Word Ladder

class Solution {
public:
    int minTransforms(string startWord, string targetWord, vector<string>& dictionary) {
        int wordLength = startWord.length();
        unordered_map<string, vector<string>> transformations;
        for (string word : dictionary) {
            for (int i = 0; i < wordLength; i++) {
                string transformed =
                    word.substr(0, i) + '*' + word.substr(i + 1, wordLength);
                transformations[transformed].push_back(word);
            }
        }
        queue<pair<string, int>> processQueue;
        processQueue.push({startWord, 1});
        unordered_map<string, bool> seenWords;
        seenWords[startWord] = true;
        while (!processQueue.empty()) {
            pair<string, int> current = processQueue.front();
            processQueue.pop();
            string currWord = current.first;
            int steps = current.second;
            for (int i = 0; i < wordLength; i++) {
                string intermediate =
                    currWord.substr(0, i) + '*' + currWord.substr(i + 1, wordLength);
                for (string neighbor : transformations[intermediate]) {
                    if (neighbor == targetWord) {
                        return steps + 1;
                    }
                    if (!seenWords[neighbor]) {
                        seenWords[neighbor] = true;
                        processQueue.push({neighbor, steps + 1});
                    }
                }
            }
        }
        return 0;
    }
};

The solution provided solves the "Word Ladder" problem using C++. In this problem, the goal is to find the minimum number of transformations from a start word to a target word, where each transformation changes exactly one letter, and all intermediate words must exist in a given dictionary.

Solution Overview:

• First, the solution preprocesses the dictionary by creating an intermediate state for each word. This state is generated by replacing each letter of the word with a wildcard character '*'. These transformations are stored in a map where each transformed word points to all dictionary words it can represent.

• The main part of the solution uses a breadth-first search (BFS) strategy. It involves a queue where each element is a pair consisting of the current word and the number of transformation steps taken to reach it.

• The BFS starts from the start word with an initial step count of 1. It processes each word by generating all possible intermediate states. For each intermediate state, it finds all the connected words in the transformation map that haven't been visited yet, marks them as visited, and then adds them to the queue with incremented step count.

• When the BFS reaches the target word, it returns the number of transformation steps as the result. If the queue is exhausted and the target word is not found, the function returns 0, indicating that no valid transformation path exists.

Key Points:

• A BFS is used to ensure the shortest path is found, since it explores all the nearest possibilities before moving on to more distant ones.

• An unordered_map functions as a set to keep track of visited words, avoiding revisits and endless loops.

• The solution effectively reduces the problem space by using transformation patterns, allowing quick access to all potential next words.

By employing these techniques, the provided C++ code efficiently addresses the complexities of the Word Ladder problem, offering a solution that minimizes the transformation steps from the start word to the target word within the constraints of the allowed dictionary entries.

class Solution {
    public int shortestTransformation(
        String startWord,
        String targetWord,
        List<String> dictionary
    ) {
        int wordLength = startWord.length();

        Map<String, List<String>> comboDictionary = new HashMap<>();

        dictionary.forEach(currentWord -> {
            for (int i = 0; i < wordLength; i++) {
                String transformedWord =
                    currentWord.substring(0, i) + '*' + currentWord.substring(i + 1, wordLength);
                List<String> relatedWords = comboDictionary.getOrDefault(
                    transformedWord,
                    new ArrayList<>()
                );
                relatedWords.add(currentWord);
                comboDictionary.put(transformedWord, relatedWords);
            }
        });

        Queue<Pair<String, Integer>> queue = new LinkedList<>();
        queue.add(new Pair(startWord, 1));

