# Alien Dictionary

`Graph`, `Topological Sorting`

Hard

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.

## Analysis & Solution

#### Topological Sorting - BFS

1. 如何将alien dictionary转化为邻接表
1. 循环words[]，words[i], words[i + 1]从头进行字符比较，如果出现不一样的字符 ch1 != ch2，说明存在lexicographical的关系（ch1 在 ch2之前），就可以记录ch2在ch1的邻接表中，并更新ch2的indegree加1。 因为从topological的角度，就是ch1依赖ch2，所以是ch2的indegree + 1。
2. 如何初始化indegree的计数
1. 有个tricky的地方在于如果用int indegree[]来记录，因为数组初始化默认都为0，所以不能仅仅凭借indegree[i - 'a']判断是否入度降为0，而需要额外辅助的set，记录在alien dictionary中出现过的字符。
2. 或者用一个HashMap<Character, Integer> 进行记录，这样既可以计数indegree，也满足只记录alien dict出现的字符
3. 何时更新indegree的计数
4. 如何判断是否存在topological sorting
1. 如果存在topological sorting，那么最后result的长度应该等于alien dictionary的key的个数
2. 或者检测indegree hashmap中的每个字符indegree计数，如果有一个不为0，则说明不满足

``````class Solution {
public String alienOrder(String[] words) {
String result = "";
if (words == null || words.length == 0) {
return result;
}
HashMap<Character, HashSet<Character>> adj = new HashMap<>();
HashMap<Character, Integer> indegree = new HashMap<>();

// Initiate indegree for characters in the alien dict
for (String word: words) {
for (char c: word.toCharArray()) {
indegree.put(c, 0);
}
}

// Build graph with adjacency list; update indegree
for (int i = 0; i < words.length - 1; i++) {
int j = 0;
while (j < words[i].length() && j < words[i + 1].length()) {
char a = words[i].charAt(j);
char b = words[i + 1].charAt(j);
if (a != b) {
HashSet<Character> set = new HashSet<>();
}
if (!set.contains(b)) {
indegree.put(b, indegree.get(b) + 1);
}
break;
}
j++;
}
}

// Initiate queue for topological sorting
for (Character c: indegree.keySet()) {
if (indegree.get(c) == 0) {
q.offer(c);
}
}

// Topological Sorting
while (!q.isEmpty()) {
char ch = q.poll();
result += ch;
for (char c : adj.get(ch)) {
indegree.put(c, indegree.get(c) - 1);
if (indegree.get(c) == 0) {
q.offer(c);
}
}
}
}

// No valid topological sorting
if (result.length() != indegree.size()) {
return "";
}

return result;
}

}
``````

#### Topological Sorting - DFS

The key to this problem:

A topological ordering is possible if and only if the graph has no directed cycles

Let's build a graph and perform a DFS. The following states made things easier.

1. `visited[i] = -1`. Not even exist.
2. `visited[i] = 0`. Exist. Non-visited.
3. `visited[i] = 1`. Visiting.
4. `visited[i] = 2`. Visited.
``````private final int N = 26;
public String alienOrder(String[] words) {
int[] visited = new int[N];

StringBuilder sb = new StringBuilder();
for(int i = 0; i < N; i++) {
if(visited[i] == 0) {                 // unvisited
if(!dfs(adj, visited, sb, i)) return "";
}
}
return sb.reverse().toString();
}

public boolean dfs(boolean[][] adj, int[] visited, StringBuilder sb, int i) {
visited[i] = 1;                            // 1 = visiting
for(int j = 0; j < N; j++) {
if(visited[j] == 1) return false;  // 1 => 1, cycle
if(visited[j] == 0) {              // 0 = unvisited
if(!dfs(adj, visited, sb, j)) return false;
}
}
}
visited[i] = 2;                           // 2 = visited
sb.append((char) (i + 'a'));
return true;
}

public void buildGraph(String[] words, boolean[][] adj, int[] visited) {
Arrays.fill(visited, -1);                 // -1 = not even existed
for(int i = 0; i < words.length; i++) {
for(char c : words[i].toCharArray()) visited[c - 'a'] = 0;
if(i > 0) {
String w1 = words[i - 1], w2 = words[i];
int len = Math.min(w1.length(), w2.length());
for(int j = 0; j < len; j++) {
char c1 = w1.charAt(j), c2 = w2.charAt(j);
if(c1 != c2) {
adj[c1 - 'a'][c2 - 'a'] = true;
break;
}
}
}
}
}
``````
##### Another DFS - More Modularized
``````class Solution {
public String alienOrder(String[] dict) {
// corner case
if (dict == null || dict.length == 0) {
return new String();
}

if (dict.length == 1) {
return dict[0];
}

Map<Character, Set<Character>> graph = initialGraph(dict);

}

private Map<Character, Set<Character>> initialGraph(String[] dict) {
Map<Character, Set<Character>> graph = new HashMap<>();

for (int i = 1; i < dict.length; i++) {
String one = dict[i - 1];
String two = dict[i];
}

return graph;
}

private void addEdge(String one, String two,
Map<Character, Set<Character>> graph) {
for (int i = 0; i < one.length(); i++) {
graph.putIfAbsent(one.charAt(i), new HashSet<Character>());
}

for (int i = 0; i < two.length(); i++) {
graph.putIfAbsent(two.charAt(i), new HashSet<Character>());
}

for (int i = 0; i < one.length() && i < two.length(); i++) {
if (one.charAt(i) != two.charAt(i)) {
return;
}
}

}

private String topoOrder(Map<Character, Set<Character>> graph) {
StringBuilder sb = new StringBuilder();
// recording visit status, -1 is visiting , 1 is visited
Map<Character, Integer> visited = new HashMap<>();

for (Character v : graph.keySet()) {
visited.put(v, 0);
}
for (Character v : graph.keySet()) {
if (traverse(graph, v, visited, sb)) {
return new String();
}
}

return sb.reverse().toString();
}

// traverse graph from v, return true if there is cycle.
// also record the order of visited order
private boolean traverse(Map<Character, Set<Character>> graph, Character v,
Map<Character, Integer> visited, StringBuilder sb) {
// base case
if (visited.get(v) == -1) {
return true;
}

if (visited.get(v) == 1) {
return false;
}

// mark visiting
visited.put(v, -1);
for (Character nei : graph.get(v)) {
if (traverse(graph, nei, visited, sb)) {
return true;
}
}

visited.put(v, 1);
sb.append(v);
return false;
}
}
``````