Formatted question description: https://leetcode.ca/all/1770.html

1770. Maximum Score from Performing Multiplication Operations

Level

Medium

Description

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 i-th 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

Use dynamic programming. Create a 2D array dp of m + 1 rows and n columns, where dp[i][j] represents the maximum score using the last i multiplers for the subarray starting from index j. The base case is when i = 1. The final result is dp[m][0].

To optimize memory, use a 1D array instead, which stores only the last row of the original dp array.

  • class Solution {
        public int maximumScore(int[] nums, int[] multipliers) {
            int n = nums.length, m = multipliers.length;
            int[] dp = new int[n];
            int minWindow = n - m + 1;
            for (int j = minWindow - 1; j < n; j++) {
                int start = j - minWindow + 1;
                dp[start] = Math.max(nums[start] * multipliers[m - 1], nums[j] * multipliers[m - 1]);
            }
            for (int i = 2; i <= m; i++) {
                int[] dpNew = new int[n];
                int window = n - m + i;
                for (int j = window - 1; j < n; j++) {
                    int start = j - window + 1;
                    dpNew[start] = Math.max(dp[start + 1] + nums[start] * multipliers[m - i], dp[start] + nums[j] * multipliers[m - i]);
                }
                dp = dpNew;
            }
            return dp[0];
        }
    }
    
  • // OJ: https://leetcode.com/problems/maximum-score-from-performing-multiplication-operations/
    // Time: O(M^2)
    // Space: O(M^2)
    class Solution {
        int memo[1001][1001] = {};
        int dfs(vector<int> &A, vector<int> &M, int i, int j) {
            if (j == M.size()) return 0;
            if (memo[j][i]) return memo[j][i];
            return memo[j][i] = max(A[i] * M[j] + dfs(A, M, i + 1, j + 1), A[A.size() - j + i - 1] * M[j] + dfs(A, M, i, j + 1));
        }
    public:
        int maximumScore(vector<int>& A, vector<int>& M) {
            return dfs(A, M, 0, 0);
        }
    };
    
  • # 1770. Maximum Score from Performing Multiplication Operations
    # https://leetcode.com/problems/maximum-score-from-performing-multiplication-operations/
    
    class Solution:
        def maximumScore(self, nums: List[int], muls: List[int]) -> int:
            n, m = len(nums), len(muls)
            
            @lru_cache(2000)
            def dp(l, i):
                if i == m: return 0
                pickLeft = dp(l+1, i+1) + nums[l] * muls[i] # Pick the left side
                pickRight = dp(l, i+1) + nums[n-(i-l)-1] * muls[i] # Pick the right side
                return max(pickLeft, pickRight)
            
            return dp(0, 0)
    
    

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