1770. Maximum Score from Performing Multiplication Operations| Leetcode solution

 1770. 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 <= 103
• m <= n <= 105
• -1000 <= nums[i], multipliers[i] <= 1000

solution:

class Solution {

public: 

    

    int  _max(vector<int>& num, vector<int>&mul  , int& m , int sum , int i  , int p ,int q ,  vector<vector<int>>&dp)

    {    

         

         if( i == m)

             return 0;

        if(dp[i][p] != -190 )

            return dp[i][p];

         sum +=   max(num[p]*mul[i] + _max(num , mul , m, sum , i+1  , p+1 , q, dp ) ,                             num[q]*mul[i] + _max(num , mul , m, sum , i+1  , p , q-1 , dp))  ; 

        return dp[i][p] = sum;

        

    }

    int maximumScore(vector<int>& nums, vector<int>& multipliers) {

             

         int m = multipliers.size();

         int sum = 0;

         int p = 0;

         int q = nums.size()-1;

         int i =0;

         vector<vector<int>>dp(m+1 , vector<int>(m+1 , -190));

         return _max(nums , multipliers  , m , sum , i , p , q , dp);

    }

};