Problem Description

This problem asks you to find the shortest transformation sequence between two words by changing one letter at a time, where each intermediate word must exist in a given dictionary.

Given:

• beginWord: The starting word
• endWord: The target word we want to reach
• wordList: A list of valid words that can be used as intermediate steps

Rules for transformation:

1. You can only change one letter at a time to form the next word
2. Each intermediate word must be present in wordList (except beginWord which doesn't need to be in the list)
3. The final word in the sequence must be endWord

The task is to return the number of words in the shortest transformation sequence from beginWord to endWord. If no valid transformation sequence exists, return 0.

For example, if beginWord = "hit", endWord = "cog", and wordList = ["hot","dot","dog","lot","log","cog"], one possible shortest sequence would be: "hit" -> "hot" -> "dot" -> "dog" -> "cog"

This sequence has 5 words total, so the answer would be 5.

The solution uses Breadth-First Search (BFS) to find the shortest path. It explores all possible one-letter transformations level by level, ensuring that when the target word is found, it's reached via the shortest possible path. The algorithm:

1. Starts from beginWord and explores all valid one-letter transformations
2. For each word, tries changing each character to all 26 possible letters
3. Checks if the new word exists in the word list
4. Continues until endWord is found or all possibilities are exhausted
5. Removes visited words from the set to avoid revisiting them

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• Yes: This problem can be modeled as a graph where each word is a node, and edges exist between words that differ by exactly one letter. We need to find the shortest path from beginWord to endWord through this graph.

Is it a tree?

• No: The word transformation graph is not a tree. It can have cycles (you could transform word A to B, then B to C, and potentially C back to A if they all differ by one letter), and nodes can have multiple parents.

Is the problem related to directed acyclic graphs?

• No: The graph is undirected (transformations work both ways) and can contain cycles, so it's not a DAG.

Is the problem related to shortest paths?

• Yes: We need to find the shortest transformation sequence (minimum number of steps) from beginWord to endWord.

Is the graph Weighted?

• No: Each transformation (edge) has the same cost of 1. We're counting the number of transformations, not dealing with different weights/costs for different transformations.

BFS

• Yes: Since we have an unweighted graph and need the shortest path, BFS is the perfect algorithm.

Conclusion: The flowchart leads us to use BFS (Breadth-First Search) for finding the shortest transformation sequence in this unweighted graph problem. BFS explores all words at distance k before exploring words at distance k+1, guaranteeing that when we reach endWord, we've found the shortest path.

Intuition

The key insight is recognizing that this is a shortest path problem in an unweighted graph. Each word represents a node, and two words are connected if they differ by exactly one letter. Since we want the minimum number of transformations, we need to find the shortest path from beginWord to endWord.

Why BFS? When all edges have equal weight (each transformation counts as 1 step), BFS naturally finds the shortest path. It explores all words reachable in 1 step, then all words reachable in 2 steps, and so on. This level-by-level exploration guarantees that the first time we reach endWord, we've found the shortest path.

The clever part of the implementation is how we find neighbors. Instead of comparing every word with every other word to build the graph (which would be expensive), we generate all possible one-letter transformations on the fly. For each position in the current word, we try all 26 letters and check if the resulting word exists in our word list.

To avoid revisiting words and getting stuck in cycles, we remove words from the set once visited. This serves as our "visited" marker and prevents infinite loops. Think of it like marking rooms as explored in a maze - once you've been to a room at distance d, there's no point visiting it again at distance d+1 or greater.

The algorithm maintains a counter for the number of words in the path. We start with 1 (for beginWord) and increment it each time we move to the next level of BFS. When we find endWord, we return this counter. If the queue becomes empty without finding endWord, it means no valid transformation sequence exists, so we return 0.

Solution Approach

The solution implements a standard BFS approach with some optimizations for finding neighbors efficiently.

