What is the time complexity of Word Ladder?

The optimal solution has a time complexity of O(M^2 * N).

What is the space complexity of Word Ladder?

The space complexity is O(M * N).

What data structures are used to solve Word Ladder?

This problem uses Hash Table, String, Breadth-First Search. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Word Ladder?

Insight 1: The problem can be modeled as a graph where each word is a node, and an edge exists between two words if they differ by only one letter. The shortest transformation sequence is then equivalent to the shortest path in this graph. Insight 2: Breadth-First Search (BFS) is the ideal algorithm for finding the shortest path in an unweighted graph.

Which companies ask LeetCode 127. Word Ladder in interviews?

This problem is frequently asked at Box, Juspay, LinkedIn, Navan, Nutanix and 13 more companies.

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Word Ladder - LeetCode 127 Solution

Q: How to solve Word Ladder?

The provided code implements a bidirectional Breadth-First Search (BFS) to find the shortest transformation sequence. It starts searching from both the `beginWord` and `endWord` simultaneously. In each iteration, it expands the search from the queue with fewer elements to improve performance. It keeps track of visited words for both searches using separate hash maps (or hash tables). When a word is encountered that has already been visited by the other search, it means a path has been found connecting the `beginWord` and `endWord`, and the algorithm returns the combined length of the paths.

Word Ladder - Complete Solution Guide

Word Ladder is LeetCode problem 127, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

A transformation sequence from word beginWord to word endWord using a dictionary wordList is a sequence of words beginWord -> s 1 -> s 2 -> ... -> s k such that: Every adjacent pair of words differs by a single letter. Every s i for 1 <= i <= k is in wordList . Note that beginWord does not need to be in wordList . s k == 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

Detailed Explanation

The "Word Ladder" problem asks you to find the shortest transformation sequence from a starting word (`beginWord`) to an ending word (`endWord`), given a dictionary of valid words (`wordList`). Each transformation in the sequence must change only one letter at a time, and every intermediate word must be present in the `wordList`. The goal is to return the length of the shortest such sequence (including the start and end words), or 0 if no such sequence exists. `beginWord` does not have to be present in `wordList`. The length of all words are the same. All words are unique. All inputs are lower case letters.

Solution Approach

The provided code implements a bidirectional Breadth-First Search (BFS) to find the shortest transformation sequence. It starts searching from both the `beginWord` and `endWord` simultaneously. In each iteration, it expands the search from the queue with fewer elements to improve performance. It keeps track of visited words for both searches using separate hash maps (or hash tables). When a word is encountered that has already been visited by the other search, it means a path has been found connecting the `beginWord` and `endWord`, and the algorithm returns the combined length of the paths.

Step-by-Step Algorithm

Step 1: Check if the `endWord` is present in the `wordList`. If not, return 0 because no transformation sequence is possible.
Step 2: Initialize two queues (`q_begin` and `q_end`) for BFS, starting with `beginWord` and `endWord` respectively.
Step 3: Initialize two hash maps (`visited_begin` and `visited_end`) to keep track of visited words and their levels (distance from the starting word) for each search direction.
Step 4: While both queues are not empty:
Step 5: Select the queue with the smaller size. Swap the `q_begin`, `q_end`, `visited_begin`, and `visited_end` if `q_begin` is larger than `q_end`.
Step 6: Dequeue a word (`word`) from the selected queue (`q_begin`). Get its level (`level`) from the corresponding `visited_begin` hash map.
Step 7: Iterate through each character of the current `word`:
Step 8: For each character, try replacing it with all possible lowercase letters ('a' to 'z'):
Step 9: Create a new word (`next_word`) with the replaced character.
Step 10: Check if `next_word` is in the `visited_end` hash map. If yes, it means a path connecting `beginWord` and `endWord` has been found. Return the sum of the levels from both sides (i.e., `level + visited_end[next_word]`).
Step 11: If `next_word` is in the `wordList` and not in the `visited_begin` hash map, it's a valid and unvisited word. Enqueue it into `q_begin`, update its level in `visited_begin` to `level + 1`.
Step 12: If no path is found after exhausting both queues, return 0.

