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Java-based LeetCode algorithm problem solutions, regularly updated

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package g1401_1500.s1436_destination_city;

// #Easy #String #Hash_Table #2022_06_21_Time_3_ms_(81.47%)_Space_43.7_MB_(53.89%)

import java.util.HashSet;
import java.util.List;
import java.util.Set;

/**
 * 1436 - Destination City\.
 *
 * Easy
 *
 * You are given the array `paths`, where paths[i] = [cityA_i, cityB_i] means there exists a direct path going from cityA_i to cityB_i. _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 <= cityA_i.length, cityB_i.length <= 10
 * *   cityA_i != cityB_i
 * *   All strings consist of lowercase and uppercase English letters and the space character.
**/
public class Solution {
    public String destCity(List> paths) {
        Set set = new HashSet<>();
        for (List strings : paths) {
            set.add(strings.get(0));
        }
        for (List path : paths) {
            if (!set.contains(path.get(1))) {
                return path.get(1);
            }
        }
        return "";
    }
}