98.34 Beats 🔥, Easiest  solution for beginners (O(1)) , Anyone can Understand.
Intuition\nBasically , I tried to find a pattern that would give the maximum product , I got to know that , this can only happen using prime numbers . From prim
# Intuition\nBasically , I tried to find a pattern that would give the maximum product , I got to know that , this can only happen using prime numbers . From prime numbers i understood that any universal number can be written in 2\'s or 3\'s .\n\n# Approach\nChecking for the remainder of number and according to the remainders 0,1,2 there is a code given.\n1. For Remainder 0 : \n                The highest product will be the product of all 3\'s.\n\n2. For Remainder 1:\n        The highest product will be multiplying 4 with remaining threes .\n3. For Remainder 2 : \nThe highest product will be multiplying only 2 with remaining 3\'s.\n\n# Complexity\n- Time complexity:\nO(1)\n\n- Space complexity:\nO(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nimport math\nclass Solution:\n    def integerBreak(self, n: int) -> int:\n        \n        rem1 = n % 3 \n        y = n//3\n        if n ==1 or n==2 :\n            return 1 \n        elif n==3:\n            return 2\n        elif rem1 ==1 :\n            s = pow(3,y-1)*4\n            return s\n        elif rem1==0:\n            return pow(3,y)\n        else:\n            s = pow(3,y)*2\n            return s \n              \n        \n\n\n\n        \n```
['Math', 'Python3']