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.
Example 1:
Input: paths = [["London","New York"],["New York","Lima"],["Lima","Sao Paulo"]] Output: "Sao Paulo" Explanation: Starting at "London" city you will reach "Sao Paulo" city which is the destination city. Your trip consist of: "London" -> "New York" -> "Lima" -> "Sao Paulo".
Example 2:
Input: paths = [["B","C"],["D","B"],["C","A"]] Output: "A" Explanation: All possible trips are: "D" -> "B" -> "C" -> "A". "B" -> "C" -> "A". "C" -> "A". "A". Clearly the destination city is "A".
Example 3:
Input: paths = [["A","Z"]] Output: "Z"
Constraints:
1 <= paths.length <= 100
paths[i].length == 2
1 <= cityAi.length, cityBi.length <= 10
cityAi != cityBi
- All strings consist of lowercase and uppercase English letters and the space character.
Solution: Count in / out degree for each node
Note: this is a more general solution to this type of problems.
Time complexity: O(n)
Space complexity: O(n)
Destination City: in_degree == 1 and out_degree == 0
C++
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// Author: Huahua class Solution { public: string destCity(vector<vector<string>>& paths) { unordered_map<string, int> in, out; for (const auto& path : paths) { ++out[path[0]]; ++in[path[1]]; } for (const auto& [city, degree] : in) if (degree == 1 && out[city] == 0) return city; return ""; } }; |
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