LeetCode //C - 127. Word Ladder

127. Word Ladder

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:

Example 2:

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.

From: LeetCode
Link: 127. Word Ladder

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Solution:

Ideas：

1. Start with beginWord “hit” and level 1.

• Queue = [“hit”]

1. Dequeue “hit” and explore its neighbors:

• Check each word in wordList to find neighbors of “hit” (words that are one letter different and unvisited).
• “hot” is a neighbor. Mark “hot” as visited and enqueue it.
• Queue = [“hot”]
• Increase level to 2.

1. Dequeue “hot” and explore its neighbors:

• Find neighbors of “hot” (unvisited words in wordList that are one letter different).
• “dot” and “lot” are neighbors. Mark them as visited and enqueue them.
• Queue = [“dot”, “lot”]
• Increase level to 3.

1. Dequeue “dot” and explore its neighbors:

• Find neighbors of “dot”.
• “dog” is a neighbor. Mark “dog” as visited and enqueue it.
• Queue = [“lot”, “dog”]

1. Dequeue “lot” and explore its neighbors:

• Find neighbors of “lot”.
• “log” is a neighbor. Mark “log” as visited and enqueue it.
• Queue = [“dog”, “log”]
• Increase level to 4.

1. Dequeue “dog” and explore its neighbors:

• Find neighbors of “dog”.
• “cog” is a neighbor. Mark “cog” as visited and enqueue it.
• Queue = [“log”, “cog”]

1. Dequeue “log” and explore its neighbors:

• Find neighbors of “log”.
• No unvisited neighbors.
• Queue = [“cog”]

1. Dequeue “cog” and explore its neighbors:

• “cog” is the endWord.
• Return level which is 5, representing the shortest transformation sequence: “hit” -> “hot” -> “dot” -> “dog” -> “cog”.

Code:

bool isOneLetterDiff(char *a, char *b) {
    int diffCount = 0;
    while (*a) {
        if (*(a++) != *(b++)) diffCount++;
        if (diffCount > 1) return false;
    }
    return diffCount == 1;
}

int ladderLength(char *beginWord, char *endWord, char **wordList, int wordListSize) {
    if (!beginWord || !endWord || !wordList || wordListSize == 0) return 0;
    
    bool *visited = (bool *)calloc(wordListSize, sizeof(bool));
    char **queue = (char **)malloc(sizeof(char *) * (wordListSize + 1));
    int front = 0, rear = 0;
    
    queue[rear++] = beginWord;
    int level = 1;
    
    while (front < rear) {
        int size = rear - front;
        for (int i = 0; i < size; i++) {
            char *word = queue[front++];
            if (strcmp(word, endWord) == 0) {
                free(visited);
                free(queue);
                return level;
            }
            for (int j = 0; j < wordListSize; j++) {
                if (!visited[j] && isOneLetterDiff(word, wordList[j])) {
                    visited[j] = true;
                    queue[rear++] = wordList[j];
                }
            }
        }
        level++;
    }
    
    free(visited);
    free(queue);
    return 0;
}