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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 \(\textit{paths}\).

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class Solution:
    def destCity(self, paths: List[List[str]]) -> str:
        s = {a for a, _ in paths}
        return next(b for _, b in paths if b not in s)

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class Solution {
    public String destCity(List<List<String>> paths) {
        Set<String> s = new HashSet<>();
        for (var p : paths) {
            s.add(p.get(0));
        }
        for (int i = 0;; ++i) {
            var b = paths.get(i).get(1);
            if (!s.contains(b)) {
                return b;
            }
        }
    }
}

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class Solution {
public:
    string destCity(vector<vector<string>>& paths) {
        unordered_set<string> s;
        for (auto& p : paths) {
            s.insert(p[0]);
        }
        for (int i = 0;; ++i) {
            auto b = paths[i][1];
            if (!s.contains(b)) {
                return b;
            }
        }
    }
};

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func destCity(paths [][]string) string {
    s := map[string]bool{}
    for _, p := range paths {
        s[p[0]] = true
    }
    for _, p := range paths {
        if !s[p[1]] {
            return p[1]
        }
    }
    return ""
}

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function destCity(paths: string[][]): string {
    const s = new Set<string>(paths.map(([a, _]) => a));
    return paths.find(([_, b]) => !s.has(b))![1];
}

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use std::collections::HashSet;

impl Solution {
    pub fn dest_city(paths: Vec<Vec<String>>) -> String {
        let s = paths
            .iter()
            .map(|p| p[0].clone())
            .collect::<HashSet<String>>();
        paths.into_iter().find(|p| !s.contains(&p[1])).unwrap()[1].clone()
    }
}

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/**
 * @param {string[][]} paths
 * @return {string}
 */
var destCity = function (paths) {
    const s = new Set(paths.map(([a, _]) => a));
    return paths.find(([_, b]) => !s.has(b))[1];
};

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