Today I have done leetcode's September LeetCoding Challenge with cpp.

September LeetCoding Challenge 16

Description

Maximum Score from Performing Multiplication Operations

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 <= 10^3
• m <= n <= 10^5``
• -1000 <= nums[i], multipliers[i] <= 1000

Solution

auto speedup = [](){
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios::sync_with_stdio(false);
  return 0;
}();
constexpr int SZ = 1001;

class Solution {
public:
  int maximumScore(vector<int>& nums, vector<int>& multipliers) {
    int m = multipliers.size();
    int len = nums.size();
    int dp[SZ][SZ] = {0};
    for(int i = 1; i <= m; ++i) {
      for(int j = 0; j <= i; ++j) {
        dp[i][j] = INT_MIN;
        if(j) dp[i][j] = max(dp[i][j], dp[i - 1][j - 1] + nums[j - 1] * multipliers[i - 1]);
        if(j != i) dp[i][j] = max(dp[i][j], dp[i - 1][j] + nums[len - i + j] * multipliers[i - 1]);
      }
    }
    
    int answer = INT_MIN;
    for(int i = 0; i <= m; ++i) answer = max(answer, dp[m][i]);
    
    return answer;
  }
};

// Accepted
// 50/50 cases passed (373 ms)
// Your runtime beats 95.59 % of cpp submissions
// Your memory usage beats 98.4 % of cpp submissions (54.7 MB)