Integer Break

Problem Description

Given an integer n, break it into the sum of k positive integers, where k >= 2, and maximize the product of those integers. Return the maximum product you can get.

Key Insights

• The problem can be optimized by recognizing that breaking numbers into smaller integers (preferably around 2 and 3) yields a higher product.
• For n = 2, the only valid break is 1 + 1, which gives a product of 1.
• For n = 3, the best break is 1 + 2, yielding a product of 2.
• For n >= 4, breaking n into parts of 3 maximizes the product, and when n is not divisible by 3, the leftover can be combined with the last part.
• The maximum product tends to favor the number 3, as larger integers are less efficient in terms of product maximization.

Space and Time Complexity

Time Complexity: O(1)
Space Complexity: O(1)

Solution

To solve the problem, we can use a mathematical approach rather than dynamic programming. The idea is to use the number 3 as much as possible because the product of 3's tends to be higher than that of larger numbers. The algorithm can be summarized as follows:

1. If n is less than 4, directly return the result based on the specific cases.
2. For n >= 4, keep breaking the number into parts of 3 until we are left with a number less than or equal to 4.
3. Combine the last parts accordingly to ensure we maximize the product. This approach uses simple arithmetic to achieve the maximum product.

Code Solutions