Word Ladder II - 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, which is well-suited for finding the shortest path in an unweighted graph. Here, each word represents a node, and an edge exists between two nodes if they differ by exactly one letter.

Naive Solution

A naive solution would involve generating all possible transformations for each word and checking if they exist in the word list. This approach is not optimal due to its high time complexity.

Optimized Solution

An optimized solution uses BFS to explore all possible transformations level by level. This ensures that the first time we reach the endWord, we have found the shortest transformation sequence.

Algorithm

1. Add the beginWord to a queue and start BFS.

2. For each word in the queue, generate all possible one-letter transformations.

3. If a transformation matches the endWord, return the current level + 1.

4. If a transformation exists in the word list, add it to the queue and remove it from the word list to avoid revisiting.

5. Continue until the queue is empty.

Code Implementation

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

using namespace std;

// Function to check if two words differ by exactly one character
bool isOneLetterDifferent(const string &word1, const string &word2) {
    int count = 0;
    for (int i = 0; i < word1.size(); ++i) {
        if (word1[i] != word2[i]) {
            count++;
            if (count > 1) return false;
        }
    }
    return count == 1;
}

// 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;

    queue<pair<string, int>> toVisit;
    toVisit.push({beginWord, 1});

    while (!toVisit.empty()) {
        string word = toVisit.front().first;
        int length = toVisit.front().second;
        toVisit.pop();

        for (int i = 0; i < word.size(); ++i) {
            string temp = word;
            for (char c = 'a'; c <= 'z'; ++c) {
                temp[i] = c;
                if (temp == endWord) return length + 1;
                if (wordSet.find(temp) != wordSet.end()) {
                    toVisit.push({temp, length + 1});
                    wordSet.erase(temp);
                }
            }
        }
    }
    return 0;
}

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^2), where N is the number of words in the word list and M is the length of each word. This is because for each word, we generate M possible transformations, and checking each transformation against the word list takes O(M) time.

The space complexity is O(N * M) due to the storage of the word list and the queue used for BFS.

Edge Cases

Potential edge cases include:

The endWord not being in the word list.
All words being the same length but no possible transformations.
Empty word list.

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

Testing

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

Simple transformations with a small word list.
Large word lists with multiple possible paths.
Edge cases such as the endWord not being in the word list.

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.
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 is crucial for developing efficient algorithms for word transformation tasks. By using BFS, we can ensure an optimal solution. Practice and exploration of similar problems will further enhance problem-solving abilities.

Additional Resources

For further reading and practice, consider the following resources: