Group Anagrams #49
Group Anagrams #49
The problem of grouping anagrams is a classic in coding interviews and algorithm design. The challenge is to group strings that are anagrams of each other, meaning they contain the same characters in a different order. While many solutions involve using a frequency map or sorting the strings, we will explore an innovative approach that does not rely on these techniques.
The Challenge
In my initial attempts to solve the problem, I considered summing the ASCII values of the characters in each string. This method worked for many cases but failed in edge cases where different strings had the same sum. For instance, “abc” and “acb” both yield the same ASCII sum of 294, but they are clearly different strings. I also experimented with binary representations to avoid collisions, yet I encountered similar issues with hash collisions.
A New Approach
One effective approach to solve the group anagrams problem without resorting to frequency maps or sorting is to use a fixed-size array to count character occurrences. Here’s a step-by-step breakdown of how to implement this:
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Character Count Array: Create an array of size 26 (for each letter in the English alphabet) initialized to zero. This will be used to count the occurrences of each character in the string.
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Index Calculation: For each character in the string, calculate its corresponding index in the array using the formula
ord(char) - ord('a'). Increment the value at that index to reflect the character’s frequency. -
Tuple as a Key: Convert the character count array into a tuple, which can be used as a key in a dictionary. This tuple uniquely represents the character composition of the string.
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Storing in a Dictionary: Use a dictionary to map each tuple to a list of strings. As you process each string, append it to the list corresponding to its character count tuple.
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Return the Results: Finally, return the values of the dictionary, which will contain the grouped anagrams.
Implementation
Here’s how the implementation looks in Python:
python def groupAnagrams(strs): anagrams = {}
for s in strs:
# Create a character count array of size 26
count = [0] * 26
# Count each character's frequency
for char in s:
count[ord(char) - ord('a')] += 1
# Convert the list to a tuple to use as a dictionary key
key = tuple(count)
# Append the string to the corresponding anagram group
if key in anagrams:
anagrams[key].append(s)
else:
anagrams[key] = [s]
return list(anagrams.values())
Time and Space Complexity
- Time Complexity: The time complexity of the above algorithm is O(N * K), where N is the number of strings and K is the maximum length of a string. This is because we iterate through each string once and count its characters.
- Space Complexity: The space complexity is O(N), as we are storing each unique group of anagrams in a dictionary.
Conclusion
This method efficiently groups anagrams without the drawbacks of ASCII summing or sorting. By leveraging a fixed-size character count array, we achieve a unique representation for each string that allows for quick grouping in a dictionary. This approach not only circumvents potential hash collisions but also maintains clarity and efficiency.
I encourage you to experiment with this implementation and consider other variations or optimizations. What other methods can you think of to solve the group anagrams problem? Let’s discuss!