LeetCode 1770. Maximum Score from Performing Multiplication Operations

DP

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:

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class Solution {
public:
    int maximumScore(vector<int>& nums, vector<int>& multipliers) {
        int n = nums.size();
        int m = multipliers.size();
        vector<vector<int>> dp(m+1, vector<int>(m+1, 0));
        
        int score = INT_MIN;
        for (int k = 1; k <= m; k++) {
            for (int l = 0; l <= k; l++) {
                int pick_left = (l == 0 ? INT_MIN : dp[l-1][k-l] + multipliers[k-1] * nums[l-1]);
                int pick_right = (l == k ? INT_MIN : dp[l][k-l-1] + multipliers[k-1] * nums[n-k+l]);
                dp[l][k-l] = max(pick_left, pick_right);

                if (k == m) {
                    score = max(score, dp[l][k-l]);
                }
            }
        }
        
        return score;
    }
};