1770. Maximum Score from Performing Multiplication Operations

Description

You are given two 0-indexed integer arrays nums and multipliersof size n and m respectively, where n >= m.

You begin with a score of 0. You want to perform exactly m operations. On the ith operation (0-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.
  • Note that multipliers[0] corresponds to the first operation, multipliers[1] to the second operation, and so on.
• Remove x from 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 <= 300
• m <= n <= 105
• -1000 <= nums[i], multipliers[i] <= 1000

 

Solutions

Solution: Dynamic Programming

• Time complexity: O(m²)
• Space complexity: O(m²)

 

JavaScript

/**
 * @param {number[]} nums
 * @param {number[]} multipliers
 * @return {number}
 */
const maximumScore = function (nums, multipliers) {
  const n = nums.length;
  const m = multipliers.length;
  const dp = Array.from({ length: m }, () => {
    return new Array(m).fill(Number.MIN_SAFE_INTEGER);
  });

  const performOperation = (start, end, index) => {
    if (index >= m) return 0;
    if (dp[index][start] !== Number.MIN_SAFE_INTEGER) return dp[index][start];
    const multiplier = multipliers[index];
    const pickStart = nums[start] * multiplier + performOperation(start + 1, end, index + 1);
    const pickEnd = nums[end] * multiplier + performOperation(start, end - 1, index + 1);
    const result = Math.max(pickStart, pickEnd);

    dp[index][start] = result;

    return result;
  };

  return performOperation(0, n - 1, 0);
};