Introduction

In this blog post, we will explore how to solve the LeetCode problem "1770" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given two 0-indexed integer arrays nums and multipliers of 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

Explanation

Here's the solution to the problem:

• Dynamic Programming: The problem can be solved using dynamic programming. We define dp[i][j] as the maximum score achievable after i operations, where j elements are taken from the left side of nums. The remaining i - j elements are taken from the right side.

• Recursive Relation: The dp[i][j] value can be calculated recursively as the maximum of two options: taking an element from the left (nums[j - 1]) or taking an element from the right (nums[n - (i - j)]). We multiply the chosen element by multipliers[i - 1] and add it to the previous best score.

• Base Case: The base case is dp[0][0] = 0, meaning no operations yield a score of 0.

• Time and Space Complexity: O(m^2), O(m^2)

Code

    def maximumScore(nums, multipliers):
    n = len(nums)
    m = len(multipliers)
    dp = [[0] * (m + 1) for _ in range(m + 1)]

    for i in range(1, m + 1):
        for j in range(i + 1):
            if j == 0:
                dp[i][j] = dp[i - 1][j] + multipliers[i - 1] * nums[n - i]
            elif j == i:
                dp[i][j] = dp[i - 1][j - 1] + multipliers[i - 1] * nums[j - 1]
            else:
                dp[i][j] = max(dp[i - 1][j] + multipliers[i - 1] * nums[n - (i - j)],
                               dp[i - 1][j - 1] + multipliers[i - 1] * nums[j - 1])

    max_score = float('-inf')
    for j in range(m + 1):
        max_score = max(max_score, dp[m][j])

    return max_score