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Maximum Score from Performing Multiplication Operations

Updated: Mar 24, 2021

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 {
    int[][] dp = new int [1000][1000];
   public int maximumScore(int[] nums, int[] multipliers) {
    return helper(nums,multipliers,0,nums.length-1,0);
    }
    public int helper(int[] a,int[] m,int start, int end, int mindex)
    {
        if(mindex==m.length) return 0;
        if(dp[mindex][start]!=0) return dp[mindex][start];
        int val1 = m[mindex]*a[start]+helper(a,m,start+1,end,mindex+1);
        int val2 = m[mindex]*a[end]+helper(a,m,start,end-1,mindex+1);
        dp[mindex][start]=Math.max(val1,val2);
        return dp[mindex][start];
    }
}

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