# 3243. Shortest Distance After Road Addition Queries I """ you are given an integer 'n' and a 2d integer array 'queries'. there are 'n' cities numbered from 0 to 'n - 1'. initially there is a unidirectional road from city 'i' to city 'i + 1' for all '0 <= i < n - 1'. 'queries[i] = [u_i, v_i]' represents the addition of a new unidirectional road from city 'u_i' to city 'v_i'. after each query, you need to find the length of the shortest path from city 0 to city 'n - 1'. return an array 'answer' where for each 'i' in the range of '[0, queries.length - 1]', 'answer[i]' is the length of the shortest path from city 0 to city 'n - 1' after processing the first 'i + 1' queries. """ class Solution(object): def shortestDistanceAfterQueries(self, n, queries): """ :type n: int :type queries: List[List[int]] :rtype: List[int] """ # ans = [] # mp = {i: [i + 1] for i in range(n - 1)} # for i in queries: # mp[i[0]].append(i[1]) # dp = [float("inf")] * n # dp[n - 1] = 0 # for j in range(n - 2, -1, -1): # for next in mp[j]: # dp[j] = min(dp[j], dp[next] + 1) # ans.append(dp[0]) # return ans sol = [] city_distances = {i: [-1] * i + list(range(n - i)) for i in range(n)} for query in queries: source, destination = query city_distances[source][destination] = 1 for i in range(destination + 1, n): city_distances[source][i] = min( city_distances[destination][i] + 1, city_distances[source][i] ) for i in range(source - 1, -1, -1): city_distances[i][destination] = min( city_distances[i][source] + 1, city_distances[i][destination] ) sol.append(city_distances[0][-1]) return sol if __name__ == "__main__": obj = Solution() print(obj.shortestDistanceAfterQueries(n=5, queries=[[2, 4], [0, 2], [0, 4]])) print(obj.shortestDistanceAfterQueries(n=4, queries=[[0, 3], [0, 2]])) print(obj.shortestDistanceAfterQueries(n=5, queries=[[1, 3], [0, 3]]))