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

Given two words (beginWord and endWord), and a dictionary's word list, find the length of shortest transformation sequence from beginWord to endWord, such that:

Only one letter can be changed at a time.
Each transformed word must exist in the word list.

Note:

Return 0 if there is no such transformation sequence.
All words have the same length.
All words contain only lowercase alphabetic characters.
You may assume no duplicates in the word list.
You may assume beginWord and endWord are non-empty and are not the same.

Example 1:

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.

Example 2:

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 existing 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:

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

Algorithm

Here is a detailed breakdown of the BFS algorithm:

Initialize a queue with the beginWord and a depth of 1.
Initialize a set for the word list and a set for visited words.
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.
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}); // Start with the beginWord and depth 1

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

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

    return 0; // No transformation sequence found
}

int main() {
    string beginWord = "hit";
    string endWord = "cog";
    vector<string> wordList = {"hot", "dot", "dog", "lot", "log", "cog"};

    int result = ladderLength(beginWord, endWord, wordList);
    cout << "Length of shortest transformation sequence: " << result << 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) for storing the word list and the queue.

Edge Cases

Potential edge cases include:

The endWord is not in the word list.
The beginWord and endWord are the same (though the problem states they are not).
All words in the word list are the same.

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 a small word list.
Cases where the endWord is not in the word list.
Cases with larger word lists to test performance.

Using a testing framework like Google Test can help automate and manage these tests effectively.

Thinking and Problem-Solving Tips

When approaching such problems, consider the following tips:

Break down the problem into smaller, manageable parts.
Think about the most efficient data structures to use.
Consider edge cases and how to handle them.
Practice similar problems to improve problem-solving skills.

Conclusion

Understanding and solving the Word Ladder problem helps improve skills in graph traversal algorithms, particularly BFS. It also enhances problem-solving abilities by requiring efficient handling of transformations and edge cases.

Practice and exploration of similar problems can further solidify these concepts and techniques.

Additional Resources

For further reading and practice, consider the following resources: