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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 <stdbool.h>
2bool isUgly(int n) {
3 if (n <= 0) return false;
4 int primes[] = {2, 3, 5};
5 for (int i = 0; i < 3; i++) {
6 while (n % primes[i] == 0) {
7 n /= primes[i];
8 }
9 }
10 return n == 1;
11}This C solution checks if a number is an ugly number by iteratively dividing it by the prime factors 2, 3, and 5. If n is reduced to 1, it returns true, indicating 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.
1function isUglyThe JavaScript version utilizes a closure helper function, recursively analyzing the prime divisibility of n until settling on its ugliness based on remaining value 1.