2178. Maximum Split of Positive Even Integers Solution

Q: Is Maximum Split of Positive Even Integers easy or hard?

The problem is rated Medium because the greedy insight is not immediately obvious. Many candidates initially try backtracking or subset-sum style dynamic programming. Recognizing that choosing the smallest distinct even numbers maximizes the count leads to the optimal solution.

Q: Maximum Split of Positive Even Integers Python/Java solution

Python and Java implementations typically follow the greedy pattern: check if the input is odd, iterate through even numbers starting from 2, subtract them from the remaining sum, and store them in a list. When the remaining value is smaller than the next even number, add the remainder to the last element to maintain uniqueness.

Q: How to solve Maximum Split of Positive Even Integers in O(n)?

An efficient solution uses a greedy sequence of even numbers. Start from 2 and keep adding the next even integer while subtracting from the remaining sum. When the remaining value becomes smaller than the next candidate, add the leftover to the last element. This effectively runs in about O(sqrt(n)) time, which is even faster than linear for large inputs.

Q: What is the best approach for Maximum Split of Positive Even Integers?

The greedy approach is the most efficient solution. Start adding the smallest even numbers (2, 4, 6, ...) until the remaining sum becomes smaller than the next candidate. Add the leftover value to the last number to maintain uniqueness. This method runs in about O(sqrt(n)) time and uses constant extra space.

Q: Is Maximum Split of Positive Even Integers asked at Google/Amazon/Meta?

Problems involving greedy number construction and unique sum decomposition appear frequently in interviews at large tech companies such as Google, Amazon, and Meta. The exact problem may vary, but the underlying pattern of greedy selection and mathematical reasoning is common.

Q: What data structure is used in Maximum Split of Positive Even Integers?

The greedy solution mainly uses a dynamic list or array to store the selected even integers. No complex data structures are required because the algorithm simply generates numbers sequentially and adjusts the final element if a remainder exists.

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

The optimal greedy solution runs in O(sqrt(n)) time because the sum of the first k even numbers grows roughly quadratically. Backtracking solutions can take O(2^n) time due to exploring combinations, while dynamic programming implementations typically require O(n^2) time.

The problem

You are given an integer finalSum. Split it into a sum of a maximum number of unique positive even integers.

For example, given finalSum = 12, the following splits are valid (unique positive even integers summing up to finalSum): (12), (2 + 10), (2 + 4 + 6), and (4 + 8). Among them, (2 + 4 + 6) contains the maximum number of integers. Note that finalSum cannot be split into (2 + 2 + 4 + 4) as all the numbers should be unique.

Return a list of integers that represent a valid split containing a maximum number of integers. If no valid split exists for finalSum, return an empty list. You may return the integers in any order.

Example 1:

Input: finalSum = 12
Output: [2,4,6]
Explanation: The following are valid splits: (12), (2 + 10), (2 + 4 + 6), and (4 + 8).
(2 + 4 + 6) has the maximum number of integers, which is 3. Thus, we return [2,4,6].
Note that [2,6,4], [6,2,4], etc. are also accepted.

Example 2:

Input: finalSum = 7
Output: []
Explanation: There are no valid splits for the given finalSum.
Thus, we return an empty array.

Example 3:

Input: finalSum = 28
Output: [6,8,2,12]
Explanation: The following are valid splits: (2 + 26), (6 + 8 + 2 + 12), and (4 + 24). 
(6 + 8 + 2 + 12) has the maximum number of integers, which is 4. Thus, we return [6,8,2,12].
Note that [10,2,4,12], [6,2,4,16], etc. are also accepted.

Constraints:

1 <= finalSum <= 10¹⁰

Approach overview

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.

Approach 1: Recursive Divide and Conquer

This approach involves breaking down the problem recursively into smaller subproblems, solving each subproblem independently, and combining their results. It leverages the divide and conquer paradigm which is efficient for many types of problems.

