## 1: U-nderstand

Understand what the interviewer is asking for by using test cases and questions about the problem.

• Established a set (2-3) of test cases to verify their own solution later.
• Established a set (1-2) of edge cases to verify their solution handles complexities.
• Have fully understood the problem and have no clarifying questions.
• Have you verified any Time/Space Constraints for this problem?
• What are the possible characters in the words? All lowercase alphabet letters.

• Do all words have the same number of characters?

• Yes, all words have the same number of characters.
• Could the beginning or end word be empty?

• No, each word must have at least one character.
• Could the beginning word be the same as the ending word?

• No, the beginning and ending word will always be different.
``````  HAPPY CASE
Input:
beginWord = "hit"
endWord   = "cog"
wordList  = ["hot","dot","dog","lot","log","cog"]

Output: 5

EDGE CASE
Input:
beginWord = "hit"
endWord   = "cog"
wordList  = ["hot","dot","dog","lot","log"]

Output: 0``````

## 2: M-atch

Match what this problem looks like to known categories of problems, e.g. Linked List or Dynamic Programming, and strategies or patterns in those categories.

``````For graphs, some of the top things we want to consider are:
``````
• BFS: We cannot use BFS to traverse the graph because we may visit exit nodes in the first traversal.
• DFS: We can use DFS to traverse the graph until we find a valid itinerary, ensuring we choose the lexicographically least itinerary.
• Adjacency List: We can use an adjacency list to store the graph, especially when the graph is sparse.
• Adjacency Matrix: We can use an adjacency matrix to store the graph, but a sparse graph will cause an unneeded worst-case runtime.
• Topological Sort: We can use topological sort when a directed graph is used and returns an array of the nodes where each node appears before all the nodes it points to. In order to have a topological sorting, the graph must not contain any cycles.
• Union Find: Are there find and union operations here? Can you perform a find operation where you can determine which subset a particular element is in? This can be used for determining if two elements are in the same subset. Can you perform a union operation where you join two subsets into a single subset? Can you check if the two subsets belong to same set? If no, then we cannot perform union.

## 3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Traverse the wordList using BFS, starting from beginWord until we either reach the end or see endWord. As we visit each node, we add all adjacent nodes by looking up if a certain word exists in wordList.

1. Create a set of all words in wordList, including endWord.
2. Create a set of visited nodes
3. Start BFS from beginWord, setting current word
• If current word is equal to endWord, return number of steps
• For each index of current word and each letter not equal to current word at index
• Create candidate word equal to currentWord
• Change candidate word at current index to new letter
• If it exists and not visited, traverse

⚠️ Common Mistakes

• Some people may attempt to find adjacent pairs of words by computing finding the number of different letters between each word, whose runtime is O(M^2). Instead, the difference can be computed via iteration over the possible letters to allow a O(M*26) runtime.

## 4: I-mplement

Implement the code to solve the algorithm.

``````  private static final char[] letters = "abcdefghijklmnopqrstuvwxyz".toCharArray();

public int ladderLength(String beginWord, String endWord, List<String> wordList) {
// create a set of all words in wordList
HashSet<String> wordsSet = new HashSet(wordList);
HashSet<String> visited = new HashSet<>();
Queue<Pair<String, Integer>> queue = new LinkedList();

// start BFS from beginWord, setting currentWord
queue.offer(new Pair<>(beginWord, 1));
while (!queue.isEmpty()) {
Pair<String, Integer> pair = queue.poll();
String word = pair.getKey();
int length = pair.getValue();
if (word.equals(endWord)) {
return length;
}

// for each index of current word and each letter not equal to current word at index
for (int i = 0; i < word.length(); i++) {
for (int j = 0; j < letters.length; j++) {
if (word.charAt(i) != letters[j]) {
String candidate = word.substring(0,i) + letters[j] + word.substring(i+1);
if (wordsSet.contains(candidate) && !visited.contains(candidate)) {
queue.offer(new Pair<>(candidate, length + 1));
}
}
}
}
}

return 0;
}``````
``````  LETTERS = set('abcdefghijklmnopqrstuvwxyz')

# create a set of all words in wordList
words = set(words)

# create a set of visited nodes
visited = set()
queue = [(start, 1)] # queue for BFS, which stores the word and distance

while len(queue) > 0:
word, length = queue.pop(0)
if word == end:
return length
for i, char in enumerate(word):
for letter in LETTERS:
if char != letter:
candidate = word[:i] + letter + word[i + 1:]
if candidate in words and candidate not in visited:
queue.append([candidate, length + 1])

return 0``````

## 5: R-eview

Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.

• Trace through your code with an input to check for the expected output
• Catch possible edge cases and off-by-one errors and verify the code works for the happy and edge cases you created in the “Understand” section

## 6: E-valuate

Evaluate the performance of your algorithm and state any strong/weak or future potential work.

Time Complexity: O(M^2 * N), where M is the length of words and N is the total number of words in the input word list
Space Complexity: O(M^2 * N) to store all MM transformations for each of the NN words in the set or dictionary