269. Alien Dictionary

Description

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:
Given the following words in dictionary,

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

The correct order is:"wertf".

Example 2:
Given the following words in dictionary,

[
  "z",
  "x"
]

The correct order is:"zx".

Example 3:
Given the following words in dictionary,

[
  "z",
  "x",
  "z"
]

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.

Discussion

After iteration, something can be gotten.

1. can construct map <smaller, set of greater>

2. can know number of chars smaller than itself

With these two data, find a way to solve the issue.

method 1

trie + map<char, set(greater ones)> + dfs

method 2

bfs (copied from leetcode discussion)

public String alienOrder(String[] words) {
    Map<Character, Set<Character>> map=new HashMap<Character, Set<Character>>();
    Map<Character, Integer> degree=new HashMap<Character, Integer>();
    String result="";
    if(words==null || words.length==0) return result;
    for(String s: words){
        for(char c: s.toCharArray()){
            degree.put(c,0);
        }
    }
    for(int i=0; i<words.length-1; i++){
        String cur=words[i];
        String next=words[i+1];
        int length=Math.min(cur.length(), next.length());
        for(int j=0; j<length; j++){
            char c1=cur.charAt(j);
            char c2=next.charAt(j);
            if(c1!=c2){
                Set<Character> set=new HashSet<Character>();
                if(map.containsKey(c1)) set=map.get(c1);
                if(!set.contains(c2)){
                    set.add(c2);
                    map.put(c1, set);
                    degree.put(c2, degree.get(c2)+1);
                }
                break;
            }
        }
    }
    Queue<Character> q=new LinkedList<Character>();
    for(char c: degree.keySet()){
        if(degree.get(c)==0) q.add(c);
    }
    while(!q.isEmpty()){
        char c=q.remove();
        result+=c;
        if(map.containsKey(c)){
            for(char c2: map.get(c)){
                degree.put(c2,degree.get(c2)-1);
                if(degree.get(c2)==0) q.add(c2);
            }
        }
    }
    if(result.length()!=degree.size()) return "";
    return result;
}