49. Group Anagrams- Detailed Explanation
Problem Statement
Description:
Given an array of strings, group the anagrams together. Two strings are anagrams if they contain the same characters in the same frequency (order does not matter).
Example 1:
- Input:
["eat", "tea", "tan", "ate", "nat", "bat"] - Output:
[["eat", "tea", "ate"], ["tan", "nat"], ["bat"]] - Explanation:
- "eat", "tea", and "ate" are anagrams of each other.
- "tan" and "nat" are anagrams of each other.
- "bat" does not have any anagram in the list.
Example 2:
- Input:
[""] - Output:
[[""]] - Explanation:
- A single empty string is trivially an anagram of itself.
Example 3:
- Input:
["a"] - Output:
[["a"]] - Explanation:
- A single character string is an anagram of itself.
Constraints:
- 1 ≤
strs.length≤ 10⁴ - 0 ≤
strs[i].length≤ 100 strs[i]consists of lowercase English letters.
Hints
-
Canonical Representation:
Consider how you can convert each string into