Data Structures Used:

• words: A set containing all valid words from wordList for O(1) lookup
• q: A deque (double-ended queue) for BFS traversal
• ans: Counter to track the number of words in the transformation sequence

Algorithm Steps:

1. Initialize the BFS:

   • Convert wordList to a set for fast membership checking
   • Add beginWord to the queue
   • Set ans = 1 (counting beginWord)

2. Level-by-level BFS traversal:

   • For each level, increment ans at the start (representing the next step)
   • Process all words at the current level using for _ in range(len(q))

3. Generate neighbors for each word:

   • Pop a word from the queue and convert it to a list of characters
   • For each position i in the word:
     • Save the original character
     • Try all 26 possible letters ('a' to 'z')
     • Form the new word by joining the modified character list
     • Check if this new word exists in our word set

4. Process valid neighbors:

   • If the new word equals endWord, we've found the shortest path - return ans
   • Otherwise, add the new word to the queue for the next level
   • Remove the word from the set to mark it as visited

5. Handle no solution:

   • If the queue becomes empty without finding endWord, return 0

Key Optimization: The reference mentions Bidirectional BFS as an advanced optimization. Instead of searching only from start to end, we search from both directions simultaneously:

• Maintain two queues: q1 (start → end) and q2 (end → start)
• Always expand the smaller queue to reduce search space
• When a word is found in both directions, we've found the shortest path
• The total distance is step1 + step2 + 1

This bidirectional approach significantly reduces the search space, especially when the shortest path is long, as we explore from both ends rather than expanding exponentially from just one direction.

Example Walkthrough

Let's walk through a small example to illustrate the BFS solution approach.

Input:

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

Step-by-step execution:

Initialization:

• Convert wordList to set: words = {"hot", "dot", "dog", "cog"}
• Queue: q = ["hit"]
• Answer counter: ans = 1 (counting "hit")

Level 1 (ans = 2):

• Process "hit" (only word in queue)
• Generate all one-letter transformations:
  • Position 0: "ait", "bit", ..., "zit" → only "hot" is in words set
  • Position 1: "hat", "hbt", ..., "hzt" → none in words set
  • Position 2: "hia", "hib", ..., "hiz" → none in words set
• Found neighbor: "hot"
• Add "hot" to queue, remove from set
• Queue: q = ["hot"], words: {"dot", "dog", "cog"}

Level 2 (ans = 3):

• Process "hot"
• Generate transformations:
  • Position 0: "aot", "bot", ..., "dot", ..., "zot" → "dot" is in words set
  • Position 1: "hat", "hbt", ..., "hzt" → none in words set
  • Position 2: "hoa", "hob", ..., "hoz" → none in words set
• Found neighbor: "dot"
• Queue: q = ["dot"], words: {"dog", "cog"}

Level 3 (ans = 4):

• Process "dot"
• Generate transformations:
  • Position 0: "aot", "bot", ..., "zot" → none in remaining words
  • Position 1: "dat", "dbt", ..., "dzt" → none in remaining words
  • Position 2: "doa", "dob", ..., "dog", ..., "doz" → "dog" is in words set
• Found neighbor: "dog"
• Queue: q = ["dog"], words: {"cog"}

Level 4 (ans = 5):

• Process "dog"
• Generate transformations:
  • Position 0: "aog", "bog", "cog", ..., "zog" → "cog" is in words set!
• Found "cog" which equals endWord
• Return ans = 5

The transformation sequence is: "hit" → "hot" → "dot" → "dog" → "cog" (5 words total)

Key observations:

1. BFS explores level by level, ensuring the first path found is the shortest
2. Each word is visited only once (removed from set after visiting)
3. We generate neighbors dynamically by trying all 26 letters at each position
4. The answer increments at each level, counting the number of words in the path

Solution Implementation

Time and Space Complexity

Time Complexity: O(M² × N), where M is the length of each word and N is the total number of words in the word list.

