Introduction

In this blog post, we will explore how to solve the LeetCode problem "593" 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

Given the coordinates of four points in 2D space p1, p2, p3 and p4, return true if the four points construct a square. The coordinate of a point pi is represented as [xi, yi]. The input is not given in any order. A valid square has four equal sides with positive length and four equal angles (90-degree angles). Example 1: Input: p1 = [0,0], p2 = [1,1], p3 = [1,0], p4 = [0,1] Output: true Example 2: Input: p1 = [0,0], p2 = [1,1], p3 = [1,0], p4 = [0,12] Output: false Example 3: Input: p1 = [1,0], p2 = [-1,0], p3 = [0,1], p4 = [0,-1] Output: true Constraints: p1.length == p2.length == p3.length == p4.length == 2 -104 <= xi, yi <= 104

Explanation

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

• Approach:

  • Calculate all possible distances between the four points. A square will have four equal sides and two equal diagonals.
  • Store these distances and check if there are exactly four equal side lengths and two equal diagonal lengths. Ensure the side length is not zero (points are not the same).
  • Use a set to efficiently count the number of distinct distances.

• Complexity:

  • Runtime Complexity: O(1) - constant time as the number of points is fixed at 4.
  • Storage Complexity: O(1) - constant space.

Code

    def validSquare(p1, p2, p3, p4):
    """
    Determines if four points form a valid square.

    Args:
        p1: Coordinates of the first point [x1, y1].
        p2: Coordinates of the second point [x2, y2].
        p3: Coordinates of the third point [x3, y3].
        p4: Coordinates of the fourth point [x4, y4].

    Returns:
        True if the four points form a valid square, False otherwise.
    """

    def dist_sq(p1, p2):
        return (p1[0] - p2[0])**2 + (p1[1] - p2[1])**2

    distances = [
        dist_sq(p1, p2),
        dist_sq(p1, p3),
        dist_sq(p1, p4),
        dist_sq(p2, p3),
        dist_sq(p2, p4),
        dist_sq(p3, p4)
    ]

    distances.sort()

    # Check for four equal sides and two equal diagonals, and that side length is not 0
    return (distances[0] > 0 and
            distances[0] == distances[1] == distances[2] == distances[3] and
            distances[4] == distances[5] and
            distances[0] * 2 == distances[4])