269 Alien Dictionary

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269. Alien Dictionary​
1. Question
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,
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[
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  "wrt",
3
  "wrf",
4
  "er",
5
  "ett",
6
  "rftt"
7
]
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The correct order is:"wertf".
Example 2:
Given the following words in dictionary,
1
[
2
  "z",
3
  "x"
4
]
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The correct order is:"zx".
Example 3:
Given the following words in dictionary,
1
[
2
  "z",
3
  "x",
4
  "z"
5
]
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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.
2. Implementation
(1) BFS
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class Solution {
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    public String alienOrder(String[] words) {
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        if (words == null || words.length == 0) {
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            return "";
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        }
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​
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        Map<Character, Integer> inDegree = new HashMap<>();
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        Map<Character, Set<Character>> adjList = new HashMap<>();
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​
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​
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        for (String word : words) {
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            for (char c : word.toCharArray()) {
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                inDegree.put(c, 0);
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            }
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        }
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​
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        for (int i = 0; i < words.length - 1; i++) {
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            String curWord = words[i];
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            String nextWord = words[i + 1];
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​
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            for (int j = 0; j < Math.min(curWord.length(), nextWord.length()); j++) {
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                char c1 = curWord.charAt(j);
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                char c2 = nextWord.charAt(j);
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​
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                // We can only know the lexicographical order from the first two different words
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                if (c1 != c2) {
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                    Set<Character> set = adjList.getOrDefault(c1, new HashSet<>());
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​
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                    if (!set.contains(c2)) {
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                        set.add(c2);
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                        adjList.put(c1, set);
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                        inDegree.put(c2, inDegree.get(c2) + 1);
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                    }
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                    break;
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                }
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                // This is not lexicographically order
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                else if (j + 1 < curWord.length() && j + 1 == nextWord.length()) {
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                    return "";
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                }
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            }
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        }
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​
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        Queue<Character> queue = new LinkedList<>();
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            for (char key: inDegree.keySet()) {
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                if (inDegree.get(key) == 0) {
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                    queue.add(key);
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                }
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            }
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​
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​
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        StringBuilder res = new StringBuilder();
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​
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        while (!queue.isEmpty()) {
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            char curC = queue.remove();
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            res.append(curC);
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​
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            if (adjList.containsKey(curC)) {
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                for (char nextC : adjList.get(curC)) {
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                    inDegree.put(nextC, inDegree.get(nextC) - 1);
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​
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                    if (inDegree.get(nextC) == 0) {
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                        queue.add(nextC);
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                    }
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                }
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            }
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        }
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        return res.length() == inDegree.size() ? res.toString() : "";
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    }
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}
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3. Time & Space Complexity
BFS:时间复杂度O(n * L), n为word的个数，L为每个word的平均长度， 空间复杂度O(k), k为不同的character个数
210 Course Schedule II
444 Sequence Reconstruction
Last modified 2yr ago
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Contents
269. Alien Dictionary
1. Question
2. Implementation
3. Time & Space Complexity
269. Alien Dictionary​

1. Question

2. Implementation

3. Time & Space Complexity

269. Alien Dictionary
