Maximum Split of Positive Even Integers - Video Solutions

Problem Overview: Given an integer finalSum, return the maximum number of unique positive even integers whose sum equals that value. If finalSum is odd, no valid split exists. The challenge is constructing the largest set of distinct even numbers while keeping the total exactly equal to the target.

Approach 1: Recursive Divide and Conquer (Backtracking) (Time: O(2^n), Space: O(n))

This approach explores combinations of even numbers using recursion. Start from the smallest even number (2) and recursively decide whether to include the next even value in the split. The recursion tracks the remaining sum and the current list of chosen numbers. If the remaining value becomes zero, a valid split is found. Because the algorithm tries many combinations, the branching factor grows quickly and results in exponential time complexity. This method is useful for understanding the search space and practicing recursion with backtracking, but it is not efficient for large inputs.

Approach 2: Iterative Dynamic Programming (Time: O(n^2), Space: O(n))

Dynamic programming builds solutions for smaller sums and uses them to construct larger ones. Create a DP array where each index represents whether a sum can be formed using unique even numbers. For each even number 2, 4, 6..., update states by checking whether adding the current number creates a valid new sum. While this avoids exploring every recursive path, it still requires iterating through many possible states and combinations. The DP approach works well when you want to explicitly track achievable sums and demonstrate subset-style reasoning often used in dynamic programming problems.

Approach 3: Greedy Mathematical Construction (Time: O(sqrt(n)), Space: O(1) excluding output)

The optimal insight comes from greedy algorithms. To maximize the count of distinct even numbers, always take the smallest possible unused even integer. Start with 2, then 4, 6, and keep subtracting from finalSum. Once the remaining value becomes smaller than the next candidate even number, add the leftover to the last element in the list. This keeps all numbers unique while preserving the total sum. If finalSum is odd, return an empty list immediately. Because the sequence of even numbers grows linearly while their sum grows quadratically, the loop runs roughly sqrt(finalSum) times, making it extremely efficient.

Recommended for interviews: The greedy construction is the expected solution. It demonstrates mathematical reasoning and efficient use of the problem constraints. Discussing the recursive search first shows you understand the full solution space, but implementing the greedy approach proves you can identify the optimal strategy quickly.