You are given two integer arrays nums and multipliersof 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: DP

dp(i, j) := max score we can get with nums[i~j] left.

k = n – (j – i + 1)
dp(i, j) = max(dp(i + 1, j) + nums[i] * multipliers[k], dp(i, j-1) + nums[j] * multipliers[k])

Time complexity: O(m*m)
Space complexity: O(m*m)

// Author: Huahua
class Solution {
public:
  int maximumScore(vector<int>& nums, vector<int>& multipliers) {
    const int m = multipliers.size();
    const int n = nums.size();
    vector<vector<int>> cache(m, vector<int>(m, INT_MIN));
    function<int(int, int)> dp = [&](int i, int j) {
      const int k = n - (j - i + 1);
      if (k == m) return 0;
      int& ans = cache[i][k];
      if (ans != INT_MIN) return ans;
      return ans = max(dp(i + 1, j) + nums[i] * multipliers[k],
                       dp(i, j - 1) + nums[j] * multipliers[k]);
    };
    return dp(0, n - 1);
  }
};

// Author: Huahua
class Solution {
public:
  int maximumScore(vector<int>& nums, vector<int>& multipliers) {
    const int m = multipliers.size();
    const int n = nums.size();
    // dp[i][j] := max score of using first i elements and last j elements
    vector<vector<int>> dp(m + 1, vector<int>(m + 1));
    for (int k = 1; k <= m; ++k)
      for (int i = 0, j = k - i; i <= k; ++i, --j)
        dp[i][j] = max((i ? dp[i - 1][j] + nums[i - 1] * multipliers[k - 1] : INT_MIN),
                       (j ? dp[i][j - 1] + nums[n - j] * multipliers[k - 1] : INT_MIN));
    int ans = INT_MIN;
    for (int i = 0; i <= m; ++i)
      ans = max(ans, dp[i][m - i]);
    return ans;
  }
};

The post 花花酱 LeetCode 1770. Maximum Score from Performing Multiplication Operations appeared first on Huahua's Tech Road.