference

1. 4的幂