🔥Why 3s why not 4s 💯🚀Explained in detail | O(1) Solution ✅| Dynamic Programming🚀
Do Upvote if you like the solution and explanation please\n\n---\n\n\n# Why 3s why not 4s?\nIn this solution, the choice of using 3s instead of 4s is based on t
# Do Upvote if you like the solution and explanation please\n\n---\n\n\n# Why 3s why not 4s?\nIn this solution, the choice of using 3s instead of 4s is based on the principle of maximizing the product of integers that sum up to `n`. \n\nWhen you break `n` into factors, using 3s as much as possible maximizes the product because:\n\n1. Three(3) is closer to the square root of 4 (which is 2) than any other integer, meaning that using 3s results in more balanced factors. Balanced factors generally lead to a higher product.\n\n2. Using 4s instead of 3s can be suboptimal because 4 can be further broken down into two 2s. So, if you use 4s, you essentially end up with more 2s, and 2 * 2 is less than 3 * 3.\n\nTherefore, breaking `n` into 3s as much as possible is a strategy to maximize the product. However, when `n` has a remainder of 1 when divided by 3, you need to introduce a 4 because 4 * 1 is greater than 3 * 1. This is why the code uses 4 in the `remainder == 1` case to handle such situations while still maximizing the product with 3s for the remaining part.\n\n---\n\n\n\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem asks us to break an integer n into the sum of positive integers such that the product of these integers is maximized. We can observe that breaking n into as many factors of 3 as possible will lead to the maximum product, with some exceptions when the remainder of n divided by 3 is 1 or 2.\n\n\n---\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. For small values of n (n <= 3), it\'s optimal to break it into (n - 1) and return (n - 1). For example, for n = 2, we can break it into 1 + 1, and the product is 1. For n = 3, we can break it into 1 + 2, and the product is 2. So, for n <= 3, we return n - 1.\n\n2. For values of n greater than 3, we aim to break n into as many factors of 3 as possible. Calculate the quotient (integer division) and remainder when dividing n by 3(Three).\n\n3. If the remainder is 0, it means n is divisible by 3. In this case, return 3 raised to the power of the quotient.\n\n4. If the remainder is 1, it means n is one more than a multiple of 3. In this case, return 4 times 3 raised to the power of (quotient - 1). This is because breaking n into one 1 and as many 3\'s as possible will maximize the product.\n\n5. If the remainder is 2, it means n is two more than a multiple of 3. In this case, return 2 times 3 raised to the power of the quotient. This is because breaking n into two 2\'s and as many 3\'s as possible will maximize the product.\n\n\n---\n\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nThe time complexity of this solution is O(1) because the computation involves only a few arithmetic operations regardless of the input value n. The maximum number of iterations is constant.\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nThe space complexity of this solution is O(1) as it uses a fixed amount of additional memory to store variables (quotient and remainder) and does not depend on the input value n.\n\n---\n\n\n# Code\n```\nclass Solution {\npublic:\n    int integerBreak(int n) {\n        if (n <= 3) {\n            return n - 1; // For n <= 3, the maximum product is achieved by breaking it into n - 1\n        }\n        \n        int quotient = n / 3;\n        int remainder = n % 3;\n        \n        if (remainder == 0) {\n            return pow(3, quotient); // n is divisible by 3\n        } else if (remainder == 1) {\n            return 4 * pow(3, quotient - 1); // n is one more than a multiple of 3\n        } else {\n            return 2 * pow(3, quotient); // n is two more than a multiple of 3\n        }\n    }\n};\n\n```
['Math', 'Dynamic Programming', 'C++']