# Solving Leetcode Interviews in Seconds with AI: Minimize Manhattan Distances


	# 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](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 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
	```python
	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
	```