You are given an integer finalSum. Split it into a sum of a maximum number of unique positive even integers.
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 <= 1010This 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.
The recursive solution divides the array into two halves and solves each half independently. It then combines the results to form the final answer.
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Time Complexity: O(n log n) for a balanced division.
Space Complexity: O(log n) due to recursion stack space.
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.
Using a DP array, the cumulative results from the array calculation are stored iteratively. This prevents recursive stack overflow and uses previous results for new calculations.
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Time Complexity: O(n)
Space Complexity: O(n) due to the DP array.
| Approach | Complexity |
|---|---|
| Recursive Divide and Conquer | Time Complexity: O(n log n) for a balanced division. |
| Iterative Dynamic Programming | Time Complexity: O(n) |
Split Array Largest Sum - Leetcode 410 - Python • NeetCode • 99,331 views views
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