Introduction

In this blog post, we will explore how to solve the LeetCode problem "2857" 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 integer array coordinates and an integer k, where coordinates[i] = [xi, yi] are the coordinates of the ith point in a 2D plane. We define the distance between two points (x1, y1) and (x2, y2) as (x1 XOR x2) + (y1 XOR y2) where XOR is the bitwise XOR operation. Return the number of pairs (i, j) such that i < j and the distance between points i and j is equal to k. Example 1: Input: coordinates = [[1,2],[4,2],[1,3],[5,2]], k = 5 Output: 2 Explanation: We can choose the following pairs: - (0,1): Because we have (1 XOR 4) + (2 XOR 2) = 5. - (2,3): Because we have (1 XOR 5) + (3 XOR 2) = 5. Example 2: Input: coordinates = [[1,3],[1,3],[1,3],[1,3],[1,3]], k = 0 Output: 10 Explanation: Any two chosen pairs will have a distance of 0. There are 10 ways to choose two pairs. Constraints: 2 <= coordinates.length <= 50000 0 <= xi, yi <= 106 0 <= k <= 100

Explanation

Here's an efficient solution to the problem:

• High-Level Approach: Iterate through all possible pairs of points (i, j) where i < j. Calculate the XOR distance between each pair and increment the count if the distance equals k.

• Complexity:

  • Runtime: O(N^2) where N is the number of coordinates.
  • Storage: O(1)

Code

    def count_pairs(coordinates, k):
    """
    Counts the number of pairs of points with a distance of k.

    Args:
        coordinates: A list of coordinate pairs.
        k: The target distance.

    Returns:
        The number of pairs with a distance of k.
    """
    count = 0
    n = len(coordinates)
    for i in range(n):
        for j in range(i + 1, n):
            distance = (coordinates[i][0] ^ coordinates[j][0]) + (coordinates[i][1] ^ coordinates[j][1])
            if distance == k:
                count += 1
    return count