Key Insights

Insight 1: The problem can be modeled as a graph where each word is a node, and an edge exists between two words if they differ by only one letter. The shortest transformation sequence is then equivalent to the shortest path in this graph.
Insight 2: Breadth-First Search (BFS) is the ideal algorithm for finding the shortest path in an unweighted graph.
Insight 3: A bidirectional BFS can significantly improve performance by searching from both the `beginWord` and the `endWord` simultaneously. The search stops when the two searches meet at a common word.

Complexity Analysis

Time Complexity: O(M^2 * N)

Space Complexity: O(M * N)

Topics

This problem involves: Hash Table, String, Breadth-First Search.

Companies

Asked at: Box, Juspay, LinkedIn, Navan, Nutanix, PhonePe, Reddit, Samsung, Snap, Tekion, The Trade Desk, TikTok, Uber, Visa, Yahoo, Yelp, ZScaler, eBay.

Word Ladder | LeetCode 127

Hard

Hash Table String Breadth-First Search

Time: O(M^2 * N)Space: O(M * N)

Solution Code

Available in:

import java.util.*;

class Solution {
    public int ladderLength(String beginWord, String endWord, List<String> wordList) {
        Set<String> wordSet = new HashSet<>(wordList);
        if (!wordSet.contains(endWord)) {
            return 0;
        }

        Queue<String> q_begin = new LinkedList<>(Collections.singletonList(beginWord));
        Queue<String> q_end = new LinkedList<>(Collections.singletonList(endWord));

        Map<String, Integer> visited_begin = new HashMap<>();
        visited_begin.put(beginWord, 1);
        Map<String, Integer> visited_end = new HashMap<>();
        visited_end.put(endWord, 1);

        while (!q_begin.isEmpty() && !q_end.isEmpty()) {
            if (q_begin.size() > q_end.size()) {
                Queue<String> temp_q = q_begin;
                q_begin = q_end;
                q_end = temp_q;

                Map<String, Integer> temp_visited = visited_begin;
                visited_begin = visited_end;
                visited_end = temp_visited;
            }

            String word = q_begin.poll();
            int level = visited_begin.get(word);

            for (int i = 0; i < word.length(); i++) {
                char original_char = word.charAt(i);
                for (int c_code = (int) 'a'; c_code <= (int) 'z'; c_code++) {
                    char char_ = (char) c_code;
                    if (char_ == original_char) {
                        continue;
                    }

                    String next_word = word.substring(0, i) + char_ + word.substring(i + 1);

                    if (visited_end.containsKey(next_word)) {
                        return level + visited_end.get(next_word);
                    }

                    if (wordSet.contains(next_word) && !visited_begin.containsKey(next_word)) {
                        visited_begin.put(next_word, level + 1);
                        q_begin.offer(next_word);
                    }
                }
            }
        }

        return 0;
    }
}

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Problem Info

Description

Problem 127: Word Ladder - Detailed explanation and solution approaches.

Topics

Hash Table String Breadth-First Search

Companies

BoxJuspayLinkedInNavanNutanixPhonePeRedditSamsung

Stats

Difficulty:Hard

Languages:4

Companies:18

Problem #:127

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

12 steps

1
Step 1: Check if the `endWord` is present in the `wordList`. If not, return 0 because no transformation sequence is possible.
2
Step 2: Initialize two queues (`q_begin` and `q_end`) for BFS, starting with `beginWord` and `endWord` respectively.
3
Step 3: Initialize two hash maps (`visited_begin` and `visited_end`) to keep track of visited words and their levels (distance from the starting word) for each search direction.
4
Step 4: While both queues are not empty:
5
Step 5: Select the queue with the smaller size. Swap the `q_begin`, `q_end`, `visited_begin`, and `visited_end` if `q_begin` is larger than `q_end`.
6
Step 6: Dequeue a word (`word`) from the selected queue (`q_begin`). Get its level (`level`) from the corresponding `visited_begin` hash map.
7
Step 7: Iterate through each character of the current `word`:
8
Step 8: For each character, try replacing it with all possible lowercase letters ('a' to 'z'):
9
Step 9: Create a new word (`next_word`) with the replaced character.
10
Step 10: Check if `next_word` is in the `visited_end` hash map. If yes, it means a path connecting `beginWord` and `endWord` has been found. Return the sum of the levels from both sides (i.e., `level + visited_end[next_word]`).
11
Step 11: If `next_word` is in the `wordList` and not in the `visited_begin` hash map, it's a valid and unvisited word. Enqueue it into `q_begin`, update its level in `visited_begin` to `level + 1`.
12
Step 12: If no path is found after exhausting both queues, return 0.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(M^2 * N)

💾

Space Complexity

O(M * N)

📝

Why?

