Introduction

In this blog post, we will explore how to solve the LeetCode problem "447" 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 n points in the plane that are all distinct, where points[i] = [xi, yi]. A boomerang is a tuple of points (i, j, k) such that the distance between i and j equals the distance between i and k (the order of the tuple matters). Return the number of boomerangs. Example 1: Input: points = [[0,0],[1,0],[2,0]] Output: 2 Explanation: The two boomerangs are [[1,0],[0,0],[2,0]] and [[1,0],[2,0],[0,0]]. Example 2: Input: points = [[1,1],[2,2],[3,3]] Output: 2 Example 3: Input: points = [[1,1]] Output: 0 Constraints: n == points.length 1 <= n <= 500 points[i].length == 2 -104 <= xi, yi <= 104 All the points are unique.

Explanation

• Iterate through each point i in the input points. For each point i, compute the distances to all other points j.
  • Use a hash map to store the frequency of each distance from point i.
  • For each distance frequency, the number of boomerangs that can be formed with point i as the central point is freq * (freq - 1). Sum up these values for all points i.

• Runtime Complexity: O(n^2), Storage Complexity: O(n)

Code

    def numberOfBoomerangs(points: list[list[int]]) -> int:
    """
    Given n points in the plane that are all distinct, where points[i] = [xi, yi].
    A boomerang is a tuple of points (i, j, k) such that the distance between i and j equals the distance between i and k (the order of the tuple matters).
    Return the number of boomerangs.

    Example 1:
    Input: points = [[0,0],[1,0],[2,0]]
    Output: 2
    Explanation: The two boomerangs are [[1,0],[0,0],[2,0]] and [[1,0],[2,0],[0,0]].

    Example 2:
    Input: points = [[1,1],[2,2],[3,3]]
    Output: 2

    Example 3:
    Input: points = [[1,1]]
    Output: 0

    Constraints:
    n == points.length
    1 <= n <= 500
    points[i].length == 2
    -104 <= xi, yi <= 104
    All the points are unique.
    """
    n = len(points)
    ans = 0
    for i in range(n):
        dist_map = {}
        for j in range(n):
            if i == j:
                continue
            dist = (points[i][0] - points[j][0])**2 + (points[i][1] - points[j][1])**2
            if dist not in dist_map:
                dist_map[dist] = 0
            dist_map[dist] += 1
        for dist in dist_map:
            freq = dist_map[dist]
            ans += freq * (freq - 1)
    return ans