Word Ladder IV - C++ Solution and Time Complexity Analysis

Input:
beginWord = "hit",
endWord = "cog",
wordList = ["hot","dot","dog","lot","log","cog"]

Output: 5

Explanation: As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length 5.

Input:
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]

Output: 0

Explanation: The endWord "cog" is not in wordList, therefore no possible transformation.

Understanding the Problem

The core challenge of this problem is to find the shortest path from the beginWord to the endWord by changing only one letter at a time, with each intermediate word being a valid word in the given word list. This problem is significant in various applications such as spell checkers, word games, and natural language processing tasks.

Potential pitfalls include not considering all possible transformations or not efficiently searching through the word list, leading to suboptimal solutions.

Approach

To solve this problem, we can use the Breadth-First Search (BFS) algorithm. BFS is ideal for finding the shortest path in an unweighted graph, which aligns with our need to find the shortest transformation sequence.

Here’s a step-by-step approach:

1. First, check if the endWord is in the word list. If not, return 0.
2. Use a queue to perform BFS. Initialize the queue with the beginWord and a depth of 1.
3. Use a set to keep track of words that have been visited to avoid cycles.
4. For each word in the queue, generate all possible transformations by changing one letter at a time.
5. If a transformation matches the endWord, return the current depth + 1.
6. If a transformation is in the word list and hasn’t been visited, add it to the queue and mark it as visited.
7. Continue the process until the queue is empty.

Algorithm

Here’s a detailed breakdown of the BFS algorithm:

1. Initialize a queue with the beginWord and a depth of 1.
2. Initialize a set for the word list and a set for visited words.
3. While the queue is not empty:
   1. Dequeue the front element (current word and depth).
   2. For each character in the current word, try changing it to every letter from 'a' to 'z'.
   3. If the new word matches the endWord, return the current depth + 1.
   4. If the new word is in the word list and hasn’t been visited, enqueue it with depth + 1 and mark it as visited.
4. If the queue is empty and no transformation sequence is found, return 0.

Code Implementation

#include <iostream>
#include <unordered_set>
#include <queue>
#include <string>
#include <vector>

using namespace std;

// Function to find the length of the shortest transformation sequence
int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
    unordered_set<string> wordSet(wordList.begin(), wordList.end());
    if (wordSet.find(endWord) == wordSet.end()) {
        return 0; // endWord is not in the word list
    }

    queue<pair<string, int>> q;
    q.push({beginWord, 1});
    unordered_set<string> visited;
    visited.insert(beginWord);

    while (!q.empty()) {
        auto [currentWord, depth] = q.front();
        q.pop();

        for (int i = 0; i < currentWord.size(); ++i) {
            string tempWord = currentWord;
            for (char c = 'a'; c <= 'z'; ++c) {
                tempWord[i] = c;
                if (tempWord == endWord) {
                    return depth + 1; // Found the endWord
                }
                if (wordSet.find(tempWord) != wordSet.end() && visited.find(tempWord) == visited.end()) {
                    q.push({tempWord, depth + 1});
                    visited.insert(tempWord);
                }
            }
        }
    }

    return 0; // No transformation sequence found
}

int main() {
    string beginWord = "hit";
    string endWord = "cog";
    vector<string> wordList = {"hot", "dot", "dog", "lot", "log", "cog"};
    cout << "Length of shortest transformation sequence: " << ladderLength(beginWord, endWord, wordList) << endl;
    return 0;
}

Complexity Analysis

The time complexity of the BFS approach is O(N * M * 26), where N is the number of words in the word list, M is the length of each word, and 26 represents the number of possible transformations for each character. The space complexity is O(N) due to the storage of the word list and visited set.

Edge Cases

Potential edge cases include:

• The endWord is not in the word list.
• The beginWord is the same as the endWord.
• All words are of length 1.

Each of these cases is handled by the algorithm, ensuring robustness.

Testing

To test the solution comprehensively, consider the following test cases:

• Basic cases with small word lists.
• Cases where no transformation is possible.
• Cases with multiple possible transformation sequences.

Using a testing framework like Google Test can help automate and validate these test cases.

Thinking and Problem-Solving Tips

When approaching such problems, consider the following tips:

• Break down the problem into smaller, manageable parts.
• Use diagrams or pseudo-code to visualize the solution.
• Practice similar problems to build familiarity with common patterns and algorithms.

Conclusion

Understanding and solving the Word Ladder problem helps improve problem-solving skills and familiarity with graph traversal algorithms like BFS. Practice and exploration of similar problems can further enhance these skills.

Additional Resources

For further reading and practice, consider the following resources:

• LeetCode - Word Ladder
• GeeksforGeeks - Word Ladder
• Coursera - Algorithms Part I