Time Complexity: In the worst case, we might visit all the words in the `wordList`. For each word, we iterate through each of its M characters and try all 26 possible replacements, leading to O(M * 26) which simplifies to O(M). Therefore, the overall time complexity can be approximated as O(N * M * M) where the last M comes from string manipulation when creating next_word. Space Complexity: The `visited_begin` and `visited_end` hash maps store, at most, all the words in the `wordList`. Therefore, the space complexity is O(N*M) because N is the number of words and M is the length of each word. Queues will also take O(N*M) space

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Unidirectional BFS

A standard BFS starting from `beginWord` and searching until `endWord` is found. This approach is simpler to implement but often less efficient than bidirectional BFS.

Time: O(M^2 * N)Space: O(M * N)

Trade-offs: Unidirectional BFS can be easier to understand and implement, but it might explore more of the search space than bidirectional BFS, especially when the shortest path is relatively short.

A* Search

A* search could be used, incorporating a heuristic function. A simple heuristic would be the Hamming distance (number of different characters) between the current word and the end word.

Time: O(M^2 * N) in the worst case, but potentially faster with a good heuristicSpace: O(M * N)

Trade-offs: A* search can be faster than BFS if a good heuristic is available, but choosing a suitable heuristic can be challenging. If the heuristic is poorly chosen, A* might perform worse than BFS.

✨

Key Insights

1Insight 1: The problem can be modeled as a graph where each word is a node, and an edge exists between two words if they differ by only one letter. The shortest transformation sequence is then equivalent to the shortest path in this graph.
2Insight 2: Breadth-First Search (BFS) is the ideal algorithm for finding the shortest path in an unweighted graph.
3Insight 3: A bidirectional BFS can significantly improve performance by searching from both the `beginWord` and the `endWord` simultaneously. The search stops when the two searches meet at a common word.

📚

Key Terms Explained

For Students

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: `endWord` is not in `wordList`. The code handles this case by returning 0 at the beginning.
Edge case 2: `beginWord` is equal to `endWord`. Although the constraints specify `beginWord != endWord`, handling this case is a good practice. The code as written now assumes they're never equal.
Edge case 3: No transformation sequence exists between `beginWord` and `endWord`. The code handles this by returning 0 after the `while` loop if no path is found.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not checking if `endWord` is in `wordList`. If `endWord` is not in the dictionary, there is no possible transformation sequence.

✗Common mistake 2: Failing to mark words as visited. This leads to infinite loops in the BFS, as the algorithm revisits previously explored words.

✗Common mistake 3: Not using a set for efficient `wordList` lookups. Checking if a word is in a list (using `in`) has O(N) complexity, while checking in a set has O(1) complexity on average. This significantly impacts the overall runtime.

✗Common mistake 4: Not handling memory allocation properly in C implementations. Memory leaks can occur if allocated strings for next_word are not freed after being used in a particular iteration of the BFS when they're no longer needed and don't get enqueued. Each of these strings has to be allocated and copied, a potential source of inefficiency.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Spell checkers can use algorithms similar to word ladder to suggest possible corrections for misspelled words. The dictionary is the `wordList`, the misspelled word is the `beginWord`, and the algorithm finds valid words close to the misspelled word.

🏦Real-world scenario 2: Bioinformatics uses similar techniques to find sequences of DNA or protein that are closely related to a given sequence. The differences between the sequences can be modeled as transformations, and the goal is to find the shortest path between them.

🎮Real-world scenario 3: Robotics path planning - finding the shortest sequence of actions for a robot to reach a goal state.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

Hash tables trade space for time. Before using one, consider if the problem constraints allow the extra memory. In interviews, always mention this trade-off.

🎯 Interview Tip

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

Don't forget to handle key collisions and null/undefined keys. Consider what happens with duplicate values.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 6, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 29

❓

Frequently Asked Questions

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Word Ladder"?

For this hard-level Hash Table problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?