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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 $O(1)$.

• Java
• C++
• Python
• Go
• TypeScript

• 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];
  }
  
  

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