# Solving Leetcode Interviews in Seconds with AI: Reach End of Array With Max Score


	# 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](https://www.chatmagic.app), 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
	```python
	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]
	```