• The algorithm uses BFS to traverse through possible word transformations
• In the worst case, we visit every word in the wordList: O(N)
• For each word visited, we iterate through all character positions: O(M)
• For each position, we try all 26 lowercase letters: O(26) = O(1)
• For each generated word, we perform string operations:
  • Converting list to string with ''.join(s): O(M)
  • Checking membership in the set t not in words: O(M) for string hashing
  • Comparing strings t == endWord: O(M)
• Total: O(N × M × 26 × M) = O(M² × N)

Space Complexity: O(M × N)

• The words set stores all words from wordList: O(M × N) where each word has length M
• The BFS queue can contain at most N words in the worst case: O(M × N)
• The temporary list s for character manipulation: O(M)
• Overall space complexity: O(M × N)

Common Pitfalls

1. Forgetting to Check if endWord Exists in wordList

One of the most common mistakes is not verifying whether endWord is present in the word list before starting the BFS. If endWord isn't in the list, it's impossible to reach it, yet the algorithm might waste time exploring all possibilities.

Problem Code:

# Missing this crucial check
available_words = set(wordList)
queue = deque([beginWord])
# Directly starts BFS without checking endWord

Solution:

available_words = set(wordList)
# Early termination check
if endWord not in available_words:
    return 0
queue = deque([beginWord])

2. Not Processing Words Level by Level

A critical mistake is processing words continuously without maintaining level boundaries. This breaks the BFS property of finding the shortest path and makes it impossible to track the correct transformation length.

Problem Code:

while queue:
    current_word = queue.popleft()
    # Incorrectly increments for each word, not each level
    transformation_length += 1
    # ... generate neighbors

Solution:

while queue:
    transformation_length += 1
    # Process all words at the current level together
    current_level_size = len(queue)
    for _ in range(current_level_size):
        current_word = queue.popleft()
        # ... generate neighbors

3. String Immutability Issues

Attempting to modify strings directly (which are immutable in Python) or inefficiently creating new strings for each character change.

Problem Code:

for position in range(len(current_word)):
    for letter in 'abcdefghijklmnopqrstuvwxyz':
        # Inefficient string concatenation
        new_word = current_word[:position] + letter + current_word[position+1:]

Solution:

# Convert to list for efficient character manipulation
word_chars = list(current_word)
for position in range(len(word_chars)):
    original_char = word_chars[position]
    for letter_index in range(26):
        word_chars[position] = chr(ord('a') + letter_index)
        new_word = ''.join(word_chars)
    word_chars[position] = original_char  # Restore for next position

4. Not Removing Visited Words from the Set

Failing to mark words as visited leads to infinite loops or redundant processing. Words might be added to the queue multiple times, causing incorrect distance calculations and performance issues.

Problem Code:

if new_word in available_words:
    if new_word == endWord:
        return transformation_length
    queue.append(new_word)
    # Forgot to remove the word from available_words

Solution:

if new_word in available_words:
    if new_word == endWord:
        return transformation_length
    queue.append(new_word)
    # Mark as visited by removing from set
    available_words.remove(new_word)

5. Including beginWord in the Available Words Set

If beginWord happens to be in wordList, not handling it properly can cause the algorithm to revisit it, creating cycles or incorrect paths.

Problem Code:

available_words = set(wordList)
# If beginWord is in wordList, it might be revisited
queue = deque([beginWord])

Solution:

available_words = set(wordList)
# Remove beginWord if it exists to prevent revisiting
available_words.discard(beginWord)  # discard() won't raise error if not present
queue = deque([beginWord])

6. Incorrect Initial Counter Value

Starting with transformation_length = 0 or incrementing at the wrong time leads to off-by-one errors in the final count.

Problem Code:

transformation_length = 0  # Wrong: should start at 1 for beginWord
while queue:
    # Or incrementing after processing instead of before
    for _ in range(len(queue)):
        # process words
    transformation_length += 1  # Wrong timing

Solution:

# Start at 1 to count beginWord
transformation_length = 1
while queue:
    # Increment at the start of each new level
    transformation_length += 1
    current_level_size = len(queue)
    for _ in range(current_level_size):
        # process words