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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]]))
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