1770. Maximum Score from Performing Multiplication Operations

Problem

You are given two integer arrays nums and multipliers** **of size n and m respectively, where n >= m. The arrays are *1-indexed*.

You begin with a score of 0. You want to perform exactly m operations. On the ith operation (1-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.

• Remove x from the array 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 <= 10^3

• m <= n <= 10^5

• -1000 <= nums[i], multipliers[i] <= 1000

Solution (Java)

class Solution {
    public int maximumScore(int[] nums, int[] mult) {
        int n = nums.length;
        int m = mult.length;
        int row = m;
        int[] dp = new int[m];
        int[] prev = new int[m + 1];
        while (--row >= 0) {
            for (int i = 0; i <= row; ++i) {
                dp[i] =
                        Math.max(
                                prev[i] + mult[row] * nums[n - row + i - 1],
                                prev[i + 1] + mult[row] * nums[i]);
            }
            prev = dp;
        }
        return dp[0];
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).