Introduction

In this blog post, we will explore how to solve the LeetCode problem "3017" 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 <= 105 1 <= x, y <= n

Explanation

Here's the solution:

• Key Idea: Calculate the shortest distance between each pair of houses using both the direct path (i to i+1) and the shortcut path (x to y). Store the counts of each distance.
• Shortest Path Calculation: For each pair (i, j), the shortest path is the minimum of abs(i - j) and abs(i - x) + 1 + abs(y - j) or abs(i - y) + 1 + abs(x - j).

• Counting Distances: Iterate through all pairs of houses, calculate their shortest distance, and increment the corresponding count.

• Runtime Complexity: O(n^2)

• Storage Complexity: O(n)

Code

    def solve():
    n, x, y = map(int, input().split())
    x -= 1
    y -= 1

    distances = [0] * n

    for i in range(n):
        for j in range(i + 1, n):
            direct_dist = abs(i - j)
            shortcut_dist1 = abs(i - x) + 1 + abs(y - j)
            shortcut_dist2 = abs(i - y) + 1 + abs(x - j)

            min_dist = min(direct_dist, shortcut_dist1, shortcut_dist2)
            distances[min_dist] += 1

    result = [0] * n
    for i in range(1, n):
        result[i-1] = distances[i] * 2
    print(*result)

solve()