How do you solve Word Ladder on LeetCode?

Word Ladder (LeetCode #127) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * m * 26) time complexity and O(n) space complexity. Key insight: Using a set for wordList allows for O(1) average time complexity for lookups.

What is the brute force solution for Word Ladder?

The brute force approach for Word Ladder is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible transformations from the beginWord to the endWord and checking if they exist in the wordList. This is inefficient as it explores many unnecessary paths. The optimal Optimal Solution approach improves this to O(n * m * 26).

What algorithm or data structure is used to solve Word Ladder?

Word Ladder uses the following concepts: Hash Table, String, Breadth-First Search. The recommended patterns to study are: Breadth-First Search, Graph Traversal.

What are the interview tips for Word Ladder?

Clearly explain your thought process and how you plan to approach the problem. Consider edge cases, such as when the endWord is not in the wordList. Practice BFS problems to get comfortable with queue operations.

What are common mistakes when solving Word Ladder?

Not checking if endWord is in wordList before starting the search. Failing to remove words from the set after processing to avoid cycles.

#127

Word Ladder

Hard

Hash TableStringBreadth-First SearchBreadth-First SearchGraph Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * m * 26)
Space	O(1)	O(n)

💡

Intuition

Time O(n * m * 26)Space O(n)

The optimal solution uses a breadth-first search (BFS) approach to explore the shortest path from the beginWord to the endWord. This ensures that we find the shortest transformation sequence efficiently.

⚙️

Algorithm

5 steps

1Step 1: Initialize a queue with the beginWord and a length of 1.
2Step 2: While the queue is not empty, dequeue the front element.
3Step 3: For each character in the current word, try replacing it with every letter from 'a' to 'z'.
4Step 4: If the new word is in the wordList, enqueue it with an incremented length and remove it from the wordList.
5Step 5: If the new word is the endWord, return the length.

solution.py19 lines

1# Full working Python code
2from collections import deque
3
4def wordLadder(beginWord, endWord, wordList):
5    if endWord not in wordList:
6        return 0
7    wordList = set(wordList)
8    queue = deque([(beginWord, 1)])
9    while queue:
10        current_word, length = queue.popleft()
11        if current_word == endWord:
12            return length
13        for i in range(len(current_word)):
14            for c in 'abcdefghijklmnopqrstuvwxyz':
15                next_word = current_word[:i] + c + current_word[i+1:]
16                if next_word in wordList:
17                    queue.append((next_word, length + 1))
18                    wordList.remove(next_word)
19    return 0

ℹ

Complexity note: The time complexity is O(n * m * 26) where n is the number of words in the wordList and m is the length of each word. We generate up to 26 new words for each character in each word.

1Using a set for wordList allows for O(1) average time complexity for lookups.
2BFS is optimal for unweighted shortest path problems like this one.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.