Introduction

In this blog post, we will explore how to solve the LeetCode problem "3282" 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 an integer array nums of length n. Your goal is to start at index 0 and reach index n - 1. You can only jump to indices greater than your current index. The score for a jump from index i to index j is calculated as (j - i) nums[i]. Return the maximum possible total score by the time you reach the last index. Example 1: Input: nums = [1,3,1,5] Output: 7 Explanation: First, jump to index 1 and then jump to the last index. The final score is 1 1 + 2 3 = 7. Example 2: Input: nums = [4,3,1,3,2] Output: 16 Explanation: Jump directly to the last index. The final score is 4 4 = 16. Constraints: 1 <= nums.length <= 105 1 <= nums[i] <= 105

Explanation

Here's a breakdown of the solution and the code:

• High-Level Approach:

  • Dynamic Programming: Store the maximum score to reach each index i.
  • Iteration: Iterate through each index i, calculating the maximum score to reach it by considering all possible jumps from previous indices.
  • Optimization: For each index i, iterate through all possible previous indices j < i, and update the dp[i] if a better score can be achieved by jumping from j to i.

• Complexity:

  • Runtime Complexity: O(n^2)
  • Storage Complexity: O(n)

Code

    def max_total_score(nums):
    """
    Calculates the maximum possible total score to reach the last index of the array.

    Args:
        nums: An integer array of length n.

    Returns:
        The maximum possible total score.
    """
    n = len(nums)
    dp = [0] * n  # dp[i] stores the maximum score to reach index i

    # Initialize the score for the first index (no jump needed)
    dp[0] = 0

    # Iterate through the array, calculating the maximum score for each index
    for i in range(1, n):
        dp[i] = 0  # Initialize the score for the current index to 0
        for j in range(i):
            score = dp[j] + (i - j) * nums[j]
            dp[i] = max(dp[i], score)

    return dp[n - 1]