Introduction

In this blog post, we will explore how to solve the LeetCode problem "3015" 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 three positive integers n, x, and y. In a city, there exist houses numbered 1 to n connected by n streets. There is a street connecting the house numbered i with the house numbered i + 1 for all 1 <= i <= n - 1 . An additional street connects the house numbered x with the house numbered y. For each k, such that 1 <= k <= n, you need to find the number of pairs of houses (house1, house2) such that the minimum number of streets that need to be traveled to reach house2 from house1 is k. Return a 1-indexed array result of length n where result[k] represents the total number of pairs of houses such that the minimum streets required to reach one house from the other is k. Note that x and y can be equal. Example 1: Input: n = 3, x = 1, y = 3 Output: [6,0,0] Explanation: Let's look at each pair of houses: - For the pair (1, 2), we can go from house 1 to house 2 directly. - For the pair (2, 1), we can go from house 2 to house 1 directly. - For the pair (1, 3), we can go from house 1 to house 3 directly. - For the pair (3, 1), we can go from house 3 to house 1 directly. - For the pair (2, 3), we can go from house 2 to house 3 directly. - For the pair (3, 2), we can go from house 3 to house 2 directly. Example 2: Input: n = 5, x = 2, y = 4 Output: [10,8,2,0,0] Explanation: For each distance k the pairs are: - For k == 1, the pairs are (1, 2), (2, 1), (2, 3), (3, 2), (2, 4), (4, 2), (3, 4), (4, 3), (4, 5), and (5, 4). - For k == 2, the pairs are (1, 3), (3, 1), (1, 4), (4, 1), (2, 5), (5, 2), (3, 5), and (5, 3). - For k == 3, the pairs are (1, 5), and (5, 1). - For k == 4 and k == 5, there are no pairs. Example 3: Input: n = 4, x = 1, y = 1 Output: [6,4,2,0] Explanation: For each distance k the pairs are: - For k == 1, the pairs are (1, 2), (2, 1), (2, 3), (3, 2), (3, 4), and (4, 3). - For k == 2, the pairs are (1, 3), (3, 1), (2, 4), and (4, 2). - For k == 3, the pairs are (1, 4), and (4, 1). - For k == 4, there are no pairs. Constraints: 2 <= n <= 100 1 <= x, y <= n

Explanation

• Compute shortest paths: Use Breadth-First Search (BFS) to compute the shortest distance between all pairs of houses, considering both the regular streets (i to i+1) and the special street (x to y).
  • Count pairs by distance: Iterate through all possible house pairs and count how many pairs have a shortest distance of k for each k from 1 to n-1.
  • Return results: Store the counts in a result array and return it.

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

Code

    from collections import deque

def count_house_pairs(n: int, x: int, y: int) -> list[int]:
    """
    Calculates the number of house pairs with a minimum distance of k for each k.

    Args:
        n: The number of houses in the city.
        x: The number of the first house connected by a special street.
        y: The number of the second house connected by a special street.

    Returns:
        A 1-indexed array where result[k] is the number of house pairs with distance k.
    """

    distances = [[float('inf')] * (n + 1) for _ in range(n + 1)]

    # Initialize distances for each house to itself
    for i in range(1, n + 1):
        distances[i][i] = 0

    # Build the graph and calculate shortest paths using BFS
    for start_node in range(1, n + 1):
        q = deque([(start_node, 0)])  # (node, distance)
        visited = {start_node}

        while q:
            node, dist = q.popleft()
            distances[start_node][node] = dist

            neighbors = []
            if node > 1:
                neighbors.append(node - 1)
            if node < n:
                neighbors.append(node + 1)
            if node == x:
                neighbors.append(y)
            if node == y:
                neighbors.append(x)

            for neighbor in neighbors:
                if neighbor not in visited:
                    visited.add(neighbor)
                    q.append((neighbor, dist + 1))

    # Count the number of pairs for each distance k
    result = [0] * n
    for i in range(1, n + 1):
        for j in range(i + 1, n + 1):
            dist = distances[i][j]
            if dist <= n-1:
                result[dist - 1] += 2  # Count (i, j) and (j, i)

    return result