127. Word Ladder

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127. Word Ladder

Description

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:

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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:

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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 .

Hints/Notes

  • 2024/12/28
  • BFS
  • No solution from 0x3F

Solution

Language: C++

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class Solution {
public:
    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
        unordered_set<string> words(wordList.begin(), wordList.end());
        if (!words.contains(endWord)) {
            return 0;
        }
        queue<string> q;
        unordered_set<string> visited;
        q.push(beginWord);
        int step = 1;
        while (!q.empty()) {
            int size = q.size();
            for (int i = 0; i < size; i++) {
                string cur = q.front();
                q.pop();
                for (int j = 0; j < cur.size(); j++) {
                    char tmp = cur[j];
                    for (int k = 0; k < 26; k++) {
                        char tmp = cur[j];
                        cur[j] = 'a' + k;
                        if (endWord == cur) {
                            return step + 1;
                        }
                        if (words.contains(cur) && !visited.contains(cur)) {
                            visited.insert(cur);
                            q.push(cur);
                        }
                    }
                    cur[j] = tmp;
                }
            }
            step++;
        }
        return 0;

    }
};