# 1262. Greatest Sum Divisible by Three

## Description

Given an integer array nums, return 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 * 104
• 1 <= nums[i] <= 104

## Solutions

Solution 1: Dynamic Programming

We define $f[i][j]$ as the maximum sum of several numbers selected from the first $i$ numbers, such that the sum modulo $3$ equals $j$. Initially, $f[0][0]=0$, and the rest are $-\infty$.

For $f[i][j]$, we can consider the state of the $i$th number $x$:

• If we do not select $x$, then $f[i][j]=f[i-1][j]$;
• If we select $x$, then $f[i][j]=f[i-1][(j-x \bmod 3 + 3)\bmod 3]+x$.

Therefore, we can get the state transition equation:

$f[i][j]=\max\{f[i-1][j],f[i-1][(j-x \bmod 3 + 3)\bmod 3]+x\}$

The final answer is $f[n][0]$.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the length of the array $nums$.

Note that the value of $f[i][j]$ is only related to $f[i-1][j]$ and $f[i-1][(j-x \bmod 3 + 3)\bmod 3]$, so we can use a rolling array to optimize the space complexity, reducing the space complexity to $O(1)$.

• class Solution {
public int maxSumDivThree(int[] nums) {
int n = nums.length;
final int inf = 1 << 30;
int[][] f = new int[n + 1][3];
f[0][1] = f[0][2] = -inf;
for (int i = 1; i <= n; ++i) {
int x = nums[i - 1];
for (int j = 0; j < 3; ++j) {
f[i][j] = Math.max(f[i - 1][j], f[i - 1][(j - x % 3 + 3) % 3] + x);
}
}
return f[n][0];
}
}

• class Solution {
public:
int maxSumDivThree(vector<int>& nums) {
int n = nums.size();
const int inf = 1 << 30;
int f[n + 1][3];
f[0][0] = 0;
f[0][1] = f[0][2] = -inf;
for (int i = 1; i <= n; ++i) {
int x = nums[i - 1];
for (int j = 0; j < 3; ++j) {
f[i][j] = max(f[i - 1][j], f[i - 1][(j - x % 3 + 3) % 3] + x);
}
}
return f[n][0];
}
};

• class Solution:
def maxSumDivThree(self, nums: List[int]) -> int:
n = len(nums)
f = [[-inf] * 3 for _ in range(n + 1)]
f[0][0] = 0
for i, x in enumerate(nums, 1):
for j in range(3):
f[i][j] = max(f[i - 1][j], f[i - 1][(j - x) % 3] + x)
return f[n][0]


• func maxSumDivThree(nums []int) int {
n := len(nums)
const inf = 1 << 30
f := make([][3]int, n+1)
f[0] = [3]int{0, -inf, -inf}
for i, x := range nums {
i++
for j := 0; j < 3; j++ {
f[i][j] = max(f[i-1][j], f[i-1][(j-x%3+3)%3]+x)
}
}
return f[n][0]
}

• function maxSumDivThree(nums: number[]): number {
const n = nums.length;
const inf = 1 << 30;
const f: number[][] = Array(n + 1)
.fill(0)
.map(() => Array(3).fill(-inf));
f[0][0] = 0;
for (let i = 1; i <= n; ++i) {
const x = nums[i - 1];
for (let j = 0; j < 3; ++j) {
f[i][j] = Math.max(f[i - 1][j], f[i - 1][(j - (x % 3) + 3) % 3] + x);
}
}
return f[n][0];
}