Leetcode 269: Alien Dictionary

Posted 2020-10-18 Keep walking

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.

 

 

 1 public class Solution {
 2     public string AlienOrder(string[] words) {
 3         if (words == null || words.Length == 0) return "";
 4         
 5         var sb = new StringBuilder();
 6         var ingree = new Dictionary<char, int>();
 7         
 8         // initialize ingree
 9         foreach (var w in words)
10         {
11             for (int i = 0; i < w.Length; i++)
12             {
13                 ingree[w[i]] = 0;
14             }
15         }
16         
17         // build adjacent list and set ingree
18         var adjacentList = new Dictionary<char, HashSet<char>>();
19         
20         for (int i = 0; i < words.Length - 1; i++)
21         {
22             int k = 0, j = i + 1;
23 
24             while (k < words[i].Length && k < words[j].Length)
25             {
26                 if (words[i][k] != words[j][k])
27                 {
28                     if (!adjacentList.ContainsKey(words[i][k]))
29                     {
30                         adjacentList[words[i][k]] = new HashSet<char>();
31                     }
32                     
33                     if (!adjacentList[words[i][k]].Contains(words[j][k]))
34                     {
35                         ingree[words[j][k]]++;
36                         adjacentList[words[i][k]].Add(words[j][k]);
37                     }
38 
39                     break;
40                 }
41 
42                 k++;
43             }
44         }
45         
46         // bfs
47         var queue = new Queue<char>();
48         
49         foreach (var pair in ingree)
50         {
51             if (pair.Value == 0)
52             {
53                 queue.Enqueue(pair.Key);
54             }
55         }
56         
57         while (queue.Count > 0)
58         {
59             var c = queue.Dequeue();
60             sb.Append(c);
61             
62             if (adjacentList.ContainsKey(c))
63             {
64                 foreach (var item in adjacentList[c])
65                 {
66                     ingree[item]--;
67 
68                     if (ingree[item] == 0)
69                     {
70                         queue.Enqueue(item);
71                     }
72                 }
73             }
74         }
75         
76         return sb.Length == ingree.Count ? sb.ToString() : "";
77     }
78 }