Data Structures and Algorithms: Word Break I and II

Welcome to our deep dive into the fascinating world of Data Structures and Algorithms! Today, we'll explore two captivating problems known as Word Break I and II. These problems are not only fun to solve but also highly relevant in real-world programming projects.

Word Break I

🎯 Objective: Given a string s and a list of words wordList, determine if it's possible to break s into a space-separated sequence of words from wordList.

Breaking it Down

Initialize an empty list result. This will store our final solution.
Iterate through the input string s. For each character, find the longest word from wordList that ends at the current index. If a word is found, add it to result and move the index to the next character.
If no word can be found, return False. If all characters have been processed, and result is not empty, return True.

💡 Pro Tip: Use a dictionary to store the indices of words in wordList. This will help us find the longest word ending at a particular index faster.

Code Example

python

def word_break(s, wordList):
    # Store indices of words in a dictionary for faster lookups
    word_dict = {word: i for i, word in enumerate(wordList)}

    result = []
    index = 0

    while index < len(s):
        word = s[index:]
        if word in word_dict and word_dict[word] > index:
            result.append(word_dict[word])
            index += word_dict[word]
        else:
            break

    # If the solution is empty, the string can't be broken
    if not result:
        return False

    # Ensure the last word is valid and the remaining string is empty
    last_word = result.pop()
    if last_word + s[last_word:] in word_dict:
        return True
    else:
        return False

Word Break II

🎯 Objective: Given a string s and a list of words wordList, find all possible ways to break s into a space-separated sequence of words from wordList.

Breaking it Down

Modify the word_break function from the previous problem to return a list of possible solutions instead of a boolean value.
To find all solutions, we can use backtracking. When we find a valid breakpoint, explore the remaining string recursively, using the remaining words in wordList.

💡 Pro Tip: Use a dictionary to store the indices of words in a reverse order to make it easier to explore words from the right end of the string.

Code Example

python

from typing import List

def word_break(s, wordList):
    # Store indices of words in a dictionary for faster lookups
    word_dict = {word: i for i, word in enumerate(wordList, 1)}

    def find_solutions(remaining_s, remaining_words, solutions):
        if not remaining_s:
            solutions.append(remaining_words)
            return

        for word in remaining_words:
            if word <= len(remaining_s) and remaning_s[:word] in word_dict:
                find_solutions(remaining_s[word:], remaining_words[word:], solutions)

    find_solutions(s, wordList, [])
    return solutions

Quiz

Now that you've learned about Word Break I and II, try solving these problems with different input strings and word lists. Happy coding! 🚀