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
):(12)
,(2 + 10)
,(2 + 4 + 6)
, and(4 + 8)
. Among them,(2 + 4 + 6)
contains the maximum number of integers. Note thatfinalSum
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 <= 1010
Solution: Greedy
The get the maximum number of elements, we must use the smallest numbers.
[2, 4, 6, …, 2k, x], where x > 2k
let s = 2 + 4 + … + 2k, x = num – s
since num is odd and s is also odd, so thus x = num – s.
Time complexity: O(sqrt(num)) for constructing outputs.
Space complexity: O(1)
C++
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// Author: Huahua class Solution { public: vector<long long> maximumEvenSplit(long long finalSum) { if (finalSum & 1) return {}; vector<long long> ans; long long s = 0; for (long long i = 2; s + i <= finalSum; s += i, i += 2) ans.push_back(i); ans.back() += (finalSum - s); return ans; } }; |
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