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:

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

Analysis & Solution

问题转化：alien单词字母表的先后顺序转化为依赖关系，那么这个字母表顺序就可以通过拓扑排序来得到。

Topological Sorting - BFS

通过input words建立邻接表，记录In-degree的数目，从in-degree为0的节点（字母）开始，放入Queue中，进行BFS。

接下来的就是用Kahn's aglorithm来做topological sort。

注意的要点：

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。
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
何时更新indegree的计数
1. 每次在queue中poll()之后，相当于要对这个ch对应的所有依赖的字符更新indegree，也就是在adjacency list邻接表中遍历对应的字符，再更新indegree的记录-1.
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

代码：

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 (adj.containsKey(a)) {
                        set = adj.get(a);
                    }
                    if (!set.contains(b)) {
                        set.add(b);
                        adj.put(a, set);
                        indegree.put(b, indegree.get(b) + 1);
                    }
                    break;
                }
                j++;
            }
        }

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

        // Topological Sorting
        while (!q.isEmpty()) {
            char ch = q.poll();
            result += ch;
            if (adj.containsKey(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

by @yavinci:

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.

visited[i] = -1. Not even exist.
visited[i] = 0. Exist. Non-visited.
visited[i] = 1. Visiting.
visited[i] = 2. Visited.

private final int N = 26;
public String alienOrder(String[] words) {
    boolean[][] adj = new boolean[N][N];
    int[] visited = new int[N];
    buildGraph(words, adj, visited);

    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(adj[i][j]) {                        // connected
            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);

        return topoOrder(graph);
    }

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

        // each adjacent pairs
        for (int i = 1; i < dict.length; i++) {
            String one = dict[i - 1];
            String two = dict[i];
            addEdge(one, two, graph);
        }

        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)) {
                graph.get(one.charAt(i)).add(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;
    }
}

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