Given an array nums of integers, we need to find the maximum possible sum of elements of the array such that it is divisible by three.
Example 1:
Input: nums = [3,6,5,1,8]
Output: 18
Explanation: Pick numbers 3, 6, 1 and 8 their sum is 18 (maximum sum divisible by 3).
Example 2:
Input: nums = [4]
Output: 0
Explanation: Since 4 is not divisible by 3, do not pick any number.
Example 3:
Input: nums = [1,2,3,4,4]
Output: 12
Explanation: Pick numbers 1, 3, 4 and 4 their sum is 12 (maximum sum divisible by 3).
Constraints:
-
1 <= nums.length <= 4 * 10^4 -
1 <= nums[i] <= 10^4
SOLUTION:
import bisect class Solution: def maxSumDivThree(self, nums: List[int]) -> int: msum = 0 n = len(nums) total = sum(nums) if total % 3 == 0: return total nums.sort() paths = [(nums[i], i) for i in range(n)] i = 0 while i < len(paths): currsum, curr = paths[i] if (total - currsum) % 3 == 0: return total - currsum for j in range(curr + 1, n): bisect.insort(paths, (currsum + nums[j], j)) i += 1 return 0 
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