Introduction

In this blog post, we will explore how to solve the LeetCode problem "3143" 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 a 2D array points and a string s where, points[i] represents the coordinates of point i, and s[i] represents the tag of point i. A valid square is a square centered at the origin (0, 0), has edges parallel to the axes, and does not contain two points with the same tag. Return the maximum number of points contained in a valid square. Note: A point is considered to be inside the square if it lies on or within the square's boundaries. The side length of the square can be zero. Example 1: Input: points = [[2,2],[-1,-2],[-4,4],[-3,1],[3,-3]], s = "abdca" Output: 2 Explanation: The square of side length 4 covers two points points[0] and points[1]. Example 2: Input: points = [[1,1],[-2,-2],[-2,2]], s = "abb" Output: 1 Explanation: The square of side length 2 covers one point, which is points[0]. Example 3: Input: points = [[1,1],[-1,-1],[2,-2]], s = "ccd" Output: 0 Explanation: It's impossible to make any valid squares centered at the origin such that it covers only one point among points[0] and points[1]. Constraints: 1 <= s.length, points.length <= 105 points[i].length == 2 -109 <= points[i][0], points[i][1] <= 109 s.length == points.length points consists of distinct coordinates. s consists only of lowercase English letters.

Explanation

Here's the approach:

• Iterate through possible side lengths: The side length of the square can be derived from the absolute values of the x and y coordinates of the given points. We consider all possible side lengths derived from max(abs(x), abs(y)) for each point.
• Check each square's validity: For each side length, iterate through the points and check if they fall within the square. If a point is within the square, check for tag conflicts. Maintain a set of used tags within the square. If a duplicate tag is found, the square is invalid.

• Maximize the count: Keep track of the maximum number of points contained in valid squares found so far and update it when a valid square with a higher count is found.

• Runtime Complexity: O(N^2), where N is the number of points. This arises from the nested loops for iterating through potential side lengths and then checking points within each square. Storage Complexity: O(N), primarily due to the set used to track tags within each square.

Code

    def max_points_in_square(points, s):
    """
    Finds the maximum number of points contained in a valid square.

    Args:
        points: A list of lists, where each inner list represents the coordinates of a point.
        s: A string representing the tag of each point.

    Returns:
        The maximum number of points contained in a valid square.
    """

    max_count = 0
    n = len(points)
    side_lengths = set()

    for x, y in points:
        side_lengths.add(max(abs(x), abs(y)))

    side_lengths = sorted(list(side_lengths))

    for side in side_lengths:
        count = 0
        used_tags = set()
        is_valid = True

        for i in range(n):
            x, y = points[i]
            if abs(x) <= side and abs(y) <= side:
                if s[i] in used_tags:
                    is_valid = False
                    break
                else:
                    used_tags.add(s[i])
                    count += 1

        if is_valid:
            max_count = max(max_count, count)

    return max_count