Integer Break - Leetcode Solution

Problem Statement:

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

Example: Input: 10 Output: 36 Explanation: 10 = 3 + 3 + 4, 3 × 3 × 4 = 36.

Solution:

This problem is related to dynamic programming. The main idea of dynamic programming is to solve a problem by breaking it down into smaller and simpler subproblems.

In this problem, we are given an integer 'n' and we have to break it down into a sum of k positive integers in such a way that the product of these integers is maximum. As the problem statement suggests, we can start by dividing the number 'n' into two positive integers 'a' and 'b'. Now, the product of these two integers can be calculated, which is a*b. Now, we can break down 'a' and 'b' again into their respective sums of positive integers and we can continue this process until we reach the desired number of positive integers.

Let's take an example to understand this more clearly. Suppose 'n' is equal to 10. We can divide it into two positive integers 'a' and 'b', in such a way that their product is maximum. The maximum product can be obtained by taking 'a' and 'b' as 5 and 5, respectively. The product is 25.

Now, we can break down 'a' and 'b' into their respective sums of positive integers. For the number 5, we can take it as 2 and 3, which gives us a product of 6. Similarly, for the other number 5, we can take it as 2 and 3, which gives us a product of 6. So, the maximum product we can get by breaking down 10 into a sum of 4 positive integers would be 2566 = 900.

The above approach can be solved using dynamic programming. We can define a 1D array dp[n] where dp[i] represents the maximum product we can get by breaking down the number 'i' into a sum of positive integers. We can fill the dp array for base cases where dp[1] = 0 and dp[2] = 1, as we can't break down 1 into any positive integer and for 2, we can only break it down into two 1's and the product is 1.

Now, we can iterate for 'i' from 3 to n and for each i, we can iterate for 'j' from 1 to i/2, as we want a sum of at least two positive integers. We can divide 'i' into two positive integers 'j' and 'i-j' and we can calculate their product. We can take the maximum product we obtain from each iteration and store it in dp[i]. At the end, dp[n] will return the maximum product we can get by breaking down 'n' into a sum of positive integers.

Pseudocode:

dp[1] = 0 dp[2] = 1 for i in range(3, n+1): for j in range(1, i//2+1): product = max(j*(i-j), j*dp[i-j]) dp[i] = max(dp[i], product) return dp[n]

Time Complexity: O(n^2) Space Complexity: O(n)

In this way, the problem can be solved by breaking down 'n' into a sum of positive integers and calculating their product using dynamic programming.

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