Maximize Integer Break: Break It Down for Maximum Product!
Intuition\nWe want to figure out how to break a number n into smaller parts to get the biggest multiplication. If n is small, it\'s easy. For bigger numbers, we
# Intuition\nWe want to figure out how to break a number n into smaller parts to get the biggest multiplication. If n is small, it\'s easy. For bigger numbers, we try to split n into groups of 3 because it often gives us the largest result. But sometimes, when n doesn\'t nicely divide by 3, we have to make some special choices.\n# Approach\n1) First, we handle the simple cases:\n\n    - If n is less than or equal to 3, we return n - 1 because it\'s the maximum product in those cases.\n\n2) For larger values of n, we want to break it down into smaller numbers. We aim to break n into as many groups of 3 as possible because multiplying by 3 tends to maximize the product.\n\n  3) We have three different scenarios based on the remainder when dividing n by 3:\n\n- If n is a multiple of 3, we calculate how many groups of 3 can be formed and return 3 raised to that power.\n- If n is one more than a multiple of 3, we calculate one fewer group of 3 and multiply the result by 4.\n- If n is two more than a multiple of 3, we calculate how many groups of 3 can be formed and multiply the result by 2.\n\n4) This approach takes advantage of the properties of numbers divisible by 3 and handles the special cases when n is not a multiple of 3.\n# Complexity\n# - **Time complexity**: **O(1)**\nThe code performs a fixed number of operations regardless of the input size, resulting in constant time complexity.\n\n\n# - Space complexity: **O(1)**\nThe code uses a constant amount of memory, as it only uses a few variables that do not depend on the input size, resulting in constant space complexity.\n# Code\n```\n/**\n * @param {number} n\n * @return {number}\n */\nvar integerBreak = function(n) {\n     // If n is less than or equal to 3, return n - 1.\n    if (n <= 3) {\n    // For n <= 3, the maximum product is obtained by breaking it into n-1.\n        return n - 1;\n    } \n    // If n is a multiple of 3, break it into as many 3s as possible.\n    else if (n % 3 === 0) {\n        // Calculate the number of 3s in the decomposition.\n        const numThrees = Math.floor(n / 3);\n        // The maximum product is 3 to the power of the number of 3s.\n        return Math.pow(3, numThrees);\n    } \n    // If n is of the form 3k + 1, break it into (3^(k-1)) * 4.\n    else if (n % 3 === 1) {\n        // Calculate the number of 3s in the decomposition (minus 1).\n        const numThrees = Math.floor(n / 3) - 1;\n        // The maximum product is 3 to the power of (number of 3s) multiplied by 4.\n        return Math.pow(3, numThrees) * 4;\n    } \n    // If n is of the form 3k + 2, break it into (3^k) * 2.\n    else {\n        // Calculate the number of 3s in the decomposition.\n        const numThrees = Math.floor(n / 3);\n        // The maximum product is 3 to the power of (number of 3s) multiplied by 2.\n        return Math.pow(3, numThrees) * 2;\n    }\n}\n```
['JavaScript']