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This approach leverages the fact that if a number n is a power of three, it should be divisible by 3 repeatedly until it becomes 1. If at any step, n is not divisible by 3 and it's greater than 1, it cannot be a power of three.
Time Complexity: O(log n)
Space Complexity: O(1)
1def isPowerOfThree(n: int) -> bool:
2 if n < 1:
3 return False
4 while n % 3 == 0:
5 n //= 3
6 return n == 1The Python function checks divisibility by 3 until n becomes 1 or stops being divisible by 3, indicating it's not a power of 3.
This approach uses logarithms to determine if a number is a power of three. By taking the logarithm of the number and comparing it to the logarithm base 3, we can check if they form an integer ratio.
Time Complexity: O(1)
Space Complexity: O(1)
1#include <cmath>
2
bool isPowerOfThree(int n) {
if (n < 1) return false;
double logRes = log10(n) / log10(3);
return (logRes - (int)logRes) == 0;
}Similar to the C solution, this C++ function uses log base 10 to determine if the division produces an integer, indicating a power of three.