Foreign Dictionary

There is a foreign language which uses the latin alphabet, but the order among letters is not “a”, “b”, “c” … “z” as in English.

You receive a list of non-empty strings words from the dictionary, where the words are sorted lexicographically based on the rules of this new language.

Derive the order of letters in this language. If the order is invalid, return an empty string. If there are multiple valid order of letters, return any of them.

A string a is lexicographically smaller than a string b if either of the following is true:

The first letter where they differ is smaller in a than in b.
There is no index i such that a[i] != b[i] and a.length < b.length.

Example 1:

Input: ["z","o"]Output: "zo"

Explanation:
From “z” and “o”, we know ‘z’ < ‘o’, so return “zo”.

Example 2:

Input: ["hrn","hrf","er","enn","rfnn"]Output: "hernf"

Explanation:

from “hrn” and “hrf”, we know ‘n’ < ‘f’
from “hrf” and “er”, we know ‘h’ < ‘e’
from “er” and “enn”, we know get ‘r’ < ‘n’
from “enn” and “rfnn” we know ‘e’<‘r’
so one possibile solution is “hernf”

Constraints:

The input words will contain characters only from lowercase ‘a’ to ‘z’.

1 <= words.length <= 100
1 <= words[i].length <= 100

Solution

From the order of the words, we can get the priority between 2 characters, which can be represented as a directed edge between them. When we organize all of the edges, we can get a DAG (Directed Acyclic Graph). Then we just need to do a topological sort with this DAG and get the answer.

Code

import queueclass Solution:def foreignDictionary(self, words: List[str]) -> str:edges = {c: set() for word in words for c in word}in_deg = {c: 0 for word in words for c in word}for i in range(len(words) - 1):w1, w2 = words[i], words[i + 1]minLen = min(len(w1), len(w2))if len(w1) > len(w2) and w1[:minLen] == w2[:minLen]:return ''for j in range(minLen):if w1[j] != w2[j]:edges[w1[j]].add(w2[j])breakfor value in edges.values():for c in value:in_deg[c] += 1print(edges)print(in_deg)que = queue.Queue()for key, value in in_deg.items():if value == 0:que.put(key)ans = []while not que.empty():cur = que.get()ans.append(cur)for nxt in edges.get(cur, []):in_deg[nxt] -= 1if in_deg[nxt] == 0:que.put(nxt)print(ans)if len(ans) != len(in_deg):return ''return ''.join(ans)