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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