Understanding the Word Ladder Problem

The Word Ladder problem is a classic puzzle frequently posed in coding interviews. It requires transforming one word into another by altering a single character at each step and ensuring every intermediate word exists in a given dictionary (word list).

Problem Statement

You are given:

1. A beginWord.
2. An endWord.
3. A list of valid words called wordList.

You need to find the shortest transformation sequence from beginWord to endWord, where each transformation:

• Changes exactly one character in the current word.
• Produces a new word that must exist in the wordList.

If no valid transformation sequence exists, you must return 0.

Formally:

• A transformation sequence from beginWord to endWord is:

  beginWord -> s1 -> s2 -> ... -> sk  (where sk == endWord)
  

• Each pair of adjacent words in this sequence differs by exactly one letter.
• Each intermediate word sᵢ must belong to wordList.
• beginWord itself does not have to be in wordList.
• The function returns the number of steps in the shortest sequence (i.e., k + 1, because beginWord counts as the first step). If no path is found, return 0.

Example

beginWord = "hit"
endWord   = "cog"
wordList  = ["hot","dot","dog","lot","log","cog"]

Shortest transformation:
hit -> hot -> dot -> dog -> cog

Answer: 5

Approach Overview

We'll discuss four methods in detail:

1. Pre-Build Adjacency List (Naive)
2. On-the-Fly BFS (No Preprocessing)
3. Wildcard Mapping (More Efficient)
4. Bidirectional BFS (Advanced Optimization)

Approach 1: Pre-Build the Adjacency List (Naive)

Method Description

1. Compare every pair of words in the dictionary to see if they differ by a single letter.
2. If they do, record that connection in an adjacency list (a graph structure).
3. Perform a BFS from beginWord until endWord is reached.

Time Complexity

• Graph construction: (O(N^2 \times L))
  • We have (N) words and compare each pair, needing (O(L)) time for each pair to see if they differ by one letter.
• BFS: (O(N + E)), but (E) can be up to (O(N^2)).
• Overall can reach (O(N^2 \times L)) in practice.

Code Example

from collections import defaultdict, deque
from typing import List
 
class Solution:
    def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) -> int:
        # If the endWord is not in wordList, no valid sequence
        if endWord not in wordList:
            return 0
 
        # Convert to set and include beginWord if missing
        wordSet = set(wordList)
        wordSet.add(beginWord)
        words = list(wordSet)
 
        # Helper to check if two words differ by exactly one letter
        def differ_by_one(w1: str, w2: str) -> bool:
            diff_count = 0
            for c1, c2 in zip(w1, w2):
                if c1 != c2:
                    diff_count += 1
                    if diff_count > 1:
                        return False
            return diff_count == 1
 
        # 1) Build the graph (adjacency list) - O(N^2 * L)
        graph = defaultdict(list)
        n = len(words)
        for i in range(n):
            for j in range(i + 1, n):
                if differ_by_one(words[i], words[j]):
                    graph[words[i]].append(words[j])
                    graph[words[j]].append(words[i])
 
        # 2) BFS to find the shortest path
        queue = deque([(beginWord, 1)])
        visited = set([beginWord])
 
        while queue:
            word, steps = queue.popleft()
            # If we've reached the endWord, return the distance
            if word == endWord:
                return steps
 
            for neighbor in graph[word]:
                if neighbor not in visited:
                    visited.add(neighbor)
                    queue.append((neighbor, steps + 1))
 
        return 0

Pros: Straightforward to understand.
Cons: Can be slow when N is large.

Approach 2: On-the-Fly BFS (No Preprocessing)

Instead of building an entire adjacency list first, you directly do a BFS:

1. Start from beginWord.
2. In each BFS step, compare the current word to every remaining word in your set to find neighbors that differ by one letter.
3. Enqueue those neighbors and mark them visited.

Time Complexity

• Each time you dequeue a word, you may compare it with all words in the set.
• In the worst case, (O(N^2 \times L)).

Code Example

from collections import deque
from typing import List
 
class Solution:
    def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) -> int:
        word_set = set(wordList)
        if endWord not in word_set:
            return 0
 
        queue = deque([(beginWord, 1)])
        visited = set([beginWord])
 
        def differ_by_one(w1: str, w2: str) -> bool:
            diff_count = 0
            for c1, c2 in zip(w1, w2):
                if c1 != c2:
                    diff_count += 1
                    if diff_count > 1:
                        return False
            return diff_count == 1
 
        while queue:
            word, steps = queue.popleft()
            if word == endWord:
                return steps
 
            # Convert set to list to avoid runtime error
            for candidate in list(word_set):
                if differ_by_one(word, candidate):
                    if candidate == endWord:
                        return steps + 1
                    visited.add(candidate)
                    queue.append((candidate, steps + 1))
                    # Remove from the set to prevent re-check
                    word_set.remove(candidate)
 
        return 0

Pros: Minimal code without pre-building a graph.
Cons: Still (O(N^2 \times L)) in worst-case scenarios.

Approach 3: Wildcard Mapping (More Efficient)

To avoid the (N^2) blowup, we can:

1. Generate wildcard patterns (like h*t, *ot) for every word in the dictionary.
2. Store in a hash map: pattern -> list of words that match it.
3. Use BFS. For a given word, to find its neighbors, generate its (L) wildcard forms and look them up directly in the hash map.

Time Complexity (Detailed)

• Preprocessing:
  • For (N) words, each of length (L), generating (L) wildcard strings.
  • Each generation is (O(L)) due to slicing, so (O(N \times L^2)) total.
• BFS:
  • Potentially (O(N \times L^2)) again, because each word might generate (L) patterns in (O(L)).
• Overall typically (O(N \times L^2)), which is far better than (O(N^2 \times L)) when (N) is large, as long as each pattern doesn't map to a huge number of words.

