Skip to main content

0342 - Power of Four (Easy)

Problem Statement​

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?

Written by @wingkwong
class Solution {
public:
    bool isPowerOfFour(int n) {
	    // the idea is to use binary search to find x to see if 4 ^ x = n is true or false
        int l = 0, r = (int) log(pow(2, 31)) / log(4);
         while (l < r) {
            // get the middle one
            // for even number of elements, take the lower one
            int m = l + (r - l) / 2;
            // exclude m
            if (pow(4, m) < n) l = m + 1;
            // include m
            else r = m;
        }
		// check if 4 ^ l is n
        // if so, then n is a power of four, otherwise it is not
        return pow(4, l) == n;
    }
};

Approach 2: Bit Manipulation​

Written by @wingkwong
class Solution {
public:
    bool isPowerOfFour(int num) {
        // 4: 100
        // 16: 10000
        // observation: 
        // count of 1s is 1 and the number of trailing zeros is even
        return __builtin_popcount(num) == 1 && // only 1 bit is set
              (__builtin_ctz(num) & 1) == 0;     // with even trailing zeros 
    }
};