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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)
1public class Solution {
2 public bool IsPowerOfThree(int n) {
3 if (n < 1) return false;
4 while (n % 3 == 0) {
5 n /= 3;
6 }
7 return n == 1;
8 }
9}This C# solution uses the same iterative method to determine if n is divisible by 3 until it becomes 1.
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)
1public classThis Java method uses Math.log10() to compute the logarithmic check for powers of three, ensuring precision.