Introduction

In this blog post, we will explore how to solve the LeetCode problem "2770" 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 array nums of n integers and an integer target. You are initially positioned at index 0. In one step, you can jump from index i to any index j such that: 0 <= i < j < n -target <= nums[j] - nums[i] <= target Return the maximum number of jumps you can make to reach index n - 1. If there is no way to reach index n - 1, return -1. Example 1: Input: nums = [1,3,6,4,1,2], target = 2 Output: 3 Explanation: To go from index 0 to index n - 1 with the maximum number of jumps, you can perform the following jumping sequence: - Jump from index 0 to index 1. - Jump from index 1 to index 3. - Jump from index 3 to index 5. It can be proven that there is no other jumping sequence that goes from 0 to n - 1 with more than 3 jumps. Hence, the answer is 3. Example 2: Input: nums = [1,3,6,4,1,2], target = 3 Output: 5 Explanation: To go from index 0 to index n - 1 with the maximum number of jumps, you can perform the following jumping sequence: - Jump from index 0 to index 1. - Jump from index 1 to index 2. - Jump from index 2 to index 3. - Jump from index 3 to index 4. - Jump from index 4 to index 5. It can be proven that there is no other jumping sequence that goes from 0 to n - 1 with more than 5 jumps. Hence, the answer is 5. Example 3: Input: nums = [1,3,6,4,1,2], target = 0 Output: -1 Explanation: It can be proven that there is no jumping sequence that goes from 0 to n - 1. Hence, the answer is -1. Constraints: 2 <= nums.length == n <= 1000 -109 <= nums[i] <= 109 0 <= target <= 2 * 109

Explanation

Here's a breakdown of the solution:

• Dynamic Programming: Utilize dynamic programming to store the maximum number of jumps to reach each index. The DP array dp[i] stores the maximum jumps to reach index i.
• Iteration and Condition Check: Iterate through the array, and for each index i, iterate through the possible jump destinations j (where i < j < n). Check if the jump condition (0 <= nums[j] - nums[i] <= target) is met.

• Update DP Array: If a valid jump is found, update dp[j] with the maximum number of jumps achievable.

• Runtime Complexity: O(n^2), where n is the length of the input array nums.

• Storage Complexity: O(n), due to the dp array.

Code

    def maxJumps(nums, target):
    n = len(nums)
    dp = [-1] * n
    dp[0] = 0

    for i in range(n):
        if dp[i] != -1:
            for j in range(i + 1, n):
                if 0 <= nums[j] - nums[i] <= target:
                    if dp[j] < dp[i] + 1:
                        dp[j] = dp[i] + 1

    return dp[n - 1]