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The idea is to determine which city does not have any outgoing paths. This can be done by keeping track of all cities that appear as the starting city of a path. The destination city is the one that is not in this set but appears as an ending city in the path list.
Time Complexity: O(n^2) due to linear checks within loops.
Space Complexity: O(n), where n is the number of cities.
1import java.util.HashSet;
2import java.util.List;
3import java.util.Set;
4
5class Solution {
6 public String destCity(List<List<String>> paths) {
7 Set<String> startingCities = new HashSet<>();
8 for (List<String> path : paths) {
9 startingCities.add(path.get(0));
10 }
11 for (List<String> path : paths) {
12 if (!startingCities.contains(path.get(1))) {
13 return path.get(1);
14 }
15 }
16 return "";
17 }
18}We use a HashSet to store all starting cities, then iterate through the paths to find a city that does not exist in the set as a starting city. This city is the destination city.
This approach involves counting occurrences of each city as an endpoint. The destination city will be the one whose occurrence as a destination is not matched by an occurrence as a start.
Time Complexity: O(n^2), since it includes iterating through paths multiple times.
Space Complexity: O(n) for storing city names.
1using System;
2using System.Collections.Generic;
public class Solution {
public string DestCity(IList<IList<string>> paths) {
Dictionary<string, int> cityCount = new Dictionary<string, int>();
foreach (var path in paths) {
if (!cityCount.ContainsKey(path[0]))
cityCount[path[0]] = 0;
cityCount[path[0]]++;
if (!cityCount.ContainsKey(path[1]))
cityCount[path[1]] = 0;
}
foreach (var path in paths) {
if (cityCount[path[1]] == 0) {
return path[1];
}
}
return "";
}
}In the C# solution, a dictionary is used to count the number of times a city appears as the starting city. If a city only appears as a destination (count = 0), it is the destination city.