README.md

1770. Maximum Score from Performing Multiplication Operations

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 (Rust)

1. Solution

impl Solution {
    pub fn maximum_score(nums: Vec<i32>, multipliers: Vec<i32>) -> i32 {
        let (n, m) = (nums.len(), multipliers.len());
        let mut dp = vec![vec![i32::MIN; m + 1]; m + 1];
        dp[0][0] = 0;

        for i in 1..=m {
            for j in 0..i {
                dp[j + 1][i - 1 - j] =
                    dp[j + 1][i - 1 - j].max(dp[j][i - 1 - j] + multipliers[i - 1] * nums[j]);
                dp[j][i - j] =
                    dp[j][i - j].max(dp[j][i - 1 - j] + multipliers[i - 1] * nums[n + j - i]);
            }
        }

        (0..=m).map(|i| dp[i][m - i]).max().unwrap()
    }
}