README_EN.md

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

Python3

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:
    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]

Java

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(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];
    }
}

C++

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 {
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];
    }
};

Go

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]
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

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]
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

TypeScript

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];
}

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];
}

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