What is the time complexity of Word Ladder II?

The optimal solution has a time complexity of O(V*26^L + N).

What is the space complexity of Word Ladder II?

The space complexity is O(V^2 + N).

What data structures are used to solve Word Ladder II?

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

What are the key insights for solving Word Ladder II?

Insight 1: Breadth-First Search (BFS) is crucial for finding the *shortest* paths in an unweighted graph. The graph is implicitly defined by the word list, where edges connect words that differ by one letter. Insight 2: Backtracking (Depth-First Search, or DFS) is necessary to construct *all* the shortest paths once the graph structure and distances are known from the BFS.

Which companies ask LeetCode 126. Word Ladder II in interviews?

This problem is frequently asked at Citadel, LinkedIn, Lyft, TikTok, Yelp.

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Word Ladder II - LeetCode 126 Solution

Q: How to solve Word Ladder II?

The solution uses a combination of Breadth-First Search (BFS) and Depth-First Search (DFS) (backtracking). First, BFS is employed to build an adjacency list representing the graph of possible word transformations, along with the distance (number of transformations) from the `beginWord` to each reachable word. The BFS stops when the `endWord` is found. After BFS, DFS is used to explore the paths from `beginWord` to `endWord` using the adjacency list, constructing all shortest paths. DFS is constrained by only exploring neighbors that are exactly one step further than the current node, as determined by the distances calculated by BFS.

Word Ladder II - Complete Solution Guide

Word Ladder II is LeetCode problem 126, 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 all the shortest transformation sequences from beginWord to endWord , or an empty list if no

Detailed Explanation

The 'Word Ladder II' problem asks you to find all the shortest transformation sequences from a given `beginWord` to an `endWord`, using a provided `wordList` as a dictionary. Each word in the sequence must differ from the previous word by exactly one letter and must be present in the `wordList`. The `beginWord` does not necessarily need to be in the `wordList`. The output is a list of lists, where each inner list represents a shortest transformation sequence. If no such sequence exists, return an empty list.

Solution Approach

The solution uses a combination of Breadth-First Search (BFS) and Depth-First Search (DFS) (backtracking). First, BFS is employed to build an adjacency list representing the graph of possible word transformations, along with the distance (number of transformations) from the `beginWord` to each reachable word. The BFS stops when the `endWord` is found. After BFS, DFS is used to explore the paths from `beginWord` to `endWord` using the adjacency list, constructing all shortest paths. DFS is constrained by only exploring neighbors that are exactly one step further than the current node, as determined by the distances calculated by BFS.

Step-by-Step Algorithm

Step 1: **Initialization:** Create a set `wordSet` from the `wordList` for efficient word lookup. Check if `endWord` is present in `wordSet`. If not, return an empty list because no transformation is possible.
Step 2: **BFS (Breadth-First Search):** Perform a BFS starting from `beginWord`. Use a queue to store words to visit. Maintain a `distance` map to store the distance of each word from `beginWord`. Also, create an `adj` (adjacency list) map to represent the graph of transformations. The BFS explores words level by level.
Step 3: **BFS Exploration:** For each word in the queue, generate all possible one-letter variations. If a variation is present in `wordSet` and has not been visited yet (not in `distance`), add it to the queue, update its distance, and add it to the adjacency list of the current word. If the variation's distance is equal to the current word's distance plus one (meaning it is a shortest path neighbor), also add it to the adjacency list. Stop BFS if endWord is found.
Step 4: **DFS (Depth-First Search / Backtracking):** If the BFS found the `endWord`, start a DFS from `beginWord` to find all shortest paths to `endWord`. Use a `path` list to keep track of the current path being explored. Add the current word to the `path`.
Step 5: **DFS Exploration:** If the current word is the `endWord`, add a copy of the current `path` to the `results` list. Otherwise, for each neighbor of the current word in the `adj` list, recursively call DFS with the neighbor, adding it to the current path and removing it after the recursive call returns (backtracking).
Step 6: **Return:** Return the `results` list containing all shortest transformation sequences.

