leetcode/solution/3200-3299/3290.Maximum Multiplication Score

comments	difficulty	edit_url	rating	source	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/3200-3299/3290.Maximum%20Multiplication%20Score/README_EN.md	1692	Weekly Contest 415 Q2	Array Dynamic Programming

3290. Maximum Multiplication Score

中文文档

Description

You are given an integer array a of size 4 and another integer array b of size at least 4.

You need to choose 4 indices i0, i1, i2, and i3 from the array b such that i0 < i1 < i2 < i3. Your score will be equal to the value a[0] * b[i0] + a[1] * b[i1] + a[2] * b[i2] + a[3] * b[i3].

Return the maximum score you can achieve.

 

Example 1:

Input: a = [3,2,5,6], b = [2,-6,4,-5,-3,2,-7]

Output: 26

Explanation:
We can choose the indices 0, 1, 2, and 5. The score will be 3 * 2 + 2 * (-6) + 5 * 4 + 6 * 2 = 26.

Example 2:

Input: a = [-1,4,5,-2], b = [-5,-1,-3,-2,-4]

Output: -1

Explanation:
We can choose the indices 0, 1, 3, and 4. The score will be (-1) * (-5) + 4 * (-1) + 5 * (-2) + (-2) * (-4) = -1.

 

Constraints:

• a.length == 4
• 4 <= b.length <= 105
• -105 <= a[i], b[i] <= 105

Solutions

Solution 1: Memoization

We design a function \textit{dfs}(i, j), which represents the maximum score that can be obtained starting from the i-th element of array a and the j-th element of array b. Then the answer is \textit{dfs}(0, 0).

The function \textit{dfs}(i, j) is calculated as follows:

• If j \geq \text{len}(b), it means array b has been completely traversed. At this point, if array a has also been completely traversed, return 0; otherwise, return negative infinity.
• If i \geq \text{len}(a), it means array a has been completely traversed. Return 0.
• Otherwise, we can either skip the j-th element of array b and move to the next element, i.e., \textit{dfs}(i, j + 1); or we can choose the j-th element of array b, in which case the score is a[i] \times b[j] plus \textit{dfs}(i + 1, j + 1). We take the maximum of these two values as the return value of \textit{dfs}(i, j).

We can use memoization to avoid redundant calculations.

The time complexity is O(m \times n), and the space complexity is O(m \times n). Here, m and n are the lengths of arrays a and b, respectively.

Python3

class Solution:
    def maxScore(self, a: List[int], b: List[int]) -> int:
        @cache
        def dfs(i: int, j: int) -> int:
            if j >= len(b):
                return 0 if i >= len(a) else -inf
            if i >= len(a):
                return 0
            return max(dfs(i, j + 1), a[i] * b[j] + dfs(i + 1, j + 1))

        return dfs(0, 0)

Java

class Solution {
    private Long[][] f;
    private int[] a;
    private int[] b;

    public long maxScore(int[] a, int[] b) {
        f = new Long[a.length][b.length];
        this.a = a;
        this.b = b;
        return dfs(0, 0);
    }

    private long dfs(int i, int j) {
        if (j >= b.length) {
            return i >= a.length ? 0 : Long.MIN_VALUE / 2;
        }
        if (i >= a.length) {
            return 0;
        }
        if (f[i][j] != null) {
            return f[i][j];
        }
        return f[i][j] = Math.max(dfs(i, j + 1), 1L * a[i] * b[j] + dfs(i + 1, j + 1));
    }
}

C++

class Solution {
public:
    long long maxScore(vector<int>& a, vector<int>& b) {
        int m = a.size(), n = b.size();
        long long f[m][n];
        memset(f, -1, sizeof(f));
        auto dfs = [&](this auto&& dfs, int i, int j) -> long long {
            if (j >= n) {
                return i >= m ? 0 : LLONG_MIN / 2;
            }
            if (i >= m) {
                return 0;
            }
            if (f[i][j] != -1) {
                return f[i][j];
            }
            return f[i][j] = max(dfs(i, j + 1), 1LL * a[i] * b[j] + dfs(i + 1, j + 1));
        };
        return dfs(0, 0);
    }
};

Go

func maxScore(a []int, b []int) int64 {
	m, n := len(a), len(b)
	f := make([][]int64, m)
	for i := range f {
		f[i] = make([]int64, n)
		for j := range f[i] {
			f[i][j] = -1
		}
	}
	var dfs func(i, j int) int64
	dfs = func(i, j int) int64 {
		if j >= n {
			if i >= m {
				return 0
			}
			return math.MinInt64 / 2
		}
		if i >= m {
			return 0
		}
		if f[i][j] != -1 {
			return f[i][j]
		}
		f[i][j] = max(dfs(i, j+1), int64(a[i])*int64(b[j])+dfs(i+1, j+1))
		return f[i][j]
	}
	return dfs(0, 0)
}

TypeScript

function maxScore(a: number[], b: number[]): number {
    const [m, n] = [a.length, b.length];
    const f: number[][] = Array.from({ length: m }, () => Array(n).fill(-1));
    const dfs = (i: number, j: number): number => {
        if (j >= n) {
            return i >= m ? 0 : -Infinity;
        }
        if (i >= m) {
            return 0;
        }
        if (f[i][j] !== -1) {
            return f[i][j];
        }
        return (f[i][j] = Math.max(dfs(i, j + 1), a[i] * b[j] + dfs(i + 1, j + 1)));
    };
    return dfs(0, 0);
}