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.

• Java
• C++
• Python

• 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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