269. Alien Dictionary

# Time:  O(n)
# Space: O(|V|+|E|) = O(26 + 26^2) = O(1)

import collections


# BFS solution.
class Solution(object):
    def alienOrder(self, words):
        """
        :type words: List[str]
        :rtype: str
        """
        result, in_degree, out_degree = [], {}, {}
        zero_in_degree_queue = collections.deque()
        nodes = set()
        for word in words:
            for c in word:
                nodes.add(c)

        for i in xrange(1, len(words)):
            if (len(words[i-1]) > len(words[i]) and
                    words[i-1][:len(words[i])] == words[i]):
                return ""
            self.findEdges(words[i - 1], words[i], in_degree, out_degree)

        for node in nodes:
            if node not in in_degree:
                zero_in_degree_queue.append(node)

        while zero_in_degree_queue:
            precedence = zero_in_degree_queue.popleft()
            result.append(precedence)

            if precedence in out_degree:
                for c in out_degree[precedence]:
                    in_degree[c].discard(precedence)
                    if not in_degree[c]:
                        zero_in_degree_queue.append(c)

                del out_degree[precedence]

        if out_degree:
            return ""

        return "".join(result)

    # Construct the graph.
    def findEdges(self, word1, word2, in_degree, out_degree):
        str_len = min(len(word1), len(word2))
        for i in xrange(str_len):
            if word1[i] != word2[i]:
                if word2[i] not in in_degree:
                    in_degree[word2[i]] = set()
                if word1[i] not in out_degree:
                    out_degree[word1[i]] = set()
                in_degree[word2[i]].add(word1[i])
                out_degree[word1[i]].add(word2[i])
                break


# DFS solution.
class Solution2(object):
    def alienOrder(self, words):
        """
        :type words: List[str]
        :rtype: str
        """
        # Find ancestors of each node by DFS.
        nodes, ancestors = set(), {}
        for i in xrange(len(words)):
            for c in words[i]:
                nodes.add(c)
        for node in nodes:
            ancestors[node] = []
        for i in xrange(1, len(words)):
            if (len(words[i-1]) > len(words[i]) and
                    words[i-1][:len(words[i])] == words[i]):
                return ""
            self.findEdges(words[i - 1], words[i], ancestors)

        # Output topological order by DFS.
        result = []
        visited = {}
        for node in nodes:
            if self.topSortDFS(node, node, ancestors, visited, result):
                return ""

        return "".join(result)

    # Construct the graph.
    def findEdges(self, word1, word2, ancestors):
        min_len = min(len(word1), len(word2))
        for i in xrange(min_len):
            if word1[i] != word2[i]:
                ancestors[word2[i]].append(word1[i])
                break

    # Topological sort, return whether there is a cycle.
    def topSortDFS(self, root, node, ancestors, visited, result):
        if node not in visited:
            visited[node] = root
            for ancestor in ancestors[node]:
                if self.topSortDFS(root, ancestor, ancestors, visited, result):
                    return True
            result.append(node)
        elif visited[node] == root:
            # Visited from the same root in the DFS path.
            # So it is cyclic.
            return True
        return False

Solution from kamyu104/LeetCode-Solutions · MIT