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

Thinking

We want the maximum subsequence sum divisible by \(3\). \(n \le 4\times 10^4\) forbids subsets. A sum has only three remainders modulo \(3\), and taking or skipping the current value moves among those three.

\(f[i][j]\) is the best sum of the first \(i\) numbers with remainder \(j\). Skip inherits; take adds \(x\) from remainder \(j-x\). The answer is \(f[n][0]\). Remainders collapse the constraint to a constant number of states.

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 \(nums\).

Python3JavaC++GoTypeScript

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: Dynamic Programming (Rolling Array)

Thinking

Solution 1 stores \(O(n)\) rows. Row \(i\) reads only three previous values, so a length-\(3\) rolling array suffices. Extra space is constant; the transition is unchanged.

\(f[i][j]\) depends only on the previous row's three residues, so an array of length \(3\) is enough. The space complexity is \(O(1)\).

Python3JavaC++GoTypeScript

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];
}

1262. Greatest Sum Divisible by Three

Description

Solutions

Solution 1: Dynamic Programming

Solution 2: Dynamic Programming (Rolling Array)

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