127. Word Ladder I

• Link: https://leetcode.com/problems/word-ladder/description/

Question

A transformation sequence from word beginWord to word endWord using a dictionary wordList is a sequence of words beginWord -> s1 -> s2 -> ... -> sk such that:

• Every adjacent pair of words differs by a single letter.
• Every si for 1 <= i <= k is in wordList. Note that beginWord does not need to be in wordList.
• sk == endWord

Given two words, beginWord and endWord, and a dictionary wordList, return the number of words in the shortest transformation sequence from beginWord to endWord, or 0 if no such sequence exists.

Example 1:

Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"]
Output: 5
Explanation: One shortest transformation sequence is "hit" -> "hot" -> "dot" -> "dog" -> cog", which is 5 words long.

Example 2:

Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log"]
Output: 0
Explanation: The endWord "cog" is not in wordList, therefore there is no valid transformation sequence.

Constraints:

• 1 <= beginWord.length <= 10
• endWord.length == beginWord.length
• 1 <= wordList.length <= 5000
• wordList[i].length == beginWord.length
• beginWord, endWord, and wordList[i] consist of lowercase English letters.
• beginWord != endWord
• All the words in wordList are unique.

Approach 1: Optimal

Intuition

• BFS
• Transform each char one at a time and check if present in the wordlist

Algorithm

Code

class Solution {
public:
    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
        queue<pair<string, int>> q;
        q.push({beginWord, 1});

        unordered_set<string> s(wordList.begin(), wordList.end());

        s.erase(beginWord);

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

            if (word == endWord) return steps;

            for (int i = 0 ; i < word.size() ; i++) {
                char originalChar = word[i];
                for (char c = 'a' ; c <= 'z' ; c++) {
                    word[i] = c;
                    if (s.find(word) != s.end()) {
                        q.push({word, steps+1});
                        s.erase(word);
                    }
                }
                word[i] = originalChar;
            }
        }
        return 0;
    }
};

Complexity Analysis

• Time Complexity: $O(26\ast n\ast m)$
• Space Complexity: $O(n)$
  • Where
    • n = size of wordList Array and
    • m = word length of words present in the wordList..