Follow

Follow

# 326. Power of Three

## How to identify if an integer is a power of three

Nhut Nguyen
·Dec 19, 2022·

### 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

Thanks for reading. Feel free to share your thought about my content and check out my FREE book “10 Classic Coding Challenges”.