Package g1701_1800.s1770_maximum_score_from_performing_multiplication_operations

Class Solution

java.lang.Object
g1701_1800.s1770_maximum_score_from_performing_multiplication_operations.Solution
public class Solution extends Object
1770 - Maximum Score from Performing Multiplication Operations\. Medium 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`

  • Constructor Details

    • Solution

      public Solution()

  • Method Details

    • maximumScore

      public int maximumScore(int[] nums, int[] mult)