# 3017. Count the Number of Houses at a Certain Distance II from itertools import accumulate """ you are given three positive integers 'n', 'x', and 'y'. in a city, there exists houses numbered 1 to 'n' connected by 'n' streets. there is a street cnnecting 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 '(house_i, house_j)' such that the minimum number of streets that need to be traveled to reach 'house_i' from 'house_j' is 'k'. return a 1-indexed array of length 'n' where 'ans[k]' represents the total number of pairs of houses such that the minimum streets required to reach one house from the other is 'k'. """ class Solution(object): def countOfPairs(self, n, x, y): """ :type n: int :type x: int :type y: int :rtype: List[int] """ x, y = min(x, y), max(x, y) ans = [0] * n for i in range(1, n + 1): ans[0] += 2 ans[min(i - 1, abs(i - y) + x)] -= 1 ans[min(n - i, abs(i - x) + 1 + n - y)] -= 1 ans[min(abs(i - x), abs(y - i) + 1)] += 1 ans[min(abs(i - x) + 1, abs(y - i))] += 1 r = max(x - i, 0) + max(i - y, 0) ans[r + (y - x + 0) // 2] -= 1 ans[r + (y - x + 1) // 2] -= 1 return list(accumulate(ans)) if __name__ == "__main__": obj = Solution() print(obj.countOfPairs(n=3, x=1, y=3)) print(obj.countOfPairs(n=5, x=2, y=4)) print(obj.countOfPairs(n=4, x=1, y=1))