Introduction

In this blog post, we will explore how to solve the LeetCode problem "1306" 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

Given an array of non-negative integers arr, you are initially positioned at start index of the array. When you are at index i, you can jump to i + arr[i] or i - arr[i], check if you can reach any index with value 0. Notice that you can not jump outside of the array at any time. Example 1: Input: arr = [4,2,3,0,3,1,2], start = 5 Output: true Explanation: All possible ways to reach at index 3 with value 0 are: index 5 -> index 4 -> index 1 -> index 3 index 5 -> index 6 -> index 4 -> index 1 -> index 3 Example 2: Input: arr = [4,2,3,0,3,1,2], start = 0 Output: true Explanation: One possible way to reach at index 3 with value 0 is: index 0 -> index 4 -> index 1 -> index 3 Example 3: Input: arr = [3,0,2,1,2], start = 2 Output: false Explanation: There is no way to reach at index 1 with value 0. Constraints: 1 <= arr.length <= 5 * 104 0 <= arr[i] < arr.length 0 <= start < arr.length

Explanation

Here's a breakdown of the approach and the Python code:

• Approach:

  • Use Breadth-First Search (BFS) to explore possible jump paths.
  • Mark visited indices to prevent cycles and redundant exploration.
  • If we encounter an index with value 0, return True. If BFS completes without finding a 0, return False.

• Complexity:

  • Runtime: O(N), where N is the length of the array.
  • Storage: O(N) in the worst case (for the queue and visited set).

Code

    from collections import deque

def canReach(arr, start):
    """
    Determines if it's possible to reach an index with value 0 from a given starting index in an array.

    Args:
        arr (list[int]): An array of non-negative integers.
        start (int): The starting index.

    Returns:
        bool: True if it's possible to reach an index with value 0, False otherwise.
    """

    n = len(arr)
    q = deque([start])
    visited = {start}

    while q:
        curr = q.popleft()

        if arr[curr] == 0:
            return True

        next_indices = [curr + arr[curr], curr - arr[curr]]

        for next_index in next_indices:
            if 0 <= next_index < n and next_index not in visited:
                q.append(next_index)
                visited.add(next_index)

    return False