# 326. Power of Three

## How to identify if an integer is a 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: Recursion

#### 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

#### 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 <= x <= y.

Because the constraint of the problem is n <= 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

#### Complexity

• Runtime: O(1).

• Extra space: O(1).

### References

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