Introduction

In this blog post, we will explore how to solve the LeetCode problem "3464" 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 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] = [xi, yi] 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. The Manhattan Distance between two cells (xi, yi) and (xj, yj) is |xi - xj| + |yi - yj|. Example 1: Input: side = 2, points = [[0,2],[2,0],[2,2],[0,0]], k = 4 Output: 2 Explanation: Select all four points. Example 2: Input: side = 2, points = [[0,0],[1,2],[2,0],[2,2],[2,1]], k = 4 Output: 1 Explanation: Select the points (0, 0), (2, 0), (2, 2), and (2, 1). Example 3: Input: side = 2, points = [[0,0],[0,1],[0,2],[1,2],[2,0],[2,2],[2,1]], k = 5 Output: 1 Explanation: Select the points (0, 0), (0, 1), (0, 2), (1, 2), and (2, 2). Constraints: 1 <= side <= 109 4 <= points.length <= min(4 side, 15 103) points[i] == [xi, yi] The input is generated such that: points[i] lies on the boundary of the square. All points[i] are unique. 4 <= k <= min(25, points.length)

Explanation

Here's the solution to the problem, incorporating the explanations, complexities, and Python code:

• Binary Search: Perform a binary search on the possible Manhattan distances between 0 and 2 * side. The search aims to find the largest minimum Manhattan distance for which we can select k points.
• Greedy Selection: For a given minimum distance during the binary search, greedily select points. Iterate through the sorted points. If the current point maintains the minimum distance from all previously selected points, then select it.

• Feasibility Check: The binary search checks if it is feasible to select k points for the guessed minimum distance.

• Runtime Complexity: O(N log N log S), where N is the number of points and S is the side length of the square. The log N is for sorting. The log S factor comes from the binary search on distance. The N within is for the greedy selection. Storage Complexity: O(N), primarily for storing the sorted points and selected points.

Code

    def solve():
    def max_min_manhattan_distance(side, points, k):
        def check(min_dist):
            selected = []
            for point in points:
                if not selected:
                    selected.append(point)
                else:
                    valid = True
                    for sel_point in selected:
                        if abs(point[0] - sel_point[0]) + abs(point[1] - sel_point[1]) < min_dist:
                            valid = False
                            break
                    if valid:
                        selected.append(point)
            return len(selected) >= k

        points.sort()
        low = 0
        high = 2 * side
        ans = 0

        while low <= high:
            mid = (low + high) // 2
            if check(mid):
                ans = mid
                low = mid + 1
            else:
                high = mid - 1
        return ans

    return max_min_manhattan_distance

def main():
    side = int(input())
    points_str = input().strip("[]").split("],[")
    points = []
    for p in points_str:
      x, y = map(int, p.split(","))
      points.append([x, y])

    k = int(input())

    solver = solve()
    result = solver(side, points, k)
    print(result)

if __name__ == "__main__":
    main()