花花酱 LeetCode 1262. Greatest Sum Divisible by Three

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:

Solution: DP

dp[i] := max sum that has a remainder i when mod 3.

dp[(i + num) % 3] = max( dp[(i + num) % 3] , dp[i] + num)

ans: dp[0]

C++

// Author: Huahua

class Solution {

public:

int maxSumDivThree(vector<int>& nums) {

vector<int> dp(3);

for (int num : nums) {

vector<int> tmp(dp);

for (int s : tmp)

dp[(s + num) % 3] = max(dp[(s + num) % 3], s + num);

}

return dp[0];

}

};

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