Alien Dictionary

There is a new alien language which uses the latin alphabet. However, the order among letters are unknown to you. You receive a list of non-empty words from the dictionary, where words are sorted lexicographically by the rules of this new language. Derive the order of letters in this language.

1. You may assume all letters are in lowercase.

2. The dictionary is invalid, if a is prefix of b and b is appear before a.

3. If the order is invalid, return an empty string.

4. There may be multiple valid order of letters, return the smallest in normal lexicographical order

Example

Example 1:

Input：["wrt","wrf","er","ett","rftt"]
Output："wertf"
Explanation：
from "wrt"and"wrf" ,we can get 't'<'f'
from "wrt"and"er" ,we can get 'w'<'e'
from "er"and"ett" ,we can get 'r'<'t'
from "ett"and"rftt" ,we can get 'e'<'r'
So return "wertf"

Example 2:

Input：["z","x"]
Output："zx"
Explanation：
from "z" and "x"，we can get 'z' < 'x'
So return "zx"

public class Solution {
    public String alienOrder(String[] words) {
        ArrayList<Integer>[] graph = new ArrayList[26];
        // Creating graph
        for (int i = 0; i < words.length; i++) {
            // Only creating those nodes which exists
            for (int j = 0; j < words[i].length(); j++)
                if (graph[words[i].charAt(j) - 'a'] == null)
                    graph[words[i].charAt(j) - 'a'] = new ArrayList<>();

            if (i != 0) {
                int j = 0;
                while (j < Math.min(words[i].length(), words[i - 1].length())
                        && words[i].charAt(j) == words[i - 1].charAt(j))
                    j++;
                // Direction graph from smaller to bigger char(alien lexically)
                if (j != Math.min(words[i].length(), words[i - 1].length()))
                    graph[words[i - 1].charAt(j) - 'a'].add(words[i].charAt(j) - 'a');
            }
        }
        // Creating indegree array
        int[] indegree = new int[26];
        Arrays.fill(indegree, -1);
        int count = 0;
        // Indegree =-1 implies that nodes does not exists
        for (int i = 0; i < 26; i++) {
            if (graph[i] != null) {
                indegree[i] = 0;
                count++;
            }
        }
        for (int i = 0; i < 26; i++) {
            if (graph[i] != null) {
                for (int x : graph[i])
                    indegree[x]++;
            }
        }
        // Topological Sort
        PriorityQueue<Integer> q = new PriorityQueue<>();
        for (int i = 0; i < 26; i++)
            if (indegree[i] == 0)
                q.add(i);
        StringBuilder ans = new StringBuilder();
        while (q.size() != 0) {
            char c = (char) (q.poll() + 'a');
            ans.append(c);
            for (int x : graph[c - 'a']) {
                indegree[x]--;
                if (indegree[x] == 0)
                    q.add(x);
            }
        }
        // If a cycle exists then return empty string
        return ans.length() == count ? ans.toString() : "";
    }
}