Power of Four#

Problem statement#

[1]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?

Solution 1: Division by four#

Code#

#include <iostream>
using namespace std;
bool isPowerOfFour(int n) {
    // perform the divison by 4 repeatedly
    while (n % 4 == 0 && n > 0) {
        n /= 4;
    }
    // if n % 4 != 0, then n > 1
    return n == 1;
}
int main() 
{
    cout << isPowerOfFour(16) << endl;
    cout << isPowerOfFour(5) << endl;
    cout << isPowerOfFour(1) << endl;
}

Output:
1
0
1

This solution repeatedly divides the given number n by 4 until n becomes either 1 or a number that is not divisible by 4. If n becomes 1 after this process, it means that n was originally a power of 4.

Complexity#

  • Runtime: O(logn).

  • Extra space: O(1).

Solution 2: Binary representation#

You can write down the binary representation of the powers of four to find the pattern.

1   : 1
4   : 100
16  : 10000
64  : 1000000
256 : 100000000
...

You might notice the patterns are n is a positive integer having only one bit 1 in its binary representation and it is located at the odd positions (starting from the right).

How can you formulate those conditions?

If n has only one bit 1 in its binary representation 10...0, then n - 1 has the complete opposite binary representation 01...1.

You can use the bit operator AND to formulate this condition

n & (n - 1) == 0

Let A be the number whose binary representation has only bits 1 at all odd positions, then n & A is never 0.

In this problem, A < 2^31. You can chooseA = 0x55555555, the hexadecimal of 0101 0101 0101 0101 0101 0101 0101 0101.

Code#

#include <iostream>
using namespace std;
bool isPowerOfFour(int n) {
    // the condition of the pattern "n is a positive integer 
    // having only one bit 1 in its binary representation and 
    // it is located at the odd positions"
    return n > 0 && (n & (n - 1)) == 0 && (n & 0x55555555) != 0;
}
int main() {
    cout << isPowerOfFour(16) << endl;
    cout << isPowerOfFour(5) << endl;
    cout << isPowerOfFour(1) << endl;
}

Output:
1
0
1

Complexity#

  • Runtime: O(1).

  • Extra space: O(1).

Conclusion#

Recognizing the unique properties of powers of four, such as their binary representation, can lead to efficient solutions. Solution 2 leverages bitwise operations to check if a number meets the criteria of being a power of four.

By examining the binary representation and ensuring that the only set bit is located at an odd position, Solution 2 effectively determines whether the number is a power of four in constant time complexity, without the need for division operations.

But in term of readable code, Solution 2 is not easy to understand like Solution 1, where complexity of O(logn) is not too bad.

Exercise#

  • Power of Two.[2].