# Word Ladder

## 1. Clarify the problem:

Before we begin, let's clarify the problem statement. The "Word Ladder" problem asks us to find the length of the shortest transformation sequence from the start word to the end word. We can only make one change to the word at a time, and each intermediate word must be a valid word in the given word list.

## 2. Analyze the problem:

Let's analyze the problem to identify the input, output, and constraints.

**Input:**

- Two strings,
`beginWord`

and`endWord`

, representing the start and end words. - A list of strings,
`wordList`

, containing the valid words.

**Output:**

- An integer representing the length of the shortest transformation sequence from
`beginWord`

to`endWord`

. If there is no such transformation sequence, return 0.

**Constraints:**

- All words in
`wordList`

and the two given words consist of lowercase English letters. - The length of
`wordList`

is at most 5,000. - The length of
`beginWord`

and`endWord`

is at least 1 and at most 100. `beginWord`

and`endWord`

are distinct.

## 3. Design an algorithm:

To solve this problem, we can use a breadth-first search (BFS) approach. Here's the general outline of our algorithm:

- Create a set,
`wordSet`

, containing all the words from`wordList`

. This will help us quickly check if a word is valid. - Create a queue,
`queue`

, and enqueue the`beginWord`

along with its level, starting from 1. - Create a set,
`visited`

, to keep track of the visited words. - While the
`queue`

is not empty:- Dequeue a word and its level from the front of the
`queue`

. - If the dequeued word is the same as the
`endWord`

, return the level as the result. - Iterate over each character of the dequeued word:
- Replace the current character with all lowercase letters from 'a' to 'z' one by one.
- If the new word is in
`wordSet`

and has not been visited before:- Enqueue the new word along with the level incremented by 1.
- Add the new word to the
`visited`

set.

- Dequeue a word and its level from the front of the
- If we exhaust all possibilities and cannot reach the
`endWord`

, return 0.

## 4. Explain your approach:

Our approach involves using a breadth-first search (BFS) to find the shortest transformation sequence from the start word to the end word. We start from the `beginWord`

and explore all possible valid words that can be formed by changing one character at a time. By keeping track of the visited words and their levels, we can find the shortest path to the `endWord`

.

## 5. Write clean and readable code:

Let's implement the algorithm in Python:

python

```
def ladderLength(beginWord, endWord, wordList):
wordSet = set(wordList)
if endWord not in wordSet:
return 0
queue = [(beginWord, 1)]
visited = set()
while queue:
word, level = queue.pop(0)
if word == endWord:
return level
for i in range(len(word)):
for char in 'abcdefghijklmnopqrstuvwxyz':
new_word = word[:i] + char + word[i+1:]
if new_word in wordSet and new_word not in visited:
queue.append((new_word, level + 1))
visited.add(new_word)
return 0
```

## 6. Test your code:

Let's test our code with different test cases to ensure its correctness. We'll consider the following cases:

Case 1:

python

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

The expected output is 5.

Case 2:

python

```
beginWord = "a"
endWord = "c"
wordList = ["a", "b", "c"]
```

The expected output is 2.

Case 3:

python

`beginWord = "hot" endWord = "dog" wordList = ["hot", "dog"]`

The expected output is 0.

## 7. Optimize if necessary:

The current solution is already efficient, and further optimization is not necessary.

## 8. Handle error cases:

Our code handles the case where the `endWord`

is not in the `wordList`

by returning 0.

## 9. Discuss complexity analysis:

The time complexity of our solution is O(M^2 * N), where M is the length of each word and N is the total number of words in the word list. This is because, in the worst case, we need to explore all possible transformations for each word, which takes O(M^2) time. Additionally, we may visit each word in the word list once, resulting in a total time complexity of O(N).

The space complexity is O(M^2 * N) as well. This is because we use sets to store the word list, visited words, and the queue for BFS, each of which can contain up to O(N) words with a length of O(M).