Maximum Score from Performing Multiplication Operations

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 <= 103
m <= n <= 105
-1000 <= nums[i], multipliers[i] <= 1000

Solution:

tricky part is calculating right boundary from left boundary and numbers used.

class Solution {
    Integer[][] dp;

    public int maximumScore(int[] nums, int[] multipliers) {
        int m = multipliers.length;
        dp = new Integer[m][m];
        return helper(nums, multipliers, 0, 0);
    }
    
    private int helper(int[] nums, int[] m, int i, int j) {
        if (i == m.length) return 0;
        if (dp[i][j] != null) return dp[i][j];
        int mul = m[i];
        int num1 = nums[j];
        int num2 = nums[nums.length - 1 - (i - j)];
        int res = Math.max(mul * num1 + helper(nums, m, i + 1, j + 1), mul * num2 + helper(nums, m, i + 1, j));
        return dp[i][j] = res;
    }
}