Word Ladder
Detailed guide and Python implementation for the 'Word Ladder' problem.
1. Concept Overview
The 'Word Ladder' problem is a key challenge in the Graphs section.
This implementation focuses on easy-level logic in Python.
We prioritize technical accuracy and code readability in our provided solutions.
2. Real-World Applications
3. Visual Intuition
Visualizing the logic flow for Word Ladder.
4. Prerequisites
5. Step-by-Step Thinking
1. Understand the problem
Read the problem statement for Word Ladder carefully.
2. Formulate brute force
Draft a simple iterative solution.
3. Identify inefficiency
Look for redundant calculations.
4. Optimize search path
Use hashing or sorting to speed up the process.
5. Final Implementation
Clean up the code for production standards.
Problem Statement
A transformation sequence from word beginWord to word endWord using a dictionary wordList is a sequence of words beginWord -> s1 -> s2 -> ... -> sk such that:
- Every adjacent pair of words differs by a single letter.
- Every si for 1 <= i <= k is in wordList. Note that beginWord does not need to be in wordList.
- sk == endWord.
Given two words, beginWord and endWord, and a dictionary wordList, return the number of words in the shortest transformation sequence from beginWord to endWord, or 0 if no such sequence exists.
Write a function ladderLength(beginWord: str, endWord: str, wordList: List[str]) -> int.
- •1 <= len(beginWord) <= 10
- •endWord.length == beginWord.length
- •1 <= len(wordList) <= 5000
- •wordList[i].length == beginWord.length
- •beginWord, endWord, and wordList[i] consist of lowercase English letters
- •All the words in wordList are unique
Examples
beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"]
5
One shortest transformation sequence is "hit" -> "hot" -> "dot" -> "dog" -> "cog", which is 5 words long.
beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log"]
0
The endWord "cog" is not in wordList, so there is no valid transformation sequence.
Need a Hint?
Edge Cases to Watch
- Empty input structures
- Single element inputs
- Large numerical bounds
Ready to Solve?
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Interview Insights & Variations
Complexity Analysis Breakdown
Why Time: Directly evaluates all possibilities.
Why Space: Uses standard local memory.
Why Time: Optimized paths reduce total operations.
Why Space: May trade memory for speed.
Optimized Solution Python Code
Optimized Solution Python Code
from collections import deque, defaultdict
def ladder_length_opt(beginWord, endWord, wordList):
if endWord not in wordList: return 0
nei = defaultdict(list)
wordList.append(beginWord)
for word in wordList:
for j in range(len(word)):
pattern = word[:j] + "*" + word[j + 1 :]
nei[pattern].append(word)
visit = {beginWord}
q = deque([beginWord])
res = 1
while q:
for i in range(len(q)):
word = q.popleft()
if word == endWord: return res
for j in range(len(word)):
pattern = word[:j] + "*" + word[j + 1 :]
for neighbor in nei[pattern]:
if neighbor not in visit:
visit.add(neighbor)
q.append(neighbor)
res += 1
return 0Brute Force Code (Spoiler Guarded)
Brute Force Code (Spoiler Guarded)
def ladder_length_brute(beginWord, endWord, wordList):
if endWord not in wordList: return 0
def is_diff_one(w1, w2):
diff = 0
for i in range(len(w1)):
if w1[i] != w2[i]: diff += 1
return diff == 1
q = [(beginWord, 1)]
visit = {beginWord}
while q:
word, dist = q.pop(0)
if word == endWord: return dist
for w in wordList:
if w not in visit and is_diff_one(word, w):
visit.add(w)
q.append((w, dist + 1))
return 0Algorithm Pattern Checklist
When dealing with Graphs data patterns.
Core Prerequisites
Revision Key Notes
Common Mistakes & Pitfalls
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