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Formatted question description: https://leetcode.ca/all/2178.html
2178. Maximum Split of Positive Even Integers (Medium)
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 tofinalSum):(2 + 10),(2 + 4 + 6), and(4 + 8). Among them,(2 + 4 + 6)contains the maximum number of integers. Note thatfinalSumcannot 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 some valid splits:(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 some 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 <= 1010
Solution 1. Backtracking
If finalSum is odd, return empty array.
Otherwise, we keep trying subtracting 2, 4, 6, 8, ... from finalSum via backtracking.
-
// OJ: https://leetcode.com/problems/maximum-split-of-positive-even-integers/ // Time: O(sqrt(N)) // Space: O(sqrt(N)) class Solution { public: vector<long long> maximumEvenSplit(long long s) { if (s % 2) return {}; vector<long long> ans; function<bool(long, long)>dfs = [&](long i, long target) { if (target == 0) return true; if (target < i) return false; ans.push_back(i); if (dfs(i + 2, target - i)) return true; // use `i` ans.pop_back(); return dfs(i + 2, target); // skip `i` }; dfs(2, s); return ans; } }; -
class Solution: def maximumEvenSplit(self, finalSum: int) -> List[int]: if finalSum % 2: return [] i = 2 ans = [] while i <= finalSum: ans.append(i) finalSum -= i i += 2 ans[-1] += finalSum return ans ############ # 2178. Maximum Split of Positive Even Integers # https://leetcode.com/problems/maximum-split-of-positive-even-integers/ class Solution: def maximumEvenSplit(self, s: int) -> List[int]: if s & 1: return [] res = [] x = 2 while x <= s: res.append(x) s -= x x += 2 res[-1] += s return res -
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; } } -
func maximumEvenSplit(finalSum int64) []int64 { ans := []int64{} if finalSum%2 == 1 { return ans } for i := int64(2); i <= finalSum; i += 2 { ans = append(ans, i) finalSum -= i } ans[len(ans)-1] += finalSum return ans } -
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; } -
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; } }
Solution 2. Greedy
In solution 1, in fact we only backtrack at most once.
We can keep trying subtracting 2, 4, 6, 8, ... from finalSum. We stop the loop when subtracting the current number i is invalid – s - i < i + 2 (the reminder after the subtraction is less than the next even number). And we push the reminder into the answer.
-
// OJ: https://leetcode.com/problems/maximum-split-of-positive-even-integers/ // Time: O(sqrt(N)) // Space: O(1) class Solution { public: vector<long long> maximumEvenSplit(long long s) { if (s % 2) return {}; vector<long long> ans; for (int i = 2; s - i >= i + 2; i += 2) { ans.push_back(i); s -= i; } ans.push_back(s); return ans; } }; -
class Solution: def maximumEvenSplit(self, finalSum: int) -> List[int]: if finalSum % 2: return [] i = 2 ans = [] while i <= finalSum: ans.append(i) finalSum -= i i += 2 ans[-1] += finalSum return ans ############ # 2178. Maximum Split of Positive Even Integers # https://leetcode.com/problems/maximum-split-of-positive-even-integers/ class Solution: def maximumEvenSplit(self, s: int) -> List[int]: if s & 1: return [] res = [] x = 2 while x <= s: res.append(x) s -= x x += 2 res[-1] += s return res -
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; } } -
func maximumEvenSplit(finalSum int64) []int64 { ans := []int64{} if finalSum%2 == 1 { return ans } for i := int64(2); i <= finalSum; i += 2 { ans = append(ans, i) finalSum -= i } ans[len(ans)-1] += finalSum return ans } -
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; } -
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; } }
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