Introduction

In this blog post, we will explore how to solve the LeetCode problem "1696" 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 a 0-indexed integer array nums and an integer k. You are initially standing at index 0. In one move, you can jump at most k steps forward without going outside the boundaries of the array. That is, you can jump from index i to any index in the range [i + 1, min(n - 1, i + k)] inclusive. You want to reach the last index of the array (index n - 1). Your score is the sum of all nums[j] for each index j you visited in the array. Return the maximum score you can get. Example 1: Input: nums = [1,-1,-2,4,-7,3], k = 2 Output: 7 Explanation: You can choose your jumps forming the subsequence [1,-1,4,3] (underlined above). The sum is 7. Example 2: Input: nums = [10,-5,-2,4,0,3], k = 3 Output: 17 Explanation: You can choose your jumps forming the subsequence [10,4,3] (underlined above). The sum is 17. Example 3: Input: nums = [1,-5,-20,4,-1,3,-6,-3], k = 2 Output: 0 Constraints: 1 <= nums.length, k <= 105 -104 <= nums[i] <= 104

Explanation

Here's the breakdown of the solution:

• Dynamic Programming with Optimization: We'll use dynamic programming to store the maximum score achievable at each index. To avoid an O(n*k) time complexity due to iterating through possible jumps, we'll optimize by using a deque to maintain a sliding window of potential best previous indices.
• Deque for Sliding Window Maximum: The deque will store indices in decreasing order of their maximum score. This ensures that the best previous index is always at the front of the deque.

• Efficiently Update DP and Deque: As we iterate, we update the dynamic programming table and the deque, maintaining the optimal sliding window for finding the maximum score.

• Runtime Complexity: O(n), Storage Complexity: O(n)

Code

    from collections import deque

def max_result(nums, k):
    n = len(nums)
    dp = [0] * n
    dp[0] = nums[0]
    dq = deque([0])

    for i in range(1, n):
        while dq and dq[0] < i - k:
            dq.popleft()

        dp[i] = dp[dq[0]] + nums[i]

        while dq and dp[dq[-1]] <= dp[i]:
            dq.pop()

        dq.append(i)

    return dp[n - 1]