Easy
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
import java.util.HashSet;
import java.util.List;
import java.util.Set;
public class Solution {
public String destCity(List<List<String>> paths) {
Set<String> set = new HashSet<>();
for (List<String> strings : paths) {
set.add(strings.get(0));
}
for (List<String> path : paths) {
if (!set.contains(path.get(1))) {
return path.get(1);
}
}
return "";
}
}