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Problem solution · Python

Alien Dictionary

Alien Dictionary: a Python solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
54 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Direct simulation

For Alien Dictionary, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 54 lines of Python from the credited upstream file 269.py.
  • The implementation visibly relies on sequence storage, hash lookup, ordered lookup, work queue.
  • No explicit loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeAlien Dictionary · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def alienOrder(self, words: list[str]) -> str:    graph = {}    inDegrees = [0] * 26     self._buildGraph(graph, words, inDegrees)    return self._topology(graph, inDegrees)   def _buildGraph(      self,      graph: dict[str, set[str]],      words: list[str],      inDegrees: list[int],  ) -> None:    # Create a node for each character in each word.    for word in words:      for c in word:        if c not in graph:          graph[c] = set()     for first, second in zip(words, words[1:]):      length = min(len(first), len(second))      for j in range(length):        u = first[j]        v = second[j]        if u != v:          if v not in graph[u]:            graph[u].add(v)            inDegrees[ord(v) - ord('a')] += 1          break  # The order of characters after this are meaningless.        # First = 'ab', second = 'a' . invalid        if j == length - 1 and len(first) > len(second):          graph.clear()          return   def _topology(self, graph: dict[str, set[str]], inDegrees: list[int]) -> str:    s = ''    q = collections.deque()     for c in graph:      if inDegrees[ord(c) - ord('a')] == 0:        q.append(c)     while q:      u = q.pop()      s += u      for v in graph[u]:        inDegrees[ord(v) - ord('a')] -= 1        if inDegrees[ord(v) - ord('a')] == 0:          q.append(v)     # Words = ['z', 'x', 'y', 'x']    return s if len(s) == len(graph) else ''

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