-2³¹ <= n <= 2³¹ - 1

23:T6e4,**Supper cool O(1) one-liner** from [chuilinux](https://leetcode.com/problems/ugly-number/discuss/69232/Python%3A-1-line-solution)\n```python\nclass Solution:\n def isUgly(self, n: int) -> bool:\n return False if n<=0 else (2*3*5)**32 % n == 0\n```\n\n**A normal O(log N) algorithm:**\n(1) ```n``` needs to be a positive integer, so return False ```if n<=0```.\n(2) divide ```n``` by ```2,3,5``` as much as possible to see if there are any other prime factor left.\n**Time complexity:** O (log n) since there are at most log n divisions.\n\n```python\nclass Solution:\n def isUgly(self, n: int) -> bool:\n if n<=0: return False\n while n%2==0: n/=2\n while n%3==0: n/=3\n while n%5==0: n/=5\n return n==1\n```\n\n**Skip the iteration for 2**, and check if ```n``` is the power of 2 at the end.\n```python\nclass Solution:\n def isUgly(self, n: int) -> bool:\n if n<=0: return False\n while n%3==0: n//=3\n while n%5==0: n//=5\n return ( n & (n-1) ) == 0\n```\n\n**O(logN\\*logN\\*logN)** To make it more inefficient and longer.\n```\nclass Solution:\n def isUgly(self, n: int) -> bool:\n if n<=0: return False\n @cache\n def dfs(cur):\n if cur==n: return True\n if cur>n: return False\n for k in (5,3,2):\n if dfs(cur*k): return True\n return dfs(1)\n```\n**Upvote** if you like this post.\n\n**Connect with me on [LinkedIn](https://www.linkedin.com/in/meida-chen-938a265b/)** if you\'d like to discuss other related topics\n\nJust in case if you are working on **ML/DL 3D data-related projects** or are interested in the topic, please check out our project **[HERE](https://github.com/meidachen/STPLS3D)**24:T855,

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