Word Ladder

The given problem is a classic example of a word transformation problem where we have to transform one word into another by changing a single letter at a time. The words that can be formed at each step must be present in the given wordList. We can use Breadth-First Search (BFS) to solve this problem.

Python code:

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from collections import deque

class Solution:
    def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) -> int:
        # Convert wordList into a set for faster lookup
        wordSet = set(wordList)

        # If the endWord is not in the wordList, return 0
        if endWord not in wordSet:
            return 0

        # Initialize a queue for BFS
        queue = deque([(beginWord, 1)])

        # Iterate until the queue is empty
        while queue:
            word, length = queue.popleft()

            # If the current word is equal to the endWord, return the length of the sequence
            if word == endWord:
                return length

            # Iterate through the current word and try all possible single-letter transformations
            for i in range(len(word)):
                for c in 'abcdefghijklmnopqrstuvwxyz':
                    next_word = word[:i] + c + word[i+1:]

                    # If the transformed word is in the wordSet, add it to the queue
                    if next_word in wordSet:
                        wordSet.remove(next_word)
                        queue.append((next_word, length + 1))

        return 0

Explanation:

  1. We use a set to store the words in the wordList to allow for fast lookups.
  2. If the endWord is not in the set, we return 0 as no transformation is possible.
  3. We use a queue to perform a Breadth-First Search starting from the beginWord.
  4. We explore all possible single-letter transformations of the current word. If the transformed word is in the wordSet, we add it to the queue with an incremented length.
  5. If we reach the endWord, we return the length of the transformation sequence.
  6. If no transformation is found, we return 0.

The time complexity of this solution is O(N * K^2), where N is the number of words in the wordList, and K is the maximum length of a word. The space complexity is O(N), as we store the words in a set.

Identifying Problem Isomorphism

“Word Ladder” has an approximate isomorph: “Minimum Genetic Mutation”.

In “Word Ladder”, you’re given two words (beginWord and endWord) and a dictionary’s word list, and you have to find the length of the shortest transformation sequence from beginWord to endWord, where only one letter can be changed at a time and each transformed word must exist in the word list.

In “Minimum Genetic Mutation”, you’re given two genes (start and end) and a gene bank, and you have to determine the minimum number of mutations needed to transform the start gene into the end gene, where only one letter can be changed at a time and each intermediate gene must be in the gene bank.

In both problems, you’re transforming one string into another, one character at a time, and each intermediate string must exist in a given list of valid strings. You can approach both problems with a breadth-first search (BFS) strategy to find the shortest transformation sequence.

Hence, the isomorphism between these two problems is that they both model a BFS on a graph, where vertices are possible states (words or genes) and edges are valid transformations (changing one character in a word or gene). The specifics of the implementation may vary according to the details of each problem, but the core concept and strategy remain the same.

10 Prerequisite LeetCode Problems

“127. Word Ladder” is a classic problem that involves graph theory and breadth-first search. Here are 10 problems to prepare:

  1. “Symmetric Tree” (LeetCode Problem #101): This is a good introduction to the concept of tree traversal, which shares similarities with graph traversal.

  2. “Binary Tree Level Order Traversal” (LeetCode Problem #102): This problem will help you familiarize with level order traversal which is essentially a BFS on a tree.

  3. “Course Schedule” (LeetCode Problem #207): This problem will help you understand topological sorting and detecting cycles in a directed graph, both of which are useful for understanding graph problems.

  4. “Course Schedule II” (LeetCode Problem #210): This is a continuation of the previous problem but this time you have to return the ordering of tasks.

  5. “Clone Graph” (LeetCode Problem #133): This problem helps you understand the concept of graph traversal and cloning (or making a deep copy) of a graph.

  6. “Open the Lock” (LeetCode Problem #752): It is very similar to the Word Ladder problem, where you need to reach from one state to another.

  7. “Number of Islands” (LeetCode Problem #200): This problem helps you understand the concepts of depth-first search (DFS) and connected components in a grid, which can be thought of as a form of a graph.

  8. “Network Delay Time” (LeetCode Problem #743): This problem helps you understand single-source shortest path algorithms in a weighted graph (like Dijkstra’s algorithm).

  9. “Perfect Squares” (LeetCode Problem #279): This problem is a classic dynamic programming problem that also can be solved using a BFS approach, similar to Word Ladder.

  10. “Valid Anagram” (LeetCode Problem #242): This problem deals with strings and frequency counts, which are useful concepts for Word Ladder.

  11. “Subsets” (LeetCode Problem #78): This problem helps in understanding the concept of generating all possible combinations, which can be applied in Word Ladder for generating all possible transformations.

  12. “Word Break” (LeetCode Problem #139): This problem is about transforming a string into words from a given list. While the problem context is different, it still provides a good exercise for dealing with string transformations.

  13. “Permutations” (LeetCode Problem #46): The problem teaches about generating all possible permutations, similar to generating all possible transformations in the Word Ladder problem.

  14. “Implement Trie (Prefix Tree)” (LeetCode Problem #208): Understanding Trie data structure can be beneficial in optimizing the Word Ladder problem.

  15. “Word Pattern” (LeetCode Problem #290): This problem is about matching patterns in strings, and understanding it can be helpful in solving the Word Ladder problem.

  16. “Number of Islands” (LeetCode Problem #200): This problem helps you understand the concept of depth-first search (DFS) in a grid, which can be thought of as a form of a graph.

