Question

1262. Greatest Sum Divisible by Three

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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:
  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]

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];
  }
};

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];
}

Solution 2

class Solution:
  def maxSumDivThree(self, nums: List[int]) -> int:
    f = [0, -inf, -inf]
    for x in nums:
      g = f[:]
      for j in range(3):
        g[j] = max(f[j], f[(j - x) % 3] + x)
      f = g
    return f[0]

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

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

func maxSumDivThree(nums []int) int {
  const inf = 1 << 30
  f := [3]int{0, -inf, -inf}
  for _, x := range nums {
    g := [3]int{}
    for j := range f {
      g[j] = max(f[j], f[(j-x%3+3)%3]+x)
    }
    f = g
  }
  return f[0]
}

function maxSumDivThree(nums: number[]): number {
  const inf = 1 << 30;
  const f: number[] = [0, -inf, -inf];
  for (const x of nums) {
    const g = [...f];
    for (let j = 0; j < 3; ++j) {
      f[j] = Math.max(g[j], g[(j - (x % 3) + 3) % 3] + x);
    }
  }
  return f[0];
}