LeetCode 5365. can be divisible by three of the largest and Greatest Sum Divisible by Three

Address  https://www.acwing.com/solution/leetcode/content/6340/

Title Description
give you an array of integers nums, and returns you to identify the elements that can be divisible by three and maximum.

Example 1 : 

Input: the nums = [ 3 , 6 , . 5 , 1 , 8 ] 
output: 18 
Explanation: select number 3 , 6 , 1 and 8 , and they are 18 (which may be 3 divisible and maximum). 
Example 2 : 

Input: the nums = [ 4 ] 
Output: 0 
Explanation: 4 can not be 3 divisible numbers can not be selected, returns 0 . 
Example 3 : 

Input: the nums = [ . 1 , 2, 3 , 4 , 4 ] 
Output: 12 
Explanation: selected number 1 , 3 , 4 and 4 , and they are 12 (which may be 3 divisible and maximum). 
 

Tip: 

. 1 <= nums.length <= . 4 * 10 ^ . 4 
. 1 <= the nums [I] <= 10 ^ . 4

Algorithm 1
last arrays, and only three cases
1 divided by 3 0 direct return of more than
2 divided by more than 31 then either a reduction of more than 31 digital divide or reduce the number more than two divided by 32 is
3 is divided by 3 2 then divided by either a reduction of more than 32 digital or reduce the number more than two divided by 3 1

class Solution {
public:
    vector<int> v[3];
int Check(int singleIdx,int doubleIdx,int sum)
{
    if (v[doubleIdx].size() < 2) {
            return sum - v[singleIdx][0];
        }
        else if (v[singleIdx].size() == 0) {
            return sum - v[doubleIdx][0] - v[doubleIdx][1];
        }
        else {
            int rem = v[singleIdx][0];
            if (rem > (v[doubleIdx][0] + v[doubleIdx][1]))  rem = (v[doubleIdx][0] + v[doubleIdx][1]);

            return sum - rem;
        }
}

int maxSumDivThree(vector<int>& nums) {
    int sum = 0;
    for (int i = 0; i < nums.size(); i++)
    {
        sum += nums[i];
        if (nums[i] % 3 == 1) {
            v[1].push_back(nums[i]);
        }
        else if(nums[i] % 3 == 2){
            v[2].push_back(nums[i]);
        }
    }

    sort(v[1].begin(), v[1].end());
    sort(v[2].begin(), v[2].end());
    int sum_n = sum % 3;

    if (sum_n == 0) return sum;
    if (sum_n == 1) {
        // reduce a two v1 and v2 select 
        return the Check ( . 1 , 2 , SUM); 
    } 

    IF (sum_n == 2 ) {
         return the Check ( 2 , . 1 , SUM); 
    } 

    return - . 1 ; 
} 

};