Scale Engineer

Maximum Split of Positive Even Integers

Med
#1983Time: O(2^N) or higher (exponential)Space: O(sqrt(finalSum))
Patterns
MathGreedyBacktracking

Prompt

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:

(12)

Example 2:

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

Example 3:

(2 + 26)

 

Constraints:

  • 1 <= finalSum <= 1010

Approaches

2 approaches with complexity analysis and trade-offs.

This approach attempts to find the solution by exploring all possible combinations of unique positive even integers that sum up to finalSum. It uses a recursive function to build potential splits. At each step, the function decides which even number to add next and recurses with the remaining sum. This brute-force method will eventually find the split with the maximum number of elements by checking every valid partition.

Algorithm

  1. Handle the base case: if finalSum is odd, it's impossible to form it from even numbers, so return an empty list.
  2. Use a recursive backtracking function, say solve(remainingSum, startEvenNum, currentSplit), to explore all possibilities.
  3. remainingSum: The target sum left to achieve.
  4. startEvenNum: The next smallest even number we can consider adding to ensure uniqueness.
  5. currentSplit: The list of numbers forming the current partial solution.
  6. Base Case: If remainingSum is 0, a valid split has been found. Compare its size with the maximum size found so far and update the global result if it's larger.
  7. Recursive Step: For each even number i starting from startEvenNum up to remainingSum: a. Add i to currentSplit. b. Make a recursive call: solve(remainingSum - i, i + 2, currentSplit). c. Backtrack by removing i from currentSplit to explore other branches.
  8. This process explores all possible partitions into unique even numbers, and we keep the one with the most elements.

Walkthrough

The backtracking algorithm systematically generates all partitions of finalSum into unique, positive, even integers. We define a recursive helper function that takes the remaining sum to be made, the starting even number to consider (to enforce uniqueness and order), and the current list of numbers in the split.

The function has two main cases:

  1. Base Case: If the remaining sum is zero, it means we've successfully found a valid split. We then check if this new split has more numbers than the best one we've found so far. If it does, we update our answer.
  2. Recursive Step: We iterate through all possible even numbers i that can be added (i.e., i <= remainingSum). For each i, we add it to our current split and make a recursive call for the new remaining sum (remainingSum - i). To ensure the numbers are unique, the next recursive call can only use even numbers greater than i. After the recursive call returns, we backtrack by removing i to explore other possibilities.

While correct in theory, this approach is computationally expensive as the number of partitions can grow exponentially with finalSum.

// NOTE: This is a conceptual implementation and will Time Limit Exceed.class Solution {    List<Long> maxSplit = new ArrayList<>();     public List<Long> maximumEvenSplit(long finalSum) {        if (finalSum % 2 != 0) {            return new ArrayList<>();        }        find(finalSum, 2L, new ArrayList<>());        return maxSplit;    }     private void find(long target, long startNum, List<Long> currentSplit) {        if (target == 0) {            if (currentSplit.size() > maxSplit.size()) {                maxSplit = new ArrayList<>(currentSplit);            }            return;        }         if (target < startNum) {            return;        }         // Explore all possible next numbers        for (long i = startNum; i <= target; i += 2) {            currentSplit.add(i);            find(target - i, i + 2, currentSplit);            currentSplit.remove(currentSplit.size() - 1); // Backtrack        }    }}

Complexity

Time

O(2^N) or higher (exponential)

Space

O(sqrt(finalSum))

Trade-offs

Pros

  • It's a straightforward brute-force method that is guaranteed to find the optimal solution if given enough time.

Cons

  • Extremely inefficient due to its exponential time complexity.

  • Will result in a 'Time Limit Exceeded' (TLE) error for the given constraints.

  • The recursion depth can be large, potentially leading to a stack overflow for very large finalSum, although the number of elements is limited by sqrt(finalSum).

Solutions

public class Solution {    public IList < long > MaximumEvenSplit(long finalSum) {        IList < long > ans = new List < long > ();        if (finalSum % 2 == 1) {            return ans;        }        for (long i = 2; i <= finalSum; i += 2) {            ans.Add(i);            finalSum -= i;        }        ans[ans.Count - 1] += finalSum;        return ans;    }}

Video walkthrough

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