README_EN.md

1770. Maximum Score from Performing Multiplication Operations

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Description

You are given two 0-indexed integer arrays nums and multipliers of size n and m respectively, where n >= m.

You begin with a score of 0. You want to perform exactly m operations. On the ith operation (0-indexed) you will:

• Choose one integer x from either the start or the end of the array nums.
• Add multipliers[i] * x to your score.
  • Note that multipliers[0] corresponds to the first operation, multipliers[1] to the second operation, and so on.
• Remove x from nums.

Return the maximum score after performing m operations.

 

Example 1:

Input: nums = [1,2,3], multipliers = [3,2,1]
Output: 14
Explanation: An optimal solution is as follows:
- Choose from the end, [1,2,3], adding 3 * 3 = 9 to the score.
- Choose from the end, [1,2], adding 2 * 2 = 4 to the score.
- Choose from the end, [1], adding 1 * 1 = 1 to the score.
The total score is 9 + 4 + 1 = 14.

Example 2:

Input: nums = [-5,-3,-3,-2,7,1], multipliers = [-10,-5,3,4,6]
Output: 102
Explanation: An optimal solution is as follows:
- Choose from the start, [-5,-3,-3,-2,7,1], adding -5 * -10 = 50 to the score.
- Choose from the start, [-3,-3,-2,7,1], adding -3 * -5 = 15 to the score.
- Choose from the start, [-3,-2,7,1], adding -3 * 3 = -9 to the score.
- Choose from the end, [-2,7,1], adding 1 * 4 = 4 to the score.
- Choose from the end, [-2,7], adding 7 * 6 = 42 to the score. 
The total score is 50 + 15 - 9 + 4 + 42 = 102.

 

Constraints:

• n == nums.length
• m == multipliers.length
• 1 <= m <= 300
• m <= n <= 105
• -1000 <= nums[i], multipliers[i] <= 1000

Solutions

Python3

class Solution:
    def maximumScore(self, nums: List[int], multipliers: List[int]) -> int:
        @cache
        def f(i, j, k):
            if k >= m or i >= n or j < 0:
                return 0
            a = f(i + 1, j, k + 1) + nums[i] * multipliers[k]
            b = f(i, j - 1, k + 1) + nums[j] * multipliers[k]
            return max(a, b)

        n = len(nums)
        m = len(multipliers)
        return f(0, n - 1, 0)

Java

class Solution {
    private Integer[][] f;
    private int[] multipliers;
    private int[] nums;
    private int n;
    private int m;

    public int maximumScore(int[] nums, int[] multipliers) {
        n = nums.length;
        m = multipliers.length;
        f = new Integer[m][m];
        this.nums = nums;
        this.multipliers = multipliers;
        return dfs(0, 0);
    }

    private int dfs(int i, int j) {
        if (i >= m || j >= m || (i + j) >= m) {
            return 0;
        }
        if (f[i][j] != null) {
            return f[i][j];
        }
        int k = i + j;
        int a = dfs(i + 1, j) + nums[i] * multipliers[k];
        int b = dfs(i, j + 1) + nums[n - 1 - j] * multipliers[k];
        f[i][j] = Math.max(a, b);
        return f[i][j];
    }
}

C++

class Solution {
public:
    int maximumScore(vector<int>& nums, vector<int>& multipliers) {
        int n = nums.size(), m = multipliers.size();
        int f[m][m];
        memset(f, 0x3f, sizeof f);
        function<int(int, int)> dfs = [&](int i, int j) -> int {
            if (i >= m || j >= m || (i + j) >= m) return 0;
            if (f[i][j] != 0x3f3f3f3f) return f[i][j];
            int k = i + j;
            int a = dfs(i + 1, j) + nums[i] * multipliers[k];
            int b = dfs(i, j + 1) + nums[n - j - 1] * multipliers[k];
            return f[i][j] = max(a, b);
        };
        return dfs(0, 0);
    }
};

Go

func maximumScore(nums []int, multipliers []int) int {
	n, m := len(nums), len(multipliers)
	f := make([][]int, m)
	for i := range f {
		f[i] = make([]int, m)
		for j := range f[i] {
			f[i][j] = 1 << 30
		}
	}
	var dfs func(i, j int) int
	dfs = func(i, j int) int {
		if i >= m || j >= m || i+j >= m {
			return 0
		}
		if f[i][j] != 1<<30 {
			return f[i][j]
		}
		k := i + j
		a := dfs(i+1, j) + nums[i]*multipliers[k]
		b := dfs(i, j+1) + nums[n-j-1]*multipliers[k]
		f[i][j] = max(a, b)
		return f[i][j]
	}
	return dfs(0, 0)
}

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

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