Leetcode 127. Word Ladder

This problem is asking us to find the shortest path between two nodes in an unweighted graph. Each word represents a node, and there's an edge between two words if they differ by exactly one character.

The transformation sequence represents a path through this graph, where we can only step on nodes (words) that exist in our wordList. We need to find the shortest such path from beginWord to endWord.

The constraint that each consecutive pair differs by exactly one letter means we're working with a very specific graph structure - each word can connect to at most 26 * word_length other words (changing each position to each possible letter).

The most straightforward approach would be to use DFS with backtracking to explore all possible transformation paths from beginWord to endWord. We would recursively try changing each character of the current word to every possible letter (a-z), check if the resulting word exists in wordList, and continue the search.

This approach would explore every possible path in the transformation graph, potentially visiting the same word multiple times through different paths. We'd need to track visited words in each path to avoid cycles, but we couldn't globally mark words as visited since they might be part of different valid paths.

The major inefficiency comes from exploring longer paths even after we've found shorter ones, and from the exponential nature of trying all possible character combinations at each step without any optimization for finding the shortest path.

def ladderLength(beginWord, endWord, wordList):
    if endWord not in wordList:
        return 0
    
    def can_transform(word1, word2):
        if len(word1) != len(word2):
            return False
        diff = 0
        for i in range(len(word1)):
            if word1[i] != word2[i]:
                diff += 1
                if diff > 1:
                    return False
        return diff == 1
    
    def dfs(current, target, words, visited, depth):
        if current == target:
            return depth
        
        min_length = float('inf')
        for i in range(len(words)):
            if not visited[i] and can_transform(current, words[i]):
                visited[i] = True
                result = dfs(words[i], target, words, visited, depth + 1)
                if result != 0:
                    min_length = min(min_length, result)
                visited[i] = False
        
        return 0 if min_length == float('inf') else min_length
    
    visited = [False] * len(wordList)
    return dfs(beginWord, endWord, wordList, visited, 1)

The breakthrough insight is recognizing this as a shortest path problem in an unweighted graph, where BFS guarantees optimality. The graph doesn't need to be explicitly built - we can generate valid neighbors dynamically by trying all single-character changes and checking against our word set.

After seeing the brute force limitations of exploring all paths, BFS emerges as the perfect algorithm because it guarantees finding the shortest path in unweighted graphs.

BFS explores nodes level by level - all nodes at distance 1, then distance 2, then distance 3, and so on. This means the first time we encounter our target (endWord), we've found it via the shortest possible path.

The graph structure is implicit but well-defined: we can generate neighbors on-the-fly by trying all single-character modifications. Using a set for wordList gives us O(1) neighbor validation, making the graph traversal efficient.

BFS also naturally handles the constraint that intermediate words must exist in wordList - we simply don't add invalid words to our queue.

Now that we understand why BFS is perfect for finding shortest paths, here's how we implement it effectively. We treat this as an unweighted graph problem where each word is a node, and edges exist between words that differ by exactly one character.

We start BFS from beginWord, exploring all words reachable in one transformation, then all words reachable in two transformations, and so on. The key optimization is using a set for wordList to achieve O(1) lookup time when checking if a transformed word exists.

For each word in our current BFS level, we systematically try changing each character position to every letter a-z. If the resulting word exists in our word set and hasn't been visited, we add it to the next BFS level. We mark words as visited immediately to prevent revisiting them in longer paths.

The moment we encounter endWord, we can return the current path length since BFS guarantees this is the shortest possible transformation sequence.

from collections import deque
def ladderLength(beginWord, endWord, wordList):
    if endWord not in wordList:
        return 0
    
    word_set = set(wordList)
    queue = deque([(beginWord, 1)])
    visited = set([beginWord])
    
    while queue:
        word, length = queue.popleft()
        
        if word == endWord:
            return length
        
        for i in range(len(word)):
            for c in 'abcdefghijklmnopqrstuvwxyz':
                if c == word[i]:
                    continue
                
                new_word = word[:i] + c + word[i+1:]
                
                if new_word in word_set and new_word not in visited:
                    visited.add(new_word)
                    queue.append((new_word, length + 1))
    
    return 0

• BFS for shortest path: When a problem asks for the shortest/minimum number of steps between two states, think BFS - it explores all possibilities level by level, guaranteeing the first time you reach the target is the shortest path.
• Graph modeling technique: Transform word transformation into a graph problem where each word is a node and edges connect words differing by exactly one character - this converts a string manipulation problem into a well-understood graph traversal.
• Set-based neighbor generation: Instead of checking every word in the dictionary, generate all possible one-character variations of the current word and check if they exist in a HashSet - this reduces time complexity from O(N*M) to O(M²) where M is word length.
• Early termination optimization: Always check if the endWord exists in wordList before starting BFS - this simple validation prevents unnecessary computation and handles a common edge case that would otherwise require full traversal to detect.
• Bidirectional BFS pattern: For large search spaces, implement bidirectional BFS starting from both ends simultaneously - this reduces time complexity from O(b^d) to O(b^(d/2)) where b is branching factor and d is depth.
• State space exploration: This pattern applies to any problem involving minimum steps between states - DNA mutation, puzzle solving, or configuration changes - where each state has a limited set of valid transitions to neighboring states.