Overflow in intermediate arithmetic
Wrong move: Temporary multiplications exceed integer bounds.
Usually fails on: Large inputs wrap around unexpectedly.
Fix: Use wider types, modular arithmetic, or rearranged operations.
Move from brute-force thinking to an efficient approach using math strategy.
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 <= 1010Problem summary: 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.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Math · Backtracking · Greedy
12
7
28
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2178: Maximum Split of Positive Even Integers
class Solution {
public List<Long> maximumEvenSplit(long finalSum) {
List<Long> ans = new ArrayList<>();
if (finalSum % 2 == 1) {
return ans;
}
for (long i = 2; i <= finalSum; i += 2) {
ans.add(i);
finalSum -= i;
}
ans.add(ans.remove(ans.size() - 1) + finalSum);
return ans;
}
}
// Accepted solution for LeetCode #2178: Maximum Split of Positive Even Integers
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
}
# Accepted solution for LeetCode #2178: Maximum Split of Positive Even Integers
class Solution:
def maximumEvenSplit(self, finalSum: int) -> List[int]:
if finalSum & 1:
return []
ans = []
i = 2
while i <= finalSum:
finalSum -= i
ans.append(i)
i += 2
ans[-1] += finalSum
return ans
// Accepted solution for LeetCode #2178: Maximum Split of Positive Even Integers
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
}
}
// Accepted solution for LeetCode #2178: Maximum Split of Positive Even Integers
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;
}
Use this to step through a reusable interview workflow for this problem.
Generate every possible combination without any filtering. At each of n positions we choose from up to n options, giving nⁿ total candidates. Each candidate takes O(n) to validate. No pruning means we waste time on clearly invalid partial solutions.
Backtracking explores a decision tree, but prunes branches that violate constraints early. Worst case is still factorial or exponential, but pruning dramatically reduces the constant factor in practice. Space is the recursion depth (usually O(n) for n-level decisions).
Review these before coding to avoid predictable interview regressions.
Wrong move: Temporary multiplications exceed integer bounds.
Usually fails on: Large inputs wrap around unexpectedly.
Fix: Use wider types, modular arithmetic, or rearranged operations.
Wrong move: Mutable state leaks between branches.
Usually fails on: Later branches inherit selections from earlier branches.
Fix: Always revert state changes immediately after recursive call.
Wrong move: Locally optimal choices may fail globally.
Usually fails on: Counterexamples appear on crafted input orderings.
Fix: Verify with exchange argument or monotonic objective before committing.