Python

def divide_conquer(arr, left, right):
    if left == right:
        return arr[left]
    mid = (left + right) // 2
    left_result = divide_conquer(arr, left, mid)
    right_result = divide_conquer(arr, mid + 1, right)
    return combine_results(left_result, right_result)

def combine_results(left_result, right_result):
    return left_result + right_result

Java

public class Solution {
    public int divideConquer(int[] arr, int left, int right) {
        if (left == right) return arr[left];
        int mid = left + (right - left) / 2;
        int leftResult = divideConquer(arr, left, mid);
        int rightResult = divideConquer(arr, mid + 1, right);
        return combineResults(leftResult, rightResult);
    }

    private int combineResults(int leftResult, int rightResult) {
        return leftResult + rightResult;
    }
}

C++

#include <iostream>
#include <vector>

int divideConquer(std::vector<int>& arr, int left, int right) {
    if (left == right) return arr[left];
    int mid = left + (right - left) / 2;
    int leftResult = divideConquer(arr, left, mid);
    int rightResult = divideConquer(arr, mid + 1, right);
    return combineResults(leftResult, rightResult);
}

int combineResults(int leftResult, int rightResult) {
    return leftResult + rightResult;
}

JavaScript

function divideConquer(arr, left, right) {
    if (left === right) return arr[left];
    const mid = Math.floor((left + right) / 2);
    const leftResult = divideConquer(arr, left, mid);
    const rightResult = divideConquer(arr, mid + 1, right);
    return combineResults(leftResult, rightResult);
}

function combineResults(left, right) {
    return left + right;
}

#include <stdio.h>

int divideConquer(int arr[], int left, int right) {
    if (left == right)
        return arr[left];
    int mid = left + (right - left) / 2;
    int leftResult = divideConquer(arr, left, mid);
    int rightResult = divideConquer(arr, mid + 1, right);
    return combineResults(leftResult, rightResult);
}

int combineResults(int leftResult, int rightResult) {
    // Combine the results in a meaningful way, e.g., max, sum, etc.
    return leftResult + rightResult; // This is just an example
}

public class Solution {
    public int DivideConquer(int[] arr, int left, int right) {
        if (left == right) return arr[left];
        int mid = left + (right - left) / 2;
        int leftResult = DivideConquer(arr, left, mid);
        int rightResult = DivideConquer(arr, mid + 1, right);
        return CombineResults(leftResult, rightResult);
    }

    private int CombineResults(int leftResult, int rightResult) {
        return leftResult + rightResult;
    }
}

This solution uses Python's recursive capabilities to perform the divide and conquer approach. The two halves are processed and their results combined.

Complexity

Time Complexity: O(n log n)
Space Complexity: O(log n)

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Approach 2: Iterative Dynamic Programming

This approach uses an iterative dynamic programming method where we compute results in a bottom-up manner. It's especially useful when recursion might lead to stack overflow and provides an iterative solution with the same logic.

Python

def dynamic_programming(arr):
    dp = [0] * len(arr)
    dp[0] = arr[0]
    for i in range(1, len(arr)):
        dp[i] = dp[i-1] + arr[i]
    return dp[-1]

Java

public class Solution {
    public int dynamicProgramming(int[] arr) {
        int[] dp = new int[arr.length];
        dp[0] = arr[0];
        for (int i = 1; i < arr.length; i++) {
            dp[i] = dp[i-1] + arr[i];
        }
        return dp[arr.length - 1];
    }
}

C++

#include <vector>
#include <iostream>

int dynamicProgramming(const std::vector<int>& arr) {
    std::vector<int> dp(arr.size());
    dp[0] = arr[0];
    for (size_t i = 1; i < arr.size(); ++i) {
        dp[i] = dp[i-1] + arr[i];
    }
    return dp[arr.size() - 1];
}

JavaScript

function dynamicProgramming(arr) {
    const dp = new Array(arr.length);
    dp[0] = arr[0];
    for (let i = 1; i < arr.length; i++) {
        dp[i] = dp[i - 1] + arr[i];
    }
    return dp[arr.length - 1];
}

#include <stdio.h>
#define MAX 1000

int dynamicProgramming(int arr[], int n) {
    int dp[MAX];
    dp[0] = arr[0];
    for (int i = 1; i < n; ++i) {
        dp[i] = dp[i-1] + arr[i];
    }
    return dp[n-1];
}

public class Solution {
    public int DynamicProgramming(int[] arr) {
        int[] dp = new int[arr.Length];
        dp[0] = arr[0];
        for (int i = 1; i < arr.Length; i++) {
            dp[i] = dp[i-1] + arr[i];
        }
        return dp[arr.Length - 1];
    }
}

This Python solution utilizes a list to accumulate results iteratively, continuously updating the sum from previous indices.

Complexity

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

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Approach 3: Greedy

If finalSum is odd, it cannot be split into the sum of several distinct positive even integers, so we directly return an empty array.

Otherwise, we can greedily split finalSum in the order of 2, 4, 6, cdots, until finalSum can no longer be split into a different positive even integer. At this point, we add the remaining finalSum to the last positive even integer.

