Power of Three#

Problem statement#

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

An integer n is a power of three, if there exists an integer x such that n == 3^x.

Example 1#

Input: n = 27
Output: true
Explanation: 27 = 3^3.

Example 2#

Input: n = 0
Output: false
Explanation: There is no x where 3^x = 0.

Example 3#

Input: n = -1
Output: false
Explanation: There is no x where 3^x = (-1).

Constraints#

  • -2^31 <= n <= 2^31 - 1.

Follow up#

  • Could you solve it without loops/recursion?

Solution 1: Repeat the division#

Code#

#include <iostream>
using namespace std;
bool isPowerOfThree(int n) {
    while (n % 3 == 0 && n > 0) {
        n /= 3;
    }
    return n == 1;
}
int main() {
    cout << isPowerOfThree(27) << endl;
    cout << isPowerOfThree(0) << endl;
    cout << isPowerOfThree(-1) << endl;
}

Output:
1
0
0

This solution repeatedly divides the input by 3 until it either becomes 1 (indicating that it was a power of three) or cannot be divided further by 3.

Complexity#

  • Runtime: O(logn).

  • Extra space: O(1).

Solution 2: Mathematics and the constraints of the problem#

A power of three must divide another bigger one, i.e. \(3^x | 3^y\) where \(0 \leq x \leq y\).

Because the constraint of the problem is \(n \leq 2^{31} - 1\), you can choose the biggest power of three in this range to test the others.

It is \(3^{19} = 1162261467\). The next power will exceed \(2^{31} = 2147483648\).

Code#

#include <iostream>
using namespace std;
bool isPowerOfThree(int n) {
    return n > 0 && 1162261467 % n == 0;
}
int main() {
    cout << isPowerOfThree(27) << endl;
    cout << isPowerOfThree(0) << endl;
    cout << isPowerOfThree(-1) << endl;
}

Output:
1
0
0

This solution effectively checks whether n is a power of three by verifying if it is a divisor of the largest power of three that fits within 32 bits. If the condition is met, it returns true, indicating that n is a power of three; otherwise, it returns false.

Complexity#

  • Runtime: O(1).

  • Extra space: O(1).

Readable code#

Though Solution 2 offers a direct approach without the need for iteration, it is not easy to understand like Solution 1, where complexity of O(logn) is not too bad.

Exercise#