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]f[i][j] as the maximum sum of several numbers selected from the first ii numbers, such that the sum modulo 33 equals jj. Initially, f[0][0]=0f[0][0]=0, and the rest are −∞-\infty.

For f[i][j]f[i][j], we can consider the state of the iith number xx:

  • If we do not select xx, then f[i][j]=f[i−1][j]f[i][j]=f[i-1][j];
  • If we select xx, then f[i][j]=f[i−1][(j−x mod 3+3) mod 3]+xf[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 mod 3+3) mod 3]+x 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]f[n][0].

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

Note that the value of f[i][j]f[i][j] is only related to f[i−1][j]f[i-1][j] and f[i−1][(j−x mod 3+3) mod 3]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)O(1).

Python Code
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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]

Java Code
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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];
    }
}

C++ Code
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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];
    }
};

Go Code
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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]
}

TypeScript Code
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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

Python Code
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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 Code
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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++ Code
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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 Code
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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]
}

TypeScript Code
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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];
}