3290. Maximum Multiplication Score

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.

Python3JavaC++GoTypeScript

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)

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));
    }
}

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 = [&](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(dfs, i, j + 1), 1LL * a[i] * b[j] + dfs(dfs, i + 1, j + 1));
        };
        return dfs(dfs, 0, 0);
    }
};

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)
}

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);
}

3290. Maximum Multiplication Score

Description

Solutions

Solution 1: Memoization

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