LeetCode 1770. Maximum Score from Performing Multiplication Operations

原题链接在这里：https://leetcode.com/problems/maximum-score-from-performing-multiplication-operations/

题目：

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

题解：

dp(l, r, i) could be calculated as maximum of dp(l + 1, r, i + 1) + nums[l] * muls[i], dp(l , r + 1, i + 1) * nums[r] * muls[i].

Here r could be calculated by l and i.

r = n - (rightTaken) - 1 

rightTaken = i - leftTaken = i - l.

r = n - (i - l) - 1 = n - i + l - 1.

Time Complexity: O(m ^ 2).

Space: O(m ^ 2).

AC Java:

class Solution {
    int n;
    int m;
    Integer [][] memo;
    int [] nums;
    int [] muls;
    public int maximumScore(int[] nums, int[] multipliers) {
        n = nums.length;
        m = multipliers.length;
        this.memo = new Integer[m][m];
        this.nums = nums;
        this.muls = multipliers;
        
        return dfs(0, 0);
    }
    
    private int dfs(int l, int i){
        if(i == m){
            return 0;
        }
        
        if(memo[l][i] != null){
            return memo[l][i];
        }
        
        int left = dfs(l + 1, i + 1) + nums[l] * muls[i];
        int right = dfs(l, i + 1) + nums[n - i + l - 1] * muls[i];
        return memo[l][i] = Math.max(left, right);
    }
}