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This approach involves dividing the number by its prime factors (2, 3, and 5) as long as it is divisible by them. If after removing all these factors, the number reduces to 1, it is an ugly number; otherwise, it is not.
Time Complexity: O(log n).
Space Complexity: O(1).
1#include <iostream>
2bool isUgly(int n) {
3 if (n <= 0) return false;
4 int primes[] = {2, 3, 5};
5 for (int prime : primes) {
6 while (n % prime == 0) {
7 n /= prime;
8 }
9 }
10 return n == 1;
11}This C++ solution uses a for-each loop to check divisibility and reduces n by dividing it with any of the primes 2, 3, and 5 until it's no longer divisible, determining if n is an ugly number.
This alternative approach involves using recursion to systematically divide the number by 2, 3, and 5. By tracing back all divisions reaching 1, this method can also verify the ugliness of a number.
Time Complexity: O(log n).
Space Complexity: O(log n), due to recursion stack.
1public class Solution {
2 private bool Helper(int n) {
if (n == 1) return true;
if (n % 2 == 0) return Helper(n / 2);
if (n % 3 == 0) return Helper(n / 3);
if (n % 5 == 0) return Helper(n / 5);
return false;
}
public bool IsUgly(int n) {
if (n <= 0) return false;
return Helper(n);
}
}This C# solution applies a recursive strategy in the helper method to efficiently decide the status of n being ugly by checking divisibility by 2, 3, and 5.