# Solving Leetcode Interviews in Seconds with AI: Jump Game II


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "45" 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 of integers nums of length n. You are initially positioned at nums[0]. Each element nums[i] represents the maximum length of a forward jump from index i. In other words, if you are at nums[i], you can jump to any nums[i + j] where:  0 <= j <= nums[i] and i + j < n  Return the minimum number of jumps to reach nums[n - 1]. The test cases are generated such that you can reach nums[n - 1].   Example 1:  Input: nums = [2,3,1,1,4] Output: 2 Explanation: The minimum number of jumps to reach the last index is 2. Jump 1 step from index 0 to 1, then 3 steps to the last index.  Example 2:  Input: nums = [2,3,0,1,4] Output: 2    Constraints:  1 <= nums.length <= 104 0 <= nums[i] <= 1000 It's guaranteed that you can reach nums[n - 1].  

	# Explanation
	Here's the solution to the minimum jumps problem:

*   **Greedy Approach:** At each step, we greedily choose the jump that allows us to reach the farthest. We keep track of the current reachable range and update it as we iterate through the array.
*   **Farthest Reachable:** Maintain `farthest` which represents the maximum index we can reach from our current position.
*   **Jump Increment:** Increment the `jumps` count when the current index `i` reaches the current `reach`. This indicates we need to make another jump.

*   **Runtime Complexity:** O(n), **Storage Complexity:** O(1)

	
	# Code
	```python
	def jump(nums):
    """
    Calculates the minimum number of jumps to reach the end of the array.

    Args:
        nums: A list of integers representing the maximum jump length from each index.

    Returns:
        The minimum number of jumps required to reach the last index.
    """

    n = len(nums)
    jumps = 0
    reach = 0
    farthest = 0

    for i in range(n - 1):
        farthest = max(farthest, i + nums[i])
        if i == reach:
            jumps += 1
            reach = farthest
            if reach >= n - 1:
                return jumps

    return jumps
	```