Given an array of strings, group the anagrams together. You can return the answer in any order.
An anagram is a word or phrase formed by rearranging the letters of a different word or phrase, using all the original letters exactly once.
Example:
Input: strings = ["eat","tea","tan","ate","nat","bat"] Output: [["bat"],["nat","tan"],["ate","eat","tea"]]
For this lesson, your algorithm should run in O(n * log n * L * log L) time and use O(n * L) extra space.
(There are faster solutions which we will discuss in future lessons)
The core challenge of this problem is to identify and group words that are anagrams of each other. An anagram is a rearrangement of the letters of a word to form another word. For example, "eat" and "tea" are anagrams because they contain the same letters.
Common applications of this problem include text analysis, cryptography, and natural language processing. A potential pitfall is not recognizing that different words with the same letters should be grouped together.
To solve this problem, we need to identify a way to group words that are anagrams. A naive solution might involve comparing each word with every other word, but this would be inefficient.
Instead, we can use a more optimized approach:
This approach is efficient because sorting each word takes O(L log L) time, and we do this for each of the n words, resulting in a total time complexity of O(n * L log L).
Here is a step-by-step breakdown of the algorithm:
def group_anagrams(strings):
# Dictionary to hold the groups of anagrams
anagrams = {}
for word in strings:
# Sort the word to form the key
sorted_word = ''.join(sorted(word))
# Add the word to the corresponding anagram group
if sorted_word in anagrams:
anagrams[sorted_word].append(word)
else:
anagrams[sorted_word] = [word]
# Return the grouped anagrams
return list(anagrams.values())
# Example usage
strings = ["eat", "tea", "tan", "ate", "nat", "bat"]
print(group_anagrams(strings))
The time complexity of this approach is O(n * L log L), where n is the number of words and L is the maximum length of a word. Sorting each word takes O(L log L) time, and we do this for each of the n words.
The space complexity is O(n * L) because we store each word in the dictionary.
Potential edge cases include:
Examples:
# Edge case: Empty input list
print(group_anagrams([])) # Output: []
# Edge case: Single word
print(group_anagrams(["word"])) # Output: [["word"]]
# Edge case: All words are anagrams
print(group_anagrams(["abc", "bca", "cab"])) # Output: [["abc", "bca", "cab"]]
To test the solution comprehensively, we should include a variety of test cases:
Using a testing framework like unittest
in Python can help automate and organize these tests.
When approaching such problems, it's important to:
Grouping anagrams is a common problem that can be efficiently solved using sorting and dictionaries. Understanding and implementing this solution helps improve problem-solving skills and prepares you for more complex challenges.
Practice and explore further to deepen your understanding and proficiency in solving such problems.
Our interactive tutorials and AI-assisted learning will help you master problem-solving skills and teach you the algorithms to know for coding interviews.
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