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Maximize the Distance Between Points on a Square

Difficulty: Hard

Problem Description

You are given an integer side, representing the edge length of a square with corners at (0, 0), (0, side), (side, 0), and (side, side) on a Cartesian plane. You are also given a positive integer k and a 2D integer array points, where points[i] = [x_i, y_i] represents the coordinate of a point lying on the boundary of the square. You need to select k elements among points such that the minimum Manhattan distance between any two points is maximized. Return the maximum possible minimum Manhattan distance between the selected k points.

Key Insights

  • The problem requires maximizing the minimum distance between selected points on the square's boundary.
  • The Manhattan distance is calculated as |x_i - x_j| + |y_i - y_j|.
  • A binary search approach can be applied to efficiently find the maximum minimum distance.
  • A greedy selection method can check if a certain minimum distance is feasible by attempting to place points with the required separation.

Space and Time Complexity

Time Complexity: O(n log(max_distance)), where n is the number of points and max_distance is the maximum possible distance on the square's boundary. Space Complexity: O(1), since we are using a constant amount of extra space.

Solution

To solve the problem, we can use a combination of binary search and greedy algorithms. The binary search will help us find the maximum possible minimum distance between selected points. We will define our search range from 0 to the maximum possible distance, which is 2 * side. For each candidate distance, we will use a greedy approach to try to place the points on the boundary of the square while maintaining at least that distance between any two points. If we can successfully place k points, we will continue to search for a larger distance; otherwise, we will reduce our search range.

Code Solutions

def maxDistance(side, points, k):
    def canPlacePoints(min_dist):
        count = 1
        last_point = points[0]
        
        for i in range(1, len(points)):
            if abs(points[i][0] - last_point[0]) + abs(points[i][1] - last_point[1]) >= min_dist:
                count += 1
                last_point = points[i]
                if count == k:
                    return True
        return False

    left, right = 0, 2 * side
    while left < right:
        mid = (left + right + 1) // 2
        if canPlacePoints(mid):
            left = mid
        else:
            right = mid - 1
    return left
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