1436. Destination City
Description
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.
Solutions
Solution 1: Hash Table
According to the problem description, the destination city will not appear in any of the $\textit{cityA}$. Therefore, we can first traverse the $\textit{paths}$ and put all $\textit{cityA}$ into a set $\textit{s}$. Then, we traverse the $\textit{paths}$ again to find the $\textit{cityB}$ that is not in $\textit{s}$.
The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of $\textit{paths}$.
1 2 3 4 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 |
|
1 2 3 4 5 6 7 8 9 10 11 12 |
|
1 2 3 4 |
|
1 2 3 4 5 6 7 8 9 10 11 |
|
1 2 3 4 5 6 7 8 |
|