# 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];
}
}
```