Approach: Strip factors 2, 3, 5

def isUgly(n):
    if n <= 0:
        return False
    for p in (2, 3, 5):
        while n % p == 0:
            n //= p
    return n == 1

Complexity: O(log n) time, O(1) space.

Full Solution — all languages

class Solution:
    def isUgly(self, n):
        if n <= 0:
            return False
        for p in (2, 3, 5):
            while n % p == 0:
                n //= p
        return n == 1

import java.util.*;
class Solution {
    public boolean isUgly(int n) {
        if (n <= 0) return false;
        int[] ps = {2, 3, 5};
        for (int p : ps) while (n % p == 0) n /= p;
        return n == 1;
    }
}

#include <bits/stdc++.h> using namespace std;
class Solution {
    public: bool isUgly(int n) {
        if (n <= 0) return false;
        for (int p : {
            2, 3, 5
        })
        while (n % p == 0) n /= p;
        return n == 1;
    }
};

/**
 * @param {number} n
 * @return {boolean}
 */ var isUgly = function(n) {
    if (n <= 0) return false;
    for (const p of [2, 3, 5]) while (n % p === 0) n /= p;
    return n === 1;
};

Ugly Number

Example 1

Example 2

Example 3

Constraints