LeetCode - 1436. Destination City (Python)

Easy - DFS, BFS, 탐색

RunTime : 11 ms / Memory : 18.25 MB

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문제

You are given the array paths, where paths[i] = [cityAi, cityBi] means there exists a direct path going from cityAi to cityBi. Return the destination city, that is, the city without any path outgoing to another city.

It is guaranteed that the graph of paths forms a line without any loop, therefore, there will be exactly one destination city.

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풀이 1. DFS 풀이

• 처음 문제를 보고 직관적으로 떠올린 풀이
• 단순하게 paths에서 연결된 노드들을 찾아 탐색한다.
• 더이상 탐색할 곳이 없다면 return

class Solution:
    def destCity(self, paths: List[List[str]]) -> str:
        visited = [0] * len(paths)
        res = ""

        def dfs(std):
            nonlocal res
            nxt = False

            for i in range(len(paths)):
                if visited[i] == 0 and paths[i][0] == std:
                    visited[i] = 1
                    dfs(paths[i][1])
                    nxt = True
            
            if not nxt:
                res = std
        
        dfs(paths[0][0])
        return res
        
        

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풀이 2. BFS 풀이

• paths가 간단한 2차원 리스트 구성이기 때문에 bfs도 효과적인 풀이가 될 수 있다.

from collections import deque, defaultdict
class Solution:
    def destCity(self, paths: List[List[str]]) -> str:
        graph = defaultdict(list)

        for start, end in paths:
            graph[start].append(end)
        
        st_city = paths[0][0]

        q = deque()
        q.append(st_city)
        dest = st_city

        while q:
            node = q.popleft()
            dest = node
            if node in graph:
                for nxt in graph[node]:
                    q.append(nxt)
        
        return dest
        
        

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풀이 3. 직관적인 풀이

• 문제의 본질을 바라보자
• 결국 A에서 출발해 B로 도착하는데,
  B에서 출발하는 경로가 없다면?
  • 정답은 당연하게 B

class Solution:
    def destCity(self, paths: List[List[str]]) -> str:
        start_cities = {a for a, _ in paths}

        for _, b in paths:
            if b not in start_cities:
                return b