Key Insights

Insight 1: Breadth-First Search (BFS) is crucial for finding the *shortest* paths in an unweighted graph. The graph is implicitly defined by the word list, where edges connect words that differ by one letter.
Insight 2: Backtracking (Depth-First Search, or DFS) is necessary to construct *all* the shortest paths once the graph structure and distances are known from the BFS.
Insight 3: Efficiently pre-compute the adjacency list. Building it during the backtracking phase would be significantly slower.
Insight 4: Optimizing memory usage in C is critical due to manual memory management. Failing to free allocated memory can lead to memory leaks, and incorrect allocation can lead to crashes.
Insight 5: The C version uses explicit string duplication (strdup) to manage strings correctly and avoid issues when modifying them. This is crucial for correctness and memory safety.

Complexity Analysis

Time Complexity: O(V*26^L + N)

Space Complexity: O(V^2 + N)

Topics

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

Companies

Asked at: Citadel, LinkedIn, Lyft, TikTok, Yelp.

Word Ladder II | LeetCode 126

Hard

Hash Table String Backtracking Breadth-First Search

Time: O(V*26^L + N)Space: O(V^2 + N)

Solution Code

Available in:

import java.util.*;

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

        Map<String, List<String>> adj = new HashMap<>();
        Map<String, Integer> distance = new HashMap<>();
        Queue<String> q = new LinkedList<>();
        distance.put(beginWord, 0);
        q.offer(beginWord);

        boolean found = false;
        int word_len = beginWord.length();

        while (!q.isEmpty() && !found) {
            int level_size = q.size();
            for (int _i = 0; _i < level_size; _i++) {
                String word = q.poll();

                for (int i = 0; i < word_len; i++) {
                    for (char c = 'a'; c <= 'z'; c++) {
                        String next_word = word.substring(0, i) + c + word.substring(i + 1);

                        if (wordSet.contains(next_word)) {
                            if (!distance.containsKey(next_word)) {
                                distance.put(next_word, distance.get(word) + 1);
                                q.offer(next_word);
                                adj.computeIfAbsent(word, k -> new ArrayList<>()).add(next_word);
                                if (next_word.equals(endWord)) {
                                    found = true;
                                }
                            } else if (distance.get(next_word).equals(distance.get(word) + 1)) {
                                adj.computeIfAbsent(word, k -> new ArrayList<>()).add(next_word);
                            }
                        }
                    }
                }
            }
        }

        List<List<String>> results = new ArrayList<>();
        if (found) {
            List<String> path = new ArrayList<>();
            path.add(beginWord);

            function dfs = new function() {
                public void apply(String current_word) {
                    if (current_word.equals(endWord)) {
                        results.add(new ArrayList<>(path));
                        return;
                    }

                    if (adj.containsKey(current_word)) {
                        for (String neighbor : adj.get(current_word)) {
                            path.add(neighbor);
                            apply(neighbor);
                            path.remove(path.size() - 1);
                        }
                    }
                }
            };

            dfs.apply(beginWord);
        }

        return results;
    }

    interface function {
        void apply(String s);
    }
}

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Problem Info

Description

Problem 126: Word Ladder II - Detailed explanation and solution approaches.

Topics

Hash Table String Backtracking Breadth-First Search

Companies

CitadelLinkedInLyftTikTokYelp

Stats

Difficulty:Hard

Languages:4

Companies:5

Problem #:126

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

6 steps

1
Step 1: **Initialization:** Create a set `wordSet` from the `wordList` for efficient word lookup. Check if `endWord` is present in `wordSet`. If not, return an empty list because no transformation is possible.
2
Step 2: **BFS (Breadth-First Search):** Perform a BFS starting from `beginWord`. Use a queue to store words to visit. Maintain a `distance` map to store the distance of each word from `beginWord`. Also, create an `adj` (adjacency list) map to represent the graph of transformations. The BFS explores words level by level.
3
Step 3: **BFS Exploration:** For each word in the queue, generate all possible one-letter variations. If a variation is present in `wordSet` and has not been visited yet (not in `distance`), add it to the queue, update its distance, and add it to the adjacency list of the current word. If the variation's distance is equal to the current word's distance plus one (meaning it is a shortest path neighbor), also add it to the adjacency list. Stop BFS if endWord is found.
4
Step 4: **DFS (Depth-First Search / Backtracking):** If the BFS found the `endWord`, start a DFS from `beginWord` to find all shortest paths to `endWord`. Use a `path` list to keep track of the current path being explored. Add the current word to the `path`.
5
Step 5: **DFS Exploration:** If the current word is the `endWord`, add a copy of the current `path` to the `results` list. Otherwise, for each neighbor of the current word in the `adj` list, recursively call DFS with the neighbor, adding it to the current path and removing it after the recursive call returns (backtracking).
6
Step 6: **Return:** Return the `results` list containing all shortest transformation sequences.

