LeetCode 127: Word Ladder Solution

Master LeetCode problem 127 (Word Ladder), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

127. Word Ladder

Hard

Hash Table String Breadth-First Search

Problem Explanation

Explanation

To solve this problem, we can use a breadth-first search (BFS) algorithm. We start by creating a graph where each word is a node, and there is an edge between two words if they differ by exactly one character. We then perform a BFS starting from the beginWord, visiting each word in the wordList to find the shortest transformation sequence to reach the endWord.

Algorithmic Idea:
1. Create a set wordSet from the wordList for faster lookup.
2. Initialize a queue q and add beginWord to the queue along with a level of 1.
3. While the queue is not empty:
  - Pop a word from the queue.
  - Generate all possible next words by changing one character at a time and check if it exists in wordSet.
  - If the generated word is the endWord, return the level.
  - Otherwise, add the generated word to the queue and remove it from wordSet.
4. If the queue becomes empty, return 0 as no transformation sequence exists.
Time Complexity:
- Building the graph: O(N * M^2), where N is the number of words in the wordList and M is the length of each word.
- BFS traversal: O(N * M), where N is the number of words in the wordList and M is the length of each word. Overall, the time complexity is O(N * M^2) + O(N * M) = O(N * M^2).
Space Complexity:
- Additional space for the graph: O(N * M^2)
- Queue and set: O(N * M) Overall, the space complexity is O(N * M^2).

Solution Code

import java.util.*;

class Solution {
    public int ladderLength(String beginWord, String endWord, List<String> wordList) {
        Set<String> wordSet = new HashSet<>(wordList);
        Queue<String> queue = new LinkedList<>();
        queue.offer(beginWord);
        int level = 1;

        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                String currentWord = queue.poll();
                if (currentWord.equals(endWord)) {
                    return level;
                }

                for (int j = 0; j < currentWord.length(); j++) {
                    char[] charArray = currentWord.toCharArray();
                    for (char ch = 'a'; ch <= 'z'; ch++) {
                        charArray[j] = ch;
                        String newWord = new String(charArray);
                        if (wordSet.contains(newWord) && !newWord.equals(currentWord)) {
                            queue.offer(newWord);
                            wordSet.remove(newWord);
                        }
                    }
                }
            }
            level++;
        }

        return 0;
    }
}

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