# 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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