        Map<String, Boolean> seen = new HashMap<>();
        seen.put(startWord, true);

        while (!queue.isEmpty()) {
            Pair<String, Integer> currentNode = queue.remove();
            String word = currentNode.getKey();
            int currentLevel = currentNode.getValue();
            for (int i = 0; i < wordLength; i++) {
                String intermediateWord =
                    word.substring(0, i) + '*' + word.substring(i + 1, wordLength);

                for (String adjacentWord : comboDictionary.getOrDefault(
                    intermediateWord,
                    new ArrayList<>()
                )) {
                    if (adjacentWord.equals(targetWord)) {
                        return currentLevel + 1;
                    }
                    if (!seen.containsKey(adjacentWord)) {
                        seen.put(adjacentWord, true);
                        queue.add(new Pair(adjacentWord, currentLevel + 1));
                    }
                }
            }
        }

        return 0;
    }
}

The provided Java solution tackles the problem of finding the shortest transformation sequence length from a startWord to a targetWord using a provided dictionary of valid intermediate words. This solution employs a breadth-first search (BFS) strategy, making it efficient by exploring the shortest paths first.

Understand how the solution operates:

1. Create a comboDictionary to group words that have the same characters, except one substituted by the placeholder '*'. This grouping aids in connecting potentially transformable words efficiently.

2. Start the BFS by adding the startWord to the queue with an initial distance of 1.

3. Utilize a seen map to track already visited words and avoid processing them more than once. This optimization reduces the amount of redundant processing.

4. For each word processed in the BFS, transform it by substituting each of its characters with '*', sequentially, and if this transformed word exists in comboDictionary, attempt to discover new words that can continue the transformation chain.

5. If the transformation directly reaches the targetWord, return the current transformation length plus one as the answer.

6. Continue until the queue is drained and check if the targetWord has been reached. If not reachable, return 0.

By intelligent grouping of words and systematic exploration via BFS, the solution ensures efficient performance with a clear method to trace the transformation path. This helps in finding the shortest path in potentially large word lists, ensuring feasibility within practical execution time limits.

// Structure for holding strings in a linked list
typedef struct StrList {
    char *data;
    struct StrList *next;
} StrList;

// Structure defining a hash table for word transformations
typedef struct HashTbl {
    char *keyword;
    StrList *transformations;
    UT_hash_handle hh;
} HashTbl;

// Node structure for BFS exploration
typedef struct BFSNode {
    char *word;
    int depth;
} BFSNode;

// Element of a queue
typedef struct QueueElement {
    BFSNode *bfsNode;
    struct QueueElement *next;
} QueueElement;

// Structure to represent a queue
typedef struct Queue {
    QueueElement *head;
    QueueElement *tail;
    int count;
} Queue;

// Function declarations
Queue *initializeQueue();
void pushQueue(Queue *queue, BFSNode *bfsNode);
BFSNode *popQueue(Queue *queue);
int isQueueEmpty(Queue *queue);
void insertTransformation(HashTbl **hashTable, char *key, char *transformation);
StrList *getTransformations(HashTbl *hashTable, char *key);

// Creating a new empty queue
Queue *initializeQueue() {
    Queue *queue = (Queue *)malloc(sizeof(Queue));
    queue->head = queue->tail = NULL;
    queue->count = 0;
    return queue;
}

// Enqueuing a BFS node to the queue
void pushQueue(Queue *queue, BFSNode *bfsNode) {
    QueueElement *newElement = (QueueElement *)malloc(sizeof(QueueElement));
    newElement->bfsNode = bfsNode;
    newElement->next = NULL;
    if (queue->tail == NULL) {
        queue->head = queue->tail = newElement;
    } else {
        queue->tail->next = newElement;
        queue->tail = newElement;
    }
    queue->count++;
}

// Dequeue a BFS node from the queue
BFSNode *popQueue(Queue *queue) {
    if (queue->head == NULL) return NULL;
    QueueElement *tmp = queue->head;
    BFSNode *node = tmp->bfsNode;
    queue->head = queue->head->next;
    if (queue->head == NULL) queue->tail = NULL;
    free(tmp);
    queue->count--;
    return node;
}

// Check if the queue is empty
int isQueueEmpty(Queue *queue) {
    return queue->head == NULL;
}

