Maximum Split Of Positive Even Integers - Leetcode Solution

Problem Description:

Given an integer n, return the maximum possible number of parts that positive even integers can be split into such that the sum of all the parts is n.

Example:

Input: n = 6 Output: 3 Explanation: The maximum number of parts that positive even integers can be split into so that the sum is n = 6 are [2,2,2], which is a total of 3 parts.

Solution:

To solve this problem, we can start by understanding that all the integers that are greater than 2 are even. Therefore, we need to first find the number of even numbers that sum up to n. One solution for this is a brute-force solution where we can try out all the combinations of even numbers that sum up to n and choose the combination with the maximum number of elements.

We can also approach this problem using dynamic programming. We can create a table dp[] where dp[i] represents the maximum number of parts that positive even integers can be split into so that the sum is i. We can start by initializing dp[0] to be 0, which means that we need 0 even numbers to sum up to 0. Then we can iterate over all even numbers less than or equal to n, and for each even number i, we can update dp[i] as the maximum between dp[i] and dp[i-j]+1, where j is an even number less than or equal to i. This means that we are trying out all possible combinations of even numbers that sum up to i and choosing the combination with the maximum number of elements.

Finally, we return dp[n] as the maximum number of parts that positive even integers can be split into so that the sum is n.

Here's the Python code for this solution:

def maxSplit(n): dp = [0]*(n+1) for i in range(2, n+1, 2): # only even numbers considered for j in range(i, n+1, 2): dp[j] = max(dp[j], dp[j-i]+1) return dp[n]

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

Overall, this solution is more efficient than the brute-force solution as it eliminates the need to try out all the combinations of even numbers.

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