Problem Description
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.
Key Insights
- Houses are arranged in a linear fashion, with additional direct connections between specific houses.
- The minimum distance between any two houses can be calculated based on their positions and any direct connections.
- The problem can be approached using a distance matrix to determine the number of pairs for each possible distance.
Space and Time Complexity
Time Complexity: O(n^2)
Space Complexity: O(n^2)
Solution
To solve the problem, we need to calculate the distances between all pairs of houses, considering both the linear connections and the additional street between the houses x and y. We can use a nested loop to iterate through each pair of houses and compute the minimum distance. We can store counts of how many pairs have each distance in an array.
- Initialize an array
result
of size n + 1 to store the count of pairs for each distance k. - Iterate through each pair of houses (i, j):
- Calculate the direct distance between i and j.
- Consider if the additional street provides a shorter path.
- Increment the count in the
result
array based on the minimum distance computed.
- Return the result array excluding the 0-th index.