Maximum Split of Positive Even Integers - Practice Coding Problems

Maximum Split of Positive Even Integers - Problem

You're given an integer finalSum and need to split it into the maximum number of unique positive even integers that sum up to finalSum.

Think of this as a mathematical puzzle: how can you break down a number into as many different even pieces as possible? For example, with finalSum = 12, you could use:

(12) - just one number
(2 + 10) - two numbers
(2 + 4 + 6) - three numbers ✨ This is optimal!
(4 + 8) - two numbers

Important: All numbers must be unique and positive even integers. You cannot use (2 + 2 + 4 + 4) because numbers repeat.

Return a list representing the split with the maximum number of integers, or an empty list if no valid split exists.

Input & Output

example_1.py — Basic Case

$ Input: finalSum = 12

› Output: [2, 4, 6]

💡 Note: We can split 12 into 2 + 4 + 6 = 12. This uses 3 unique positive even integers, which is the maximum possible. Other valid splits like (12) or (2, 10) use fewer numbers.

example_2.py — Small Even Number

$ Input: finalSum = 7

› Output: []

💡 Note: Since 7 is odd, it's impossible to split it into a sum of even integers. Even + Even + ... + Even will always be Even, never Odd.

example_3.py — Minimum Case

$ Input: finalSum = 2

› Output: [2]

💡 Note: The smallest positive even integer is 2. Since finalSum = 2, the only possible split is [2], which uses 1 number.

Constraints

1 ≤ finalSum ≤ 10⁹
All integers in the result must be unique and positive even
The result should contain the maximum possible number of integers

Visualization

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Understanding the Visualization

Start Small

Begin with the smallest even block (2)

Add Consecutively

Keep adding the next even block (4, 6, 8, ...)

Check Remaining

Stop when the next block would exceed our target

Adjust Final

Combine remaining value with the last block to hit exact target

Key Takeaway

🎯 Key Insight: Greedy selection of smallest even numbers maximizes count, and we can always adjust the final number to hit the exact target while maintaining uniqueness.

Asked in

f Meta 35 G Google 28 a Amazon 22 ⊞ Microsoft 18

The optimal solution uses a greedy approach: start with the smallest even integers (2, 4, 6, ...) and adjust the final number to reach the exact target. This maximizes the count of unique integers while ensuring they're all positive and even. Time complexity is O(√n) and space complexity is O(√n).

Common Approaches

Approach	Time	Space	Notes
✓ Greedy Algorithm (Optimal)	O(√n)	O(√n)	Greedily pick smallest even numbers first, then adjust the last number to reach target
Brute Force (Generate All Combinations)	O(2^n)	O(n)	Generate all possible combinations of unique even integers and find the one with maximum count

Greedy Algorithm (Optimal) — Algorithm Steps

Check if finalSum is odd - if so, return empty list
Start with the smallest even number (2) and keep adding consecutive even numbers
Stop when adding the next even number would exceed finalSum
Calculate the remaining sum needed
Add this remaining sum to the last number in our list
Verify that the final number doesn't create duplicates

Visualization

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Step-by-Step Walkthrough

Start with 2

Add the smallest even number: [2], remaining = finalSum - 2

Add 4

Add next even number: [2, 4], remaining = finalSum - 6

Add 6

Continue pattern: [2, 4, 6], remaining = finalSum - 12

Adjust final

When remaining < next_even, add remaining to get exact sum

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>

long long* maximumEvenSplit(long long finalSum, int* returnSize) {
    *returnSize = 0;
    if (finalSum % 2 != 0) {
        return NULL;
    }
    
    long long* result = (long long*)malloc(1000 * sizeof(long long));
    long long currentEven = 2;
    long long remaining = finalSum;
    
    while (remaining > currentEven) {
        result[(*returnSize)++] = currentEven;
        remaining -= currentEven;
        currentEven += 2;
    }
    
    if (remaining > 0) {
        if (*returnSize > 0 && remaining == result[*returnSize - 1]) {
            result[*returnSize - 1] += remaining;
        } else {
            result[(*returnSize)++] = remaining;
        }
    }
    
    return result;
}

int main() {
    long long finalSum;
    scanf("%lld", &finalSum);
    
    int returnSize;
    long long* result = maximumEvenSplit(finalSum, &returnSize);
    
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        printf("%lld", result[i]);
        if (i < returnSize - 1) printf(", ");
    }
    printf("]\n");
    
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(√n)

We iterate through even numbers up to approximately √(2*finalSum), as the sum of first k even numbers is k(k+1)

✓ Linear Growth

Space Complexity

O(√n)

We store at most √(2*finalSum) numbers in our result array

⚡ Linearithmic Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Greedy Algorithm (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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