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.
class Solution:
def destCity(self, paths: List[List[str]]) -> str:
mp = {a: b for a, b in paths}
a = paths[0][0]
while mp.get(a):
a = mp[a]
return a
class Solution {
public String destCity(List<List<String>> paths) {
Map<String, String> mp = new HashMap<>();
for (List<String> path : paths) {
mp.put(path.get(0), path.get(1));
}
String a = paths.get(0).get(0);
while (mp.get(a) != null) {
a = mp.get(a);
}
return a;
}
}
class Solution {
public:
string destCity(vector<vector<string>>& paths) {
unordered_map<string, string> mp;
for (auto& path : paths) mp[path[0]] = path[1];
string a = paths[0][0];
while (mp.find(a) != mp.end()) a = mp[a];
return a;
}
};
func destCity(paths [][]string) string {
mp := make(map[string]string)
for _, path := range paths {
mp[path[0]] = path[1]
}
a := paths[0][0]
for true {
if _, ok := mp[a]; !ok {
return a
}
a = mp[a]
}
return ""
}