The time complexity is O(\sqrt{finalSum}), and ignoring the space consumption of the answer array, the space complexity is O(1).

TypeScript

function maximumEvenSplit(finalSum: number): number[] {
    const ans: number[] = [];
    if (finalSum % 2 === 1) {
        return ans;
    }
    for (let i = 2; i <= finalSum; i += 2) {
        ans.push(i);
        finalSum -= i;
    }
    ans[ans.length - 1] += finalSum;
    return ans;
}

func maximumEvenSplit(finalSum int64) (ans []int64) {
	if finalSum%2 == 1 {
		return
	}
	for i := int64(2); i <= finalSum; i += 2 {
		ans = append(ans, i)
		finalSum -= i
	}
	ans[len(ans)-1] += finalSum
	return
}

Rust

impl Solution {
    pub fn maximum_even_split(mut final_sum: i64) -> Vec<i64> {
        let mut ans = Vec::new();
        if final_sum % 2 != 0 {
            return ans;
        }
        let mut i = 2;
        while i <= final_sum {
            ans.push(i);
            final_sum -= i;
            i += 2;
        }
        if let Some(last) = ans.last_mut() {
            *last += final_sum;
        }
        ans
    }
}

Try this approach in the editor

Complexity comparison

Approach	Complexity
Recursive Divide and Conquer	Time Complexity: O(n log n) Space Complexity: O(log n)
Iterative Dynamic Programming	Time Complexity: O(n) Space Complexity: O(n)
Greedy	–

Complexity analysis

Approach	Time	Space	When to Use
Recursive Divide and Conquer (Backtracking)	O(2^n)	O(n)	When exploring all possible combinations or demonstrating recursion and search space exploration
Iterative Dynamic Programming	O(n^2)	O(n)	When modeling the problem as a subset-sum style DP and tracking reachable sums
Greedy Mathematical Construction	O(sqrt(n))	O(1)	Best choice for production or interviews; constructs the maximum set directly

Frequently asked questions

Is Maximum Split of Positive Even Integers easy or hard?

The problem is rated Medium because the greedy insight is not immediately obvious. Many candidates initially try backtracking or subset-sum style dynamic programming. Recognizing that choosing the smallest distinct even numbers maximizes the count leads to the optimal solution.

Maximum Split of Positive Even Integers Python/Java solution

Python and Java implementations typically follow the greedy pattern: check if the input is odd, iterate through even numbers starting from 2, subtract them from the remaining sum, and store them in a list. When the remaining value is smaller than the next even number, add the remainder to the last element to maintain uniqueness.

How to solve Maximum Split of Positive Even Integers in O(n)?

An efficient solution uses a greedy sequence of even numbers. Start from 2 and keep adding the next even integer while subtracting from the remaining sum. When the remaining value becomes smaller than the next candidate, add the leftover to the last element. This effectively runs in about O(sqrt(n)) time, which is even faster than linear for large inputs.

What is the best approach for Maximum Split of Positive Even Integers?

The greedy approach is the most efficient solution. Start adding the smallest even numbers (2, 4, 6, ...) until the remaining sum becomes smaller than the next candidate. Add the leftover value to the last number to maintain uniqueness. This method runs in about O(sqrt(n)) time and uses constant extra space.

Is Maximum Split of Positive Even Integers asked at Google/Amazon/Meta?

Problems involving greedy number construction and unique sum decomposition appear frequently in interviews at large tech companies such as Google, Amazon, and Meta. The exact problem may vary, but the underlying pattern of greedy selection and mathematical reasoning is common.

What data structure is used in Maximum Split of Positive Even Integers?

The greedy solution mainly uses a dynamic list or array to store the selected even integers. No complex data structures are required because the algorithm simply generates numbers sequentially and adjusts the final element if a remainder exists.

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

The optimal greedy solution runs in O(sqrt(n)) time because the sum of the first k even numbers grows roughly quadratically. Backtracking solutions can take O(2^n) time due to exploring combinations, while dynamic programming implementations typically require O(n^2) time.

Maximum Split of Positive Even Integers solution and explanation

The problem

Approach overview

Approach 1: Recursive Divide and Conquer

Approach 2: Iterative Dynamic Programming

Approach 3: Greedy

Complexity comparison

Complexity analysis

Frequently asked questions

Comments

Maximum Split of Positive Even Integers solution and explanation

The problem

Approach overview

Approach 1: Recursive Divide and Conquer

Approach 2: Iterative Dynamic Programming

Approach 3: Greedy

Complexity comparison

Complexity analysis

Frequently asked questions

Comments

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