Lyft Technical Interview: Word Ladder Deep Dive

Lyft's technical interviews are known for focusing on algorithms and logic, often including complex graph theory and search problems. This article reconstructs a candidate's journey solving a real Lyft interview problem, demonstrating how oavoservice real-time assistance helps candidates explain their thoughts clearly and handle follow-ups confidently.

📋 Interview Question: Word Ladder

Problem Description

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.

A transformation sequence is beginWord -> s1 -> s2 -> ... -> sk such that:

1. Every adjacent pair differs by a single letter.
2. Every si is in wordList.
3. sk == endWord.

Example

Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"]
Output: 5
Explanation: "hit" -> "hot" -> "dot" -> "dog" -> "cog"

🎯 Core Concepts

This is classic LeetCode 127. Word Ladder.

1. BFS: Shortest path in unweighted graph.
2. Graph Modeling: Transforming words into graph traversal.
3. Optimization: Efficient neighbor generation.

💡 Solution Strategy (oavoservice Guidance)

Method 1: BFS + Hash Set

oavoservice Suggestion: BFS is optimal for shortest paths.

from collections import deque

def ladderLength(beginWord, endWord, wordList):
    word_set = set(wordList)
    if endWord not in word_set:
        return 0
    
    queue = deque([(beginWord, 1)])
    visited = {beginWord}
    
    while queue:
        current_word, length = queue.popleft()
        
        if current_word == endWord:
            return length
        
        for i in range(len(current_word)):
            for c in 'abcdefghijklmnopqrstuvwxyz':
                next_word = current_word[:i] + c + current_word[i+1:]
                
                if next_word in word_set and next_word not in visited:
                    visited.add(next_word)
                    queue.append((next_word, length + 1))
    
    return 0

Method 2: Bidirectional BFS (Optimization)

oavoservice Advanced Guidance: For large data, bidirectional BFS significantly improves efficiency.

def ladderLength_bidirectional(beginWord, endWord, wordList):
    word_set = set(wordList)
    if endWord not in word_set:
        return 0
    
    begin_set = {beginWord}
    end_set = {endWord}
    visited = set()
    length = 1
    
    while begin_set and end_set:
        if len(begin_set) > len(end_set):
            begin_set, end_set = end_set, begin_set
        
        next_set = set()
        for word in begin_set:
            for i in range(len(word)):
                for c in 'abcdefghijklmnopqrstuvwxyz':
                    next_word = word[:i] + c + word[i+1:]
                    
                    if next_word in end_set:
                        return length + 1
                    
                    if next_word in word_set and next_word not in visited:
                        visited.add(next_word)
                        next_set.add(next_word)
        
        begin_set = next_set
        length += 1
    
    return 0

🚀 Deep Follow-ups

Q1: Optimize character replacement?

oavoservice Suggestion: Preprocess a generic state map ("hot" -> "*ot", "h*t", "ho*").

Q2: Return all shortest paths?

oavoservice Suggestion: Use BFS to record parents, then backtrack (LeetCode 126).

Q3: Large scale dictionary?

oavoservice Suggestion: Trie tree, precomputed index, distributed processing.

💼 How oavoservice Helps with Lyft Interviews

Algorithm Selection - Identify optimal approach quickly. Code Implementation - Ensure clear logic. Optimization Discussion - Demonstrate engineering skills. Edge Cases - Avoid missing special cases.

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Tags: #Lyft #WordLadder #BFS #GraphTheory #ShortestPath #VOHelp #InterviewPrep #1point3acres

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