Introduction

In this blog post, we will explore how to solve the LeetCode problem "2849" 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 four integers sx, sy, fx, fy, and a non-negative integer t. In an infinite 2D grid, you start at the cell (sx, sy). Each second, you must move to any of its adjacent cells. Return true if you can reach cell (fx, fy) after exactly t seconds, or false otherwise. A cell's adjacent cells are the 8 cells around it that share at least one corner with it. You can visit the same cell several times. Example 1: Input: sx = 2, sy = 4, fx = 7, fy = 7, t = 6 Output: true Explanation: Starting at cell (2, 4), we can reach cell (7, 7) in exactly 6 seconds by going through the cells depicted in the picture above. Example 2: Input: sx = 3, sy = 1, fx = 7, fy = 3, t = 3 Output: false Explanation: Starting at cell (3, 1), it takes at least 4 seconds to reach cell (7, 3) by going through the cells depicted in the picture above. Hence, we cannot reach cell (7, 3) at the third second. Constraints: 1 <= sx, sy, fx, fy <= 109 0 <= t <= 109

Explanation

Here's the breakdown of the solution:

• Minimum Time Calculation: Calculate the minimum time required to reach the target cell (fx, fy) from the starting cell (sx, sy). This is equivalent to the Chebyshev distance, which is the maximum of the absolute differences in x and y coordinates.
• Comparison with Available Time: Compare the minimum time with the given time t. If the minimum time is greater than t, it's impossible to reach the target cell within the given time.

• Special Case Handling: If the start and finish locations are the same (sx == fx and sy == fy), and t == 0, return False since you must move each second. If they are at the same location and t > 0, return True, since the problem statement says we can revisit cells.

• Time Complexity: O(1), Space Complexity: O(1)

Code

    def isReachableAtTime(sx: int, sy: int, fx: int, fy: int, t: int) -> bool:
    """
    Determines if it's possible to reach (fx, fy) from (sx, sy) in exactly t seconds.

    Args:
        sx: The starting x-coordinate.
        sy: The starting y-coordinate.
        fx: The target x-coordinate.
        fy: The target y-coordinate.
        t: The available time.

    Returns:
        True if it's possible to reach the target cell in exactly t seconds, False otherwise.
    """

    min_time = max(abs(fx - sx), abs(fy - sy))

    if min_time > t:
        return False

    if sx == fx and sy == fy:
        return t != 0

    return True