How do you solve Maximum Split of Positive Even Integers on LeetCode?

Maximum Split of Positive Even Integers (LeetCode #2178) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Even sums can only be formed from even integers.

What is the time complexity of Maximum Split of Positive Even Integers?

The optimal solution for Maximum Split of Positive Even Integers runs in O(n) time and uses O(n) space. The time complexity is O(n) because we iterate through the even numbers up to finalSum, and the space complexity is O(n) due to storing the result.

What is the brute force solution for Maximum Split of Positive Even Integers?

The brute force approach for Maximum Split of Positive Even Integers is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves trying to generate all possible combinations of unique even integers that sum up to finalSum. This is straightforward but inefficient as it explores many unnecessary combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Split of Positive Even Integers?

Maximum Split of Positive Even Integers uses the following concepts: Math, Backtracking, Greedy. The recommended patterns to study are: Greedy, Backtracking.

What are the interview tips for Maximum Split of Positive Even Integers?

Always check constraints first to avoid unnecessary calculations. Think about greedy approaches when maximizing counts or minimizing values. Explain your thought process clearly; it shows your understanding.

What are common mistakes when solving Maximum Split of Positive Even Integers?

Forgetting to check if finalSum is even before proceeding. Not using unique integers, leading to invalid combinations.

#2178

Maximum Split of Positive Even Integers

Medium

MathBacktrackingGreedyGreedyBacktracking

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution leverages the fact that we want to use the smallest unique even integers possible to maximize the count. By using a greedy approach, we can directly calculate the largest unique even integers that sum to finalSum.

⚙️

Algorithm

3 steps

1Step 1: Check if finalSum is odd; if so, return an empty list.
2Step 2: Initialize an empty list to store the result and a variable to track the current even number starting from 2.
3Step 3: While finalSum is greater than or equal to the current even number, add it to the result and subtract it from finalSum, then increment the current even number by 2.

solution.py11 lines

1def maxSplit(finalSum):
2    if finalSum % 2 != 0:
3        return []
4    result = []
5    current_even = 2
6    while finalSum >= current_even:
7        result.append(current_even)
8        finalSum -= current_even
9        current_even += 2
10    return result
11

ℹ

Complexity note: The time complexity is O(n) because we iterate through the even numbers up to finalSum, and the space complexity is O(n) due to storing the result.

1Even sums can only be formed from even integers.
2Using the smallest unique integers maximizes the count of integers in the sum.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.