[LC] 269. 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.

Example 1:

Input:
[
  "wrt",
  "wrf",
  "er",
  "ett",
  "rftt"
]

Output: "wertf"

Example 2:

Input:
[
  "z",
  "x"
]

Output: "zx"

Example 3:

Input:
[
  "z",
  "x",
  "z"
] 

Output: "" 

Explanation: The order is invalid, so return "".

Note:

1. You may assume all letters are in lowercase.
2. You may assume that if a is a prefix of b, then a must appear before b in the given dictionary.
3. If the order is invalid, return an empty string.
4. There may be multiple valid order of letters, return any one of them is fine.

class Solution {
    public String alienOrder(String[] words) {
        StringBuilder sb = new StringBuilder();
        int[] indegree = new int[26];
        int count = 0;
        Map<Character, Set<Character>> map = new HashMap<>();
        for (String word: words) {
            for (char c : word.toCharArray()) {
                if (indegree[c - 'a'] == 0) {
                    indegree[c - 'a'] += 1;
                    count += 1;   
                }
            }
        }
        
        // construct graph
        for (int i = 0; i < words.length - 1; i++) {
            String cur = words[i];
            String next = words[i + 1];
            int min = Math.min(cur.length(), next.length());
            for (int j = 0; j < min; j++) {
                if (cur.charAt(j) != next.charAt(j)) {
                    if (!map.containsKey(cur.charAt(j))) {
                        map.put(cur.charAt(j), new HashSet<>());
                    }
                    if (map.get(cur.charAt(j)).add(next.charAt(j))) {
                        indegree[next.charAt(j) - 'a'] += 1;
                    }
                    break;
                }
            }
        }
        
        // initial point
        Queue<Character> queue = new LinkedList<>();
        for (int i = 0; i < 26; i++) {
            if (indegree[i] == 1) {
                queue.offer((char)('a' + i));
            }
        }
        
        while (!queue.isEmpty()) {
            Character cur = queue.poll();
            sb.append(cur);
            if (!map.containsKey(cur)) {
                continue;
            }
            for (char ch : map.get(cur)) {
                if (--indegree[ch - 'a'] == 1) {
                    queue.offer(ch);
                }
            }
        }
        if (sb.length() != count) {
            return "";
        }
        return sb.toString();
    }
}