✅ C++ (0ms) beats 100% || Arithmetic
Idea \uD83D\uDCA1:\n\nTo maximize the product of the sum of integers, it is important to make them as close to each other as possible. \uD83D\uDCCF\uD83D\uDD0D\
> Idea \uD83D\uDCA1:\n\nTo maximize the product of the sum of integers, it is important to make them as close to each other as possible. \uD83D\uDCCF\uD83D\uDD0D\n\nGiven that we don\'t know the size of k, we need to iterate through each possible k value and compute the product by distributing the closest integers.\n\n> How to find the closest integers\n\nHere\'s how we can find the closest integers: For a given k, we calculate n divided by k. This gives us the minimum integer value for the closest list. Since n = k * (n / k) + (n % k), we need to distribute the remainder (n % k) among the elements.\n\n> Example of finding closest integers\n\nFor example, if we have n = 10 and k = 3, n / k = 10 / 3 = 3. So, we have the list 3, 3, 3. Then, 10 % 3 = 1, which means we have 1 left to distribute. We add it to one of the previous elements, resulting in 4, 3, 3. This arrangement maximizes the product for k = 3 when n = 10.\n\n> Coding this approach will look like this\n\n# Code 1\n```\nclass Solution {\npublic:\n    int integerBreak(int n) {\n        int result = 0;\n        for(int k = 2; k <= n; k++){\n            int mi = n / k;\n            int rest = n % k;\n            int temps = 1;\n            for(int i = 0; i < k; i++){\n                if(rest){\n                    temps *= (mi + 1);\n                    rest--;\n                }\n                else temps *= mi;\n            }\n            result = max(result, temps);\n        }\n        return result;\n    }\n};\n```\n\n> How can we optimize this\n\nOne interesting aspect is that we know exactly how many lower elements and upper elements we will have. This allows us to shorten our code.\n\n# Code 2\n```\nclass Solution {\npublic:\n    int integerBreak(int n) {\n        int result = 0;\n        for (int k = 2; k <= n; k++) {\n            int mi = n / k;\n            int rest = n % k;\n            int product = pow(mi + 1, rest) * pow(mi, k - rest);\n            result = max(result, product);\n        }\n        return result;\n    }\n};\n```\n\nI hope this explanation was helpful to you! Feel free to watch the video for a better understanding. \uD83C\uDFA5\n\nDon\'t forget to upvote! It helps us to keep posting more content. \uD83D\uDC4D\u2728\n\nhttps://youtu.be/41sjWdNT1IY\n
['C++']