Constant space and constant time O(1) Solution
Approach\n Describe your approach to solving the problem. \n\nHere\'s an explanation of how the code works:\n\n1. The base cases:\n   - If n is less than 3, it 
# Approach\n<!-- Describe your approach to solving the problem. -->\n\nHere\'s an explanation of how the code works:\n\n1. The base cases:\n   - If `n` is less than 3, it returns 1. This is because the problem likely doesn\'t make sense for values less than 3, and returning 1 is a reasonable default in such cases.\n   - If `n` is 2 or 3, it returns 2. This is because, for these values, the maximum product is achieved by breaking them into 1 + 1 or 1 + 2, respectively.\n\n2. Calculate the remainder and quotient when dividing `n` by 3:\n   - `remainder` stores the result of `n % 3`.\n   - `quotient` stores the result of `n / 3`.\n\n3. Handle the case where `remainder` is 1:\n   - In this case, decrement `quotient` by 1 and set `remainder` to 4. This is done because breaking the integer into a 3 and a 1 would result in a product of 3, which is not as optimal as breaking it into 2 and 2 (product of 4).\n\n4. Calculate the result:\n   - Compute `result` by raising 3 to the power of `quotient` using `Math.pow`. This corresponds to the product of the maximum number of 3s that can be obtained from `n`.\n   - If `remainder` is not 0, multiply `result` by `remainder`. This accounts for the remaining part of `n` that couldn\'t be represented as a multiple of 3.\n\n5. Return the final `result`.\n\nThe code effectively calculates the maximum product that can be obtained by breaking the input integer `n` into smaller integers, with the constraint that the smaller integers should be as close to 3 as possible, as long as it results in a higher product. This is often used as a mathematical optimization problem.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(1)$$ --> \n$$O(1)$$\n \n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(1)$$ -->\n$$O(1)$$\n\n# Code\n```\nclass Solution {\n    public int integerBreak(int n) {\n        if (n < 3) return 1;\n        if (n == 2 || n == 3) return 2;\n        \n        int remainder = n % 3;\n        int quotient = n / 3;\n        if (remainder == 1) {\n            quotient--;\n            remainder = 4;\n        }\n        \n        int result = (int) Math.pow(3, quotient);\n\n        if (remainder != 0) {\n            result *= remainder;\n        }\n        \n        return result;\n    }    \n}\n\n```
['Java']