Problem Description

You need to split a given integer finalSum into a sum of unique positive even integers, aiming to maximize the count of integers used.

The problem requires:

• All integers in the split must be positive and even (2, 4, 6, 8, ...)
• All integers must be unique (no duplicates allowed)
• The sum of these integers must equal finalSum
• You want to use as many integers as possible

For example, if finalSum = 12:

• Valid splits include: (12), (2 + 10), (2 + 4 + 6), and (4 + 8)
• Among these, (2 + 4 + 6) uses 3 integers, which is the maximum possible
• (2 + 2 + 4 + 4) would sum to 12 but is invalid because it contains duplicates

If it's impossible to split finalSum into unique positive even integers (which happens when finalSum is odd), return an empty list.

The solution uses a greedy approach: start with the smallest even number (2) and keep adding consecutive even numbers (4, 6, 8, ...) until you can't add another without exceeding finalSum. Then, add any remaining value to the last number in your list to ensure the total equals finalSum.

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• No: The problem doesn't involve nodes and edges or graph traversal. We're dealing with splitting a number into a sum of integers.

Need to solve for kth smallest/largest?

• No: We're not looking for the kth smallest or largest element. We want to maximize the count of integers in our split.

Involves Linked Lists?

• No: The problem doesn't involve linked list data structures or operations.

Does the problem have small constraints?

• Yes: While the specific constraints aren't given in the problem description, the nature of splitting a number into unique positive even integers suggests we're dealing with a finite set of possibilities. We need to explore different combinations of even integers that sum to finalSum.

Brute force / Backtracking?

• Yes: Backtracking would be appropriate here because:
  • We need to find all valid combinations of unique even integers that sum to finalSum
  • We must explore different possibilities (e.g., should we include 2? 4? 6?)
  • We need to backtrack when a path doesn't lead to a valid solution
  • Among all valid solutions, we need to find the one with maximum count

Conclusion: The flowchart suggests using a backtracking approach for this problem. We would recursively try adding different even integers (2, 4, 6, ...) to our current combination, checking if they sum to finalSum, and backtracking when necessary. Among all valid combinations found, we'd return the one with the maximum number of integers.

Note: While the actual solution uses a greedy approach (which is more efficient), the flowchart correctly identifies that backtracking is a valid pattern for solving this type of combination/partition problem where we need to explore different ways to split a number.

Intuition

To maximize the number of unique positive even integers that sum to finalSum, we want to use the smallest possible even integers. This is because smaller numbers allow us to include more numbers in our sum.

Think about it this way: if we have finalSum = 12, we could use just (12) which gives us 1 number, or we could use (2 + 4 + 6) which gives us 3 numbers. The key insight is that using smaller numbers leaves more "room" for additional numbers.

The greedy strategy emerges naturally: start with the smallest even number (2), then the next smallest (4), then 6, 8, and so on. We keep adding consecutive even numbers as long as we can without exceeding finalSum.

But what happens when we can't add the next even number without going over? For example, if finalSum = 20 and we've already added 2 + 4 + 6 = 12, the next number would be 8, but adding it would give us exactly 20. That's perfect! However, if finalSum = 30 and we have 2 + 4 + 6 + 8 = 20, the next number would be 10, giving us exactly 30 - also perfect!

The tricky case is when adding the next consecutive even number would exceed finalSum. For instance, if finalSum = 28 and we have 2 + 4 + 6 + 8 = 20, the next number would be 10, but that gives us 30, which is too much. We have 8 remaining. Instead of trying to fit this remainder as a new number (which might create duplicates), we simply add it to our last number: 2 + 4 + 6 + (8 + 8) = 2 + 4 + 6 + 16 = 28.

This greedy approach guarantees we use the maximum number of integers because we're always using the smallest possible values, and we handle any remainder by adjusting the last number rather than reducing our count.

If finalSum is odd, it's impossible to express it as a sum of even integers (since the sum of even integers is always even), so we return an empty list immediately.

Learn more about Greedy, Math and Backtracking patterns.

Solution Approach

The implementation follows the greedy strategy we identified:

1. Check for odd numbers: First, we check if finalSum is odd using the bitwise AND operation finalSum & 1. If the result is 1 (meaning finalSum is odd), we immediately return an empty list since no sum of even integers can produce an odd number.

2. Initialize variables: We create an empty list ans to store our result and set i = 2 as our starting even number.

3. Greedy accumulation: We use a while loop with condition i <= finalSum. In each iteration:

   • Subtract the current even number i from finalSum: finalSum -= i
   • Add i to our answer list: ans.append(i)
   • Move to the next even number: i += 2

   This loop continues as long as we can "afford" to subtract the next even number from our remaining sum.