💡

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(V*26^L + N)

💾

Space Complexity

O(V^2 + N)

📝

Why?

The BFS visits each valid word (at most V times). Generating neighbors takes O(26*L) for each word, giving O(V * 26 * L). The DFS constructs shortest paths. In the worst case, there could be a large number of shortest paths, resulting in a space complexity proportional to the total length of all paths, which is O(N), if N is the total length of all paths. Storing the adjacency list can take O(V^2) in the worst-case scenario where almost all words are neighbors.

📖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

Bidirectional BFS

Perform BFS simultaneously from both `beginWord` and `endWord`. This can significantly reduce the search space in some cases.

Time: O(V*26^L) (potentially faster in practice due to reduced search space)Space: O(V^2)

Trade-offs: Bidirectional BFS can be faster when the search space is large and the paths are relatively short. However, it's more complex to implement, especially when constructing the paths. Can also have worse performance if the branching factor is very low early on.

✨

Key Insights

1Insight 1: Breadth-First Search (BFS) is crucial for finding the *shortest* paths in an unweighted graph. The graph is implicitly defined by the word list, where edges connect words that differ by one letter.
2Insight 2: Backtracking (Depth-First Search, or DFS) is necessary to construct *all* the shortest paths once the graph structure and distances are known from the BFS.
3Insight 3: Efficiently pre-compute the adjacency list. Building it during the backtracking phase would be significantly slower.
4Insight 4: Optimizing memory usage in C is critical due to manual memory management. Failing to free allocated memory can lead to memory leaks, and incorrect allocation can lead to crashes.
5Insight 5: The C version uses explicit string duplication (strdup) to manage strings correctly and avoid issues when modifying them. This is crucial for correctness and memory safety.

📚

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` not present in `wordList`. The code explicitly checks for this case and returns an empty list.
Edge case 2: No transformation sequence exists. If the BFS doesn't find the `endWord`, the `found` flag remains false, and an empty list is returned.
Edge case 3: `beginWord` is equal to `endWord`. Although the problem states `beginWord != endWord`, a robust solution should handle this. The provided code doesn't explicitly address this, but the BFS would terminate immediately, returning an empty list because beginWord would not have any valid neighbors other than itself.
Edge case 4: `wordList` is empty. The code handles this correctly. The `wordSet` would be empty, and if the `endWord` is not in it (which it will not be), an empty list will be returned.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where `endWord` is not in `wordList`. This will lead to incorrect results or infinite loops.

✗Common mistake 2: Failing to consider only shortest paths during backtracking. If not restricted, DFS will find all possible paths, including longer ones, exceeding time limit.

✗Common mistake 3: Constructing the adjacency list inefficiently or during the backtracking step will lead to TLE (Time Limit Exceeded).

✗Common mistake 4: Forgetting to duplicate strings properly in C. String literals and other char arrays need to be copied (using strdup or similar) before being stored in data structures. Directly storing pointers to these can lead to modification issues and memory corruption.

✗Common mistake 5: Memory leaks in C. Failing to free dynamically allocated memory (strings, queues, hash sets, adjacency lists, etc.) will cause the program to run out of memory over time.

✅

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 checking. Suggesting alternative words to a misspelled word based on a dictionary and a small edit distance.

🏦Real-world scenario 2: Biological sequence analysis. Finding evolutionary pathways between different DNA or protein sequences based on point mutations or amino acid substitutions.

🎮Real-world scenario 3: Code optimization. Finding sequences of code transformations that minimize execution time or code size.

📱Real-world scenario 4: Social Network Analysis: Finding short chains of connections between users based on shared interests or friends.

💡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

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ 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 5, 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

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

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

@InterviewProJul 7

❓

Frequently Asked Questions

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Word Ladder II"?

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?