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from collections import defaultdict
from collections import deque

class Solution:
    def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) -> int:
        """
        :type beginWord: str
        :type endWord: str
        :type wordList: List[str]
        :rtype: int
        """
        if endWord not in wordList or not endWord or not beginWord or not wordList:
            return 0
        L = len(beginWord)
        all_combo_dict = defaultdict(list)
        for word in wordList:
            for i in range(L):
                all_combo_dict[word[:i] + "*" + word[i+1:]].append(word) 
        queue = deque([(beginWord, 1)])
        visited = set()
        visited.add(beginWord)
        while queue:
            current_word, level = queue.popleft()
            for i in range(L):
                intermediate_word = current_word[:i] + "*" + current_word[i+1:]
                for word in all_combo_dict[intermediate_word]:
                    if word == endWord:
                        return level + 1
                    if word not in visited:
                        visited.add(word)
                        queue.append((word, level + 1))
        return 0

Problem Classification

This is about finding a sequence of words from a start word to an end word with each word differing by one letter from its neighbors. But this time, we only need to find the length of the shortest sequence. It’s also a Word Connection Problem.

Language Agnostic Coding Drills

  1. Basic Data Types: Understand basic data types such as integer, string, boolean, etc. Practice initializing variables, performing basic operations, and understanding the type system.

    Drill: Create and manipulate variables of different data types. Perform operations like addition, concatenation, comparison, etc.

  2. Arrays/Lists: Learn how to create, access and manipulate lists/arrays. Practice adding elements, removing elements, iterating through a list, etc.

    Drill: Create a list of numbers, write a function to add an element at a specific index, remove an element, find an element, etc.

  3. Control Flow: Understand if-else statements, for and while loops. Practice writing code that includes conditional statements and loops.

    Drill: Write a program that iterates over a list and performs a certain operation (e.g. increment a counter) based on a conditional statement.

  4. Functions: Learn how to write reusable pieces of code in the form of functions. Understand parameters, return statements and scope.

    Drill: Write a function that takes in two numbers as parameters and returns their sum.

  5. Dictionaries/Hashmaps: Learn about key-value pair data structures. Practice creating, accessing, and manipulating dictionaries/hashmaps.

    Drill: Create a hashmap of students where the key is the student’s ID and the value is the student’s name. Write functions to add a student, remove a student, and find a student by ID.

  6. Sets: Understand the set data structure and its unique properties. Learn how to add, remove, and check if an item exists within a set.

    Drill: Create a set of items, add new items to the set, remove items, and check membership of an item.

  7. Queues: Learn about the queue data structure and its properties. Understand how to enqueue, dequeue, and check if a queue is empty.

    Drill: Implement a queue using a list/array. Add elements to the end of the queue, remove elements from the front, and write a function to check if the queue is empty.

  8. Graph Traversal (BFS and DFS): Understand how to traverse a graph using breadth-first search (BFS) and depth-first search (DFS).

    Drill: Given an adjacency list representation of a graph, implement BFS and DFS.

  9. Recursion: Understand how to solve problems using recursion. Practice identifying base cases and recursive cases.

    Drill: Write a recursive function to calculate the factorial of a number.

  10. Dynamic Programming: Understand how to solve complex problems by breaking them down into simpler subproblems. Learn about memoization and bottom-up dynamic programming.

    Drill: Implement a dynamic programming solution for the Fibonacci sequence. Both top-down (with memoization) and bottom-up solutions should be attempted.

Targeted Drills in Python

1. Understanding collections.defaultdict

A defaultdict is a dictionary subclass that calls a factory function to supply missing values. In the case of this problem, it’s used to hold lists of words that share common intermediate states.

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from collections import defaultdict

# Creating a defaultdict
dd = defaultdict(list)

# Adding elements
dd["a"].append(1)
dd["a"].append(2)
dd["b"].append(3)

print(dd)  # defaultdict(<class 'list'>, {'a': [1, 2], 'b': [3]})

2. Understanding collections.deque

deque (Double Ended Queue) in Python is implemented using the module “collections“. Deque is preferred over list in the cases where we need quicker append and pop operations from both the ends of the container, as deque provides an O(1) time complexity for append and pop operations.

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from collections import deque

# Creating a deque
d = deque()

# Adding elements
d.append('a')
d.append('b')

print(d)  # deque(['a', 'b'])

3. Understanding Sets in Python

A set is an unordered collection of items. Every set element is unique (no duplicates) and must be immutable (cannot be changed).

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# Creating a set
s = set()

# Adding elements
s.add('a')
s.add('b')
s.add('a')  # This won't be added as 'a' is already in the set

print(s)  # {'a', 'b'}

4. Looping through a word’s characters

This drill is about string manipulation. You can iterate through the characters of a string and replace them.

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# Original string
s = "cat"

# Create all combinations where one letter is replaced with '*'
for i in range(len(s)):
    print(s[:i] + "*" + s[i+1:])  # Outputs: *at, c*t, ca*

5. BFS (Breadth-First Search)

BFS is a traversing algorithm where you should start traversing from a selected node (source or starting node) and traverse the graph layerwise thus exploring the neighbour nodes. Nodes are visited level by level. BFS uses a queue data structure to maintain the nodes.

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from collections import deque

def BFS(graph, start):
    visited = set()
    queue = deque([start])

    while queue:
        vertex = queue.popleft()
        print(vertex)

        for neighbour in graph[vertex]:
            if neighbour not in visited:
                visited.add(neighbour)
                queue.append(neighbour)

graph = {'A': ['B', 'C'], 'B': ['A', 'D', 'E'], 'C': ['A', 'F'], 'D': ['B'], 'E': ['B', 'F'], 'F': ['C', 'E']}
BFS(graph, 'A')