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+# 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
+
+
+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]]))