# Solving Leetcode Interviews in Seconds with AI: Maximum Number of Jumps to Reach the Last Index


	# 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](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 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
	```python
	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]
	```