Code Example

from collections import defaultdict, deque
from typing import List
 
class Solution:
    def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) -> int:
        word_set = set(wordList)
        if endWord not in word_set:
            return 0
 
        # Preprocess the wordList to build a pattern map
        pattern_map = defaultdict(list)
        L = len(beginWord)
 
        for word in word_set:
            for i in range(L):
                pattern = word[:i] + '*' + word[i+1:]
                pattern_map[pattern].append(word)
 
        # BFS initialization
        queue = deque([(beginWord, 1)])
        visited = set([beginWord])
 
        while queue:
            word, steps = queue.popleft()
 
            for i in range(L):
                wildcard = word[:i] + '*' + word[i+1:]
                neighbors = pattern_map[wildcard]
                for neighbor in neighbors:
                    if neighbor == endWord:
                        return steps + 1
                    if neighbor not in visited:
                        visited.add(neighbor)
                        queue.append((neighbor, steps + 1))
 
                # Clear the list to avoid re-processing the same neighbors in future iterations
                pattern_map[wildcard] = []
 
        return 0

Pros: Often much faster than naive methods in large dictionaries.
Cons: A bit more complex to implement, and in extreme cases where a pattern maps to many words, it can degrade.

Approach 4: Bidirectional BFS (Advanced Optimization)

Even with wildcard mapping, BFS might traverse a large portion of the graph. A powerful trick is Bidirectional BFS:

1. Start one search from beginWord and another from endWord.
2. Expand each frontier level by level.
3. If the two frontiers ever intersect, you have found the shortest path.

This effectively cuts the search space roughly in half (or more), providing an exponential speedup in many cases.

Time Complexity

• In the best or average scenario, the depth from each side shrinks from (d) to about (d/2), greatly reducing the total number of visits.
• Implementation details vary, but combining bidirectional BFS with wildcard mapping is typically the fastest known approach for large inputs.

Code Example (Bidirectional BFS with Wildcard)

from collections import defaultdict
from typing import List
 
class Solution:
    def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) -> int:
        word_set = set(wordList)
        if endWord not in word_set:
            return 0
 
        L = len(beginWord)
        # Pre-build wildcard patterns
        pattern_map = defaultdict(list)
        for word in word_set:
            for i in range(L):
                pattern = word[:i] + "*" + word[i+1:]
                pattern_map[pattern].append(word)
 
        # Initialize two "fronts": one from beginWord, one from endWord
        begin_set = {beginWord}
        end_set = {endWord}
        visited = {beginWord, endWord}
        distance = 1  # We'll count 'beginWord' as the first step
 
        # Perform bidirectional BFS
        while begin_set and end_set:
            # Always expand the smaller frontier for efficiency
            if len(begin_set) > len(end_set):
                begin_set, end_set = end_set, begin_set
 
            next_level = set()
            distance += 1  # Increment distance for each level
 
            for word in begin_set:
                for i in range(L):
                    pattern = word[:i] + "*" + word[i+1:]
                    for neighbor in pattern_map[pattern]:
                        if neighbor in end_set:
                            return distance  # Found intersection
                        if neighbor not in visited:
                            visited.add(neighbor)
                            next_level.add(neighbor)
 
            begin_set = next_level  # Update frontier for next iteration
 
        return 0  # No path found

How it works:

1. We maintain two "search fronts": one from beginWord (forward) and one from endWord (backward).
2. Each iteration, we expand the smaller set of words by one step.
3. If the new set of visited words intersects with the other frontier, it means we've found a path in distance transformations total.
4. The wildcard mapping ensures quick neighbor lookups (avoiding an (N^2) blowup).

Pros:

• Often the fastest approach, especially for large dictionaries.
• Greatly prunes the search space.

Cons:

• More code complexity due to maintaining two fronts and deciding when they intersect.

Edge Cases and Takeaways

1. Edge Cases

   • If endWord is not in the wordList, return 0 immediately.
   • If beginWord == endWord, some variations might consider the length as 1 or 0, but typically this scenario is trivially no transformation needed or is not applicable in standard statements.

2. Time Complexity Summary

   • Naive: (O(N^2 \times L)) for building or scanning adjacency.
   • Wildcard: ~(O(N \times L^2)) typical.
   • Bidirectional BFS can reduce the effective search depth in half.

3. Why Wildcard Patterns Help

   • Instead of comparing each word against all others, we transform each word into its (L) "wildcard keys" (like h*t) and store them in a hash map.
   • This offers direct neighbor lookups in BFS, usually saving a ton of comparisons.

4. Python String Slicing Cost

   • Doing word[:i] + '*' + word[i+1:] is (O(L)), so keep that in mind. Sometimes you'll see it referred to as (O(N \times L)), but strictly speaking, it's (O(N \times L^2)) considering the slicing cost.

Conclusion

Word Ladder is an excellent demonstration of how string manipulation can be reframed as a graph shortest-path problem:

• Approach 1 & 2: Build or dynamically compare words in (O(N^2 \times L)).
• Approach 3: Use wildcard patterns for an often much faster BFS ((O(N \times L^2))).
• Approach 4: Bidirectional BFS can cut search depths drastically, making it the most optimal variant when implemented with wildcard mappings.

When implementing in a real interview:

1. Start with a BFS approach.
2. If the interviewer asks about efficiency or the dictionary is large, move to wildcard or bidirectional BFS.
3. Always note edge cases, slicing costs, and potential extreme scenarios.

With these methods, you'll be well-equipped to handle Word Ladder questions confidently—and possibly adapt similar ideas for other one-letter transformation puzzles.