How do you solve Group Anagrams on LeetCode?

Group Anagrams (LeetCode #49) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * k log k) time complexity and O(n) space complexity. Key insight: Anagrams can be identified by sorting their characters.

What algorithm or data structure is used to solve Group Anagrams?

Group Anagrams uses the following concepts: Array, Hash Table, String, Sorting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Group Anagrams?

Always clarify the input constraints and edge cases. Think about space vs. time trade-offs when choosing data structures. Explain your thought process clearly while coding.

What are common mistakes when solving Group Anagrams?

Not considering empty strings or single-character strings. Forgetting to handle case sensitivity or character sets beyond lowercase letters.

Group Anagrams — LeetCode #49 (Medium)

Tags: Array, Hash Table, String, Sorting

Related patterns: Hash Map, Array

Brute Force approach

Time complexity: O(n² * k log k). Space complexity: O(1).

The brute force approach involves checking each string against every other string to see if they are anagrams. This is straightforward but inefficient for larger inputs.

The time complexity is O(n²) because we compare each string with every other string, and sorting each string takes O(k log k), where k is the length of the string.

Step 1: For each string in the input list, compare it with every other string.
Step 2: Sort both strings and check if they are equal. If they are, they are anagrams.
Step 3: Group the anagrams together in a list.

Input: strs = ["eat","tea","tan","ate","nat","bat"] 1. Compare 'eat' with 'tea' → anagram → group = ['tea'] 2. Compare 'eat' with 'tan' → not anagram → continue 3. Compare 'eat' with 'ate' → anagram → group = ['tea', 'ate'] 4. Compare 'eat' with 'nat' → not anagram → continue 5. Compare 'eat' with 'bat' → not anagram → continue 6. Final group for 'eat' = ['eat', 'tea', 'ate']

Optimal Solution approach

Time complexity: O(n * k log k). Space complexity: O(n).

The optimal solution uses a hash table to group anagrams by their sorted string representation. This way, we can efficiently categorize anagrams without comparing each string directly.

The time complexity is O(n * k log k) because we sort each string (O(k log k)) and do this for n strings. The space complexity is O(n) for storing the grouped anagrams.

Step 1: Create a hash map to store groups of anagrams, using the sorted string as the key.
Step 2: For each string, sort it and use the sorted string as the key to group the original strings.
Step 3: Collect all groups from the hash map into a list and return it.

Input: strs = ["eat","tea","tan","ate","nat","bat"] 1. 'eat' → key = 'aet' → anagrams = {'aet': ['eat']} 2. 'tea' → key = 'aet' → anagrams = {'aet': ['eat', 'tea']} 3. 'tan' → key = 'ant' → anagrams = {'aet': ['eat', 'tea'], 'ant': ['tan']} 4. 'ate' → key = 'aet' → anagrams = {'aet': ['eat', 'tea', 'ate'], 'ant': ['tan']} 5. 'nat' → key = 'ant' → anagrams = {'aet': ['eat', 'tea', 'ate'], 'ant': ['tan', 'nat']} 6. 'bat' → key = 'abt' → anagrams = {'aet': ['eat', 'tea', 'ate'], 'ant': ['tan', 'nat'], 'abt': ['bat']}

Key Insights

Anagrams can be identified by sorting their characters.
Using a hash map allows for efficient grouping of anagrams.

Common Mistakes

Not considering empty strings or single-character strings.
Forgetting to handle case sensitivity or character sets beyond lowercase letters.

Interview Tips

Always clarify the input constraints and edge cases.
Think about space vs. time trade-offs when choosing data structures.
Explain your thought process clearly while coding.

#49

Group Anagrams

Medium

Array↗Hash Table↗String↗Sorting↗Hash Map↗Array↗

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n² * k log k)	O(n * k log k)
Space	O(1)	O(n)

💡

Intuition

Time O(n * k log k)Space O(n)

The optimal solution uses a hash table to group anagrams by their sorted string representation. This way, we can efficiently categorize anagrams without comparing each string directly.

⚙️

Algorithm

3 steps

1Step 1: Create a hash map to store groups of anagrams, using the sorted string as the key.
2Step 2: For each string, sort it and use the sorted string as the key to group the original strings.
3Step 3: Collect all groups from the hash map into a list and return it.

solution.py8 lines

1from collections import defaultdict
2
3def groupAnagrams(strs):
4    anagrams = defaultdict(list)
5    for s in strs:
6        key = ''.join(sorted(s))
7        anagrams[key].append(s)
8    return list(anagrams.values())

ℹ

Complexity note: The time complexity is O(n * k log k) because we sort each string (O(k log k)) and do this for n strings. The space complexity is O(n) for storing the grouped anagrams.

1Anagrams can be identified by sorting their characters.
2Using a hash map allows for efficient grouping of anagrams.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.