Introduction

In this blog post, we will explore how to solve the LeetCode problem "3102" 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 an array points representing integer coordinates of some points on a 2D plane, where points[i] = [xi, yi]. The distance between two points is defined as their Manhattan distance. Return the minimum possible value for maximum distance between any two points by removing exactly one point. Example 1: Input: points = [[3,10],[5,15],[10,2],[4,4]] Output: 12 Explanation: The maximum distance after removing each point is the following: After removing the 0th point the maximum distance is between points (5, 15) and (10, 2), which is |5 - 10| + |15 - 2| = 18. After removing the 1st point the maximum distance is between points (3, 10) and (10, 2), which is |3 - 10| + |10 - 2| = 15. After removing the 2nd point the maximum distance is between points (5, 15) and (4, 4), which is |5 - 4| + |15 - 4| = 12. After removing the 3rd point the maximum distance is between points (5, 15) and (10, 2), which is |5 - 10| + |15 - 2| = 18. 12 is the minimum possible maximum distance between any two points after removing exactly one point. Example 2: Input: points = [[1,1],[1,1],[1,1]] Output: 0 Explanation: Removing any of the points results in the maximum distance between any two points of 0. Constraints: 3 <= points.length <= 105 points[i].length == 2 1 <= points[i][0], points[i][1] <= 108

Explanation

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

• Approach:

  • Iterate through each point in the points array, considering it as the point to be removed.
  • For each removal, calculate the maximum Manhattan distance between all remaining pairs of points.
  • Keep track of the minimum of these maximum distances found across all removals.

• Complexity:

  • Runtime: O(n^2), where n is the number of points.
  • Storage: O(1) (constant extra space)

Code

    def min_max_distance(points):
    """
    Calculates the minimum possible value for the maximum Manhattan distance between any two points
    after removing exactly one point.

    Args:
        points: A list of lists representing the integer coordinates of points on a 2D plane.

    Returns:
        The minimum possible value for the maximum distance.
    """

    n = len(points)
    min_max_dist = float('inf')

    for k in range(n):  # Iterate through each point to remove
        max_dist = 0
        for i in range(n):
            if i == k:
                continue
            for j in range(i + 1, n):
                if j == k:
                    continue
                dist = abs(points[i][0] - points[j][0]) + abs(points[i][1] - points[j][1])
                max_dist = max(max_dist, dist)
        min_max_dist = min(min_max_dist, max_dist)

    return min_max_dist