Alien Dictionary
The task of determining the order of characters in an alien language can be modeled as finding the topological sort in a directed graph. In this graph, each node represents a character from the alien language, and a directed edge from character u
to character v
means that u
comes before v
in the language.
Here’s a stepbystep approach to solving this problem:
 Create a graph: Start by creating an adjacency list to represent the graph.
 Determine edges: Compare adjacent words in the list to determine the order between characters.
 Topological Sort: Perform topological sorting to determine the ordering of characters.
 Detect Cycles: If a cycle is detected, return an empty string as the order would be invalid.
The following code implements these steps:
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 from collections import defaultdict, Counter, deque
class Solution:
def alienOrder(self, words: List[str]) > str:
# Step 1: Create a graph
adj_list = defaultdict(set)
in_degree = Counter({c: 0 for word in words for c in word})
# Step 2: Determine edges by comparing adjacent words
for i in range(len(words)  1):
word1, word2 = words[i], words[i + 1]
for j in range(min(len(word1), len(word2))):
if word1[j] != word2[j]:
if word2[j] not in adj_list[word1[j]]:
adj_list[word1[j]].add(word2[j])
in_degree[word2[j]] += 1
break
else:
if len(word1) > len(word2): return ""
# Step 3: Topological Sort using BFS
queue = deque([c for c in in_degree if in_degree[c] == 0])
result = []
while queue:
node = queue.popleft()
result.append(node)
for neighbor in adj_list[node]:
in_degree[neighbor] = 1
if in_degree[neighbor] == 0:
queue.append(neighbor)
# Step 4: Detect Cycles
if len(result) != len(in_degree):
return ""
return "".join(result)

Key Takeaways
 The problem can be solved by constructing a graph and performing topological sorting.
 If the graph contains a cycle, the ordering is invalid.
 The time complexity of this solution is (O(N \cdot M)), where (N) is the number of words and (M) is the maximum length of a word. The space complexity is (O(C)), where (C) is the number of unique characters in the input.
title: Alien Dictionary
excerpt: Graph, DFS, Adjacency List
Solution
I refactored the solution that I found online.
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 def alien_order(words)
graph = Graph.new
# Add nodes to the graph. No neighbors added at this time.
words.each do word
word.each_char do c
graph.add_node(c)
end
end
# Create graph
(words.size  1).times do i
# ['abc', 'ab'] is not valid, return ''
if words[i].size != words[i+1].size
if words[i].start_with?(words[i+1])
return ''
end
end
# Create the precedence rules
words[i].each_char.with_index do c, j
if words[i+1][j] && c != words[i+1][j]
graph.add_edge(c, words[i+1][j])
break
end
end
end
# DFS Toplogical Sort
TopSort.new(graph).sort
end
class TopSort
def initialize(graph)
@adj = graph.adj
@stack = []
@visited = {}
end
def sort
@adj.keys.each do node
return '' if @visited[node].nil? && dfs(node)
end
@stack.join('')
end
def dfs(node)
@visited[node] = false
@adj[node].each do neighbor
# We can explore neighbors only if it has not been explored
if @visited[neighbor].nil?
# Explore only if the neighbor has neighbors to be explored
if @adj[neighbor]
return true if dfs(neighbor)
end
else
return true if !@visited[neighbor]
end
end
# adds the node to the beginning of the array
# this happens when recursive calls begin to return
@stack.unshift(node)
@visited[node] = true
false
end
end
class Graph
attr_accessor :adj
def initialize
@adj = Hash.new([])
end
def add_node(node)
@adj[node] = []
end
def add_edge(source, dest)
# Nothing to do, if the precedence rule is already defined
return if @adj[source].include?(dest)
@adj[source].push(dest)
end
end

Few more lines to be cleaned up in the dfs method.