// Add a word transformation to the hash table
void insertTransformation(HashTbl **hashTable, char *key, char *transformation) {
    HashTbl *entry;
    HASH_FIND_STR(*hashTable, key, entry);
    if (entry == NULL) {
        entry = (HashTbl *)malloc(sizeof(HashTbl));
        entry->keyword = strdup(key);
        entry->transformations = NULL;
        HASH_ADD_KEYPTR(hh, *hashTable, entry->keyword, strlen(key), entry);
    }
    StrList *newEntry = (StrList *)malloc(sizeof(StrList));
    newEntry->data = strdup(transformation);
    newEntry->next = entry->transformations;
    entry->transformations = newEntry;
}

// Retrieve a list of transformations from the hash table
StrList *getTransformations(HashTbl *hashTable, char *key) {
    HashTbl *entry;
    HASH_FIND_STR(hashTable, key, entry);
    return (entry == NULL) ? NULL : entry->transformations;
}

The provided C code primarily deals with constructing and manipulating data structures for a Word Ladder problem using Breadth-First Search (BFS). Word Ladder is a transformation puzzle where you connect two words by changing one letter at a time, with each intermediate word being a valid word itself.

• Structures used:

  • StrList: A linked list for storing strings.
  • HashTbl: A hash table for mapping a word to its potential transformations.
  • BFSNode: Represents a node to be processed in BFS, tracking the word and its transformation depth.
  • QueueElement and Queue: Structures to mimic queue operations crucial for BFS processing.

• Key Functions:

  1. initializeQueue(): Creates and initializes an empty queue. This is critical to managing the BFS frontier nodes in an orderly fashion.
  2. pushQueue(Queue *queue, BFSNode *bfsNode): Adds a new node to the queue, necessary for BFS to explore each transformation level by level.
  3. popQueue(Queue *queue): Removes and retrieves the front node from the queue, signifying the exploration of the next node in BFS.
  4. isQueueEmpty(Queue *queue): Checks if there are more nodes to explore in BFS by examining whether the queue is empty.

• Hash Table Manipulation:

  1. insertTransformation(HashTbl **hashTable, char *key, char *transformation): Inserts transformations (words that can be achieved by changing one letter) of a given word into the hash table. This is essential for quick lookup of next possible transformations in the BFS process.
  2. getTransformations(HashTbl *hashTable, char *key): Retrieves a list of transformations for a particular word, facilitating the word-ladder path discovery.

Harnessing these components, the BFS approach enables sequentially exploring all potential transformations of a word, assessing the shortest path or ladder between the start word and the end word by incrementally exploring reachable transformations. Each node (word) processed must check and queue unexplored transformations, thus efficiently utilizing the queue and hash table to manage state and transitions in the word ladder exploration.

var shortestTransformation = function (startWord, targetWord, dictionary) {
    let wordLength = startWord.length;
    let comboDict = {};
    dictionary.forEach(word => {
        for (let i = 0; i < wordLength; i++) {
            let transformedWord = word.slice(0, i) + '*' + word.slice(i + 1);
            if (!comboDict[transformedWord]) comboDict[transformedWord] = [];
            comboDict[transformedWord].push(word);
        }
    });

    let queue = [[startWord, 1]];
    let visited = { [startWord]: true };

    while (queue.length) {
        let [currentWord, steps] = queue.shift();
        for (let i = 0; i < wordLength; i++) {
            let tempWord = currentWord.slice(0, i) + '*' + currentWord.slice(i + 1);
            for (let relatedWord of comboDict[tempWord] || []) {
                if (relatedWord === targetWord) {
                    return steps + 1;
                }
                if (!visited[relatedWord]) {
                    visited[relatedWord] = true;
                    queue.push([relatedWord, steps + 1]);
                }
            }
        }
    }
    return 0;
};

The Word Ladder problem aims to determine the shortest transformation sequence from a start word to a target word using a given dictionary of words. The transformation must occur in such a way that only one letter can be changed at a time and all intermediate words resulting from the change must be valid according to the dictionary.