4. Handle the remainder: After the loop exits, finalSum contains the remainder that couldn't form another distinct even number. We add this remainder to the last number in our list: ans[-1] += finalSum. This ensures:

   • All numbers remain distinct (we're modifying an existing number, not adding a duplicate)
   • The total sum equals the original finalSum
   • We maintain the maximum count of numbers

For example, with finalSum = 28:

• Start with i = 2, finalSum = 28, ans = []
• Add 2: finalSum = 26, ans = [2], i = 4
• Add 4: finalSum = 22, ans = [2, 4], i = 6
• Add 6: finalSum = 16, ans = [2, 4, 6], i = 8
• Add 8: finalSum = 8, ans = [2, 4, 6, 8], i = 10
• Can't add 10 (since 10 > 8), loop exits
• Add remainder to last element: ans = [2, 4, 6, 16]

This approach runs in O(√n) time since we're iterating through even numbers up to approximately √(2 * finalSum), and uses O(√n) space for storing the result.

Example Walkthrough

Let's walk through the solution with finalSum = 14:

Step 1: Check if odd

• 14 & 1 = 0, so 14 is even. We can proceed.

Step 2: Initialize

• ans = [] (empty list for our result)
• i = 2 (start with smallest even number)
• finalSum = 14

Step 3: Greedy accumulation

Iteration 1:

• Can we subtract 2? Yes, since 2 ≤ 14
• finalSum = 14 - 2 = 12
• ans = [2]
• i = 2 + 2 = 4

Iteration 2:

• Can we subtract 4? Yes, since 4 ≤ 12
• finalSum = 12 - 4 = 8
• ans = [2, 4]
• i = 4 + 2 = 6

Iteration 3:

• Can we subtract 6? Yes, since 6 ≤ 8
• finalSum = 8 - 6 = 2
• ans = [2, 4, 6]
• i = 6 + 2 = 8

Iteration 4:

• Can we subtract 8? No, since 8 > 2
• Loop exits

Step 4: Handle remainder

• We have finalSum = 2 remaining
• Add this to the last element: ans[-1] = 6 + 2 = 8
• Final answer: ans = [2, 4, 8]

Verification:

• All numbers are positive and even ✓
• All numbers are unique ✓
• Sum: 2 + 4 + 8 = 14 ✓
• We maximized the count (3 numbers) by starting with the smallest possible values

Solution Implementation

Time and Space Complexity

The time complexity is O(√finalSum). This is because the loop starts with i = 2 and increments by 2 in each iteration (i += 2), while subtracting i from finalSum each time. The loop continues while i <= finalSum. Since we're adding consecutive even numbers (2, 4, 6, 8, ...), and the sum of the first n even numbers is n(n+1), the maximum number of iterations occurs when n(n+1) ≈ finalSum. Solving for n, we get n ≈ √finalSum, which means the loop runs approximately O(√finalSum) times.

The space complexity is O(1) if we exclude the space used by the answer array ans. The algorithm only uses a constant amount of extra space for the variable i and modifies finalSum in place. The answer array ans grows proportionally to the number of even numbers added, which is O(√finalSum), but following the convention stated in the reference answer, we ignore the space consumption of the answer array when analyzing space complexity.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Index Error When Adding Remainder to Empty List

A critical pitfall occurs when trying to access result[-1] on an empty list. This happens when finalSum is very small (like 2), causing the while loop to exit immediately after the first iteration or even without entering at all.

Problematic scenario:

• If finalSum = 2: After subtracting 2, finalSum becomes 0, and current_even becomes 4. The loop exits with result = [2] and finalSum = 0. Adding 0 to the last element works fine.
• However, if someone modifies the loop condition incorrectly (e.g., current_even < finalSum instead of <=), the list could be empty when trying to access result[-1].

Solution: Add a safety check before modifying the last element:

if result and finalSum > 0:
    result[-1] += finalSum

2. Integer Overflow in Loop Condition

When dealing with very large values of finalSum, continuously adding even numbers and checking current_even <= finalSum could theoretically cause issues if not handled properly, especially in languages with fixed integer sizes.

Solution: Use subtraction-based checking instead of addition:

while finalSum >= current_even:
    finalSum -= current_even
    result.append(current_even)
    current_even += 2

3. Forgetting to Handle the Case Where No Remainder Exists

The code adds finalSum (the remainder) to the last element unconditionally. If finalSum is exactly 0 after the loop (a perfect triangular even sum), this still works correctly, but it's worth understanding this edge case.

Example: finalSum = 2 results in [2], not an error, because we add 0 to the last element.

4. Using Wrong Parity Check

Using modulo operator finalSum % 2 instead of bitwise AND finalSum & 1 works correctly but is slightly less efficient. However, some might incorrectly write if finalSum & 2 thinking it checks for even numbers, which is wrong.

Correct approaches:

• if finalSum & 1: (checks if odd)
• if finalSum % 2 == 1: (checks if odd)
• if finalSum % 2: (checks if odd)

5. Not Understanding Why Adding Remainder to Last Element Works

A conceptual pitfall is not understanding why adding the remainder to the last element always produces a valid unique even number. The key insight is that the remainder is always less than the next even number we would have added (otherwise we would have added it), and since we're adding it to an already large even number, the result remains unique and even.

Mathematical guarantee:

• If we stop at even number k, the remainder is less than k+2
• The last element in our list is at least k
• So k + remainder < k + (k+2) = 2k + 2
• This ensures the modified last element doesn't conflict with any other numbers in our sequence