Implement a function shortestTransformation that takes three parameters: startWord, targetWord, and dictionary. This approach utilizes breadth-first search (BFS) to explore the shortest path from the start word to the target word. Here's a breakdown of the solution:

• Prepare a Dictionary for Transformation Possibilities:

  • Create a comboDict to store all potential generic words and the words in the dictionary that conform to the generic form. A generic word is formed by replacing each letter of the word with a star (*).
  • For each word in the dictionary, and for each character in the word, replace the character with * and add the original word to the list of words in comboDict that correspond to the formed generic word.

• Implement the BFS Algorithm:

  • Initialize the queue with the startWord and the initial number of steps (1). Also, create a visited hash map to track visited words.
  • Process each item in the queue by examining its current word and current steps. For each character in the current word, replace it with * to form a temporary word.
  • Access the transformations available in comboDict corresponding to the temporary word. For each transformation:
    • If it is the target word, return the current step count incremented by one (indicating the transformation sequence is complete).
    • If the transformation has not been visited, mark it as visited, and enqueue it with the incremented step count.
  • If no transformations are found, return 0, indicating no valid transformation is possible.

This approach efficiently maps out all potential single-step transformations using a dictionary structure and then strategically explores them using BFS, ensuring that the path with the minimum steps required to transform the start word into the target word is found. The use of the visited dictionary prevents revisiting the same words, therefore, optimizing the search process and improving performance.

from collections import defaultdict, deque

class WordLadderSolver(object):
    def shortestTransformationLength(self, startWord: str, targetWord: str, wordsList: List[str]) -> int:

        if targetWord not in wordsList or not targetWord or not startWord or not wordsList:
            return 0
    
        wordLength = len(startWord)

        combosByTransformation = defaultdict(list)
        for word in wordsList:
            for index in range(wordLength):
                pattern = word[:index] + "*" + word[index + 1 :]
                combosByTransformation[pattern].append(word)

        queue = deque([(startWord, 1)])
        checkedWords = {startWord: True}
        while queue:
            current, distance = queue.popleft()
            for idx in range(wordLength):
                intermediate = current[:idx] + "*" + current[idx + 1:]

                for neighbour in combosByTransformation[intermediate]:
                    if neighbour == targetWord:
                        return distance + 1
                    if neighbour not in checkedWords:
                        checkedWords[neighbour] = True
                        queue.append((neighbour, distance + 1))
                combosByTransformation[intermediate] = []
        return 0

The provided Python script implements a solution for the problem of finding the shortest transformation length from a start word to a target word by changing one letter at a time, using an allowed list of intermediate words. The script employs the Breadth-First Search (BFS) algorithm, which is appropriate for finding the shortest path in unweighted graphs.

Here's a breakdown of how the solution works:

1. Initial conditions are checked where if the target word is not in the list or either start word or list is empty, it would immediately return 0.

2. The word length of the start word is determined for use in generating patterns—a sequence of words with one character replaced by '*'.

3. A dictionary (combosByTransformation) maps these generalized word patterns to all words in the list that match the pattern. For example, for the word "log", the patterns "log", "lg", and "og" are created and the dictionary keeps track of all words that can fit these patterns.

4. A queue is initialized with a tuple containing the start word and initial distance of 1. A dictionary checkedWords is used to keep track of words that have been processed to avoid redundant work.

5. The BFS begins by dequeuing an element, generating all possible patterns of the current word using '*' replacements at various positions, and exploring all neighboring words (that match the pattern) in the word list.

6. For each neighbor:

   • If the neighbor is the target word, the function returns the current distance incremented by 1, which signifies the transformation length.
   • If the neighbor has not been checked before, it is marked as checked, and added to the queue with incremented distance value for further exploration.

7. After exploring all possibilities, if the target word is not found, the function returns 0, meaning the transformation is not possible with the given list.

This method ensures efficient traversal and checking using BFS while leveraging the pattern-based grouping of words to drastically reduce the number of comparisons needed, especially useful with a large list of words. Additionally, by clearing out the dictionary entries for combosByTransformation once visited, memory usage is optimized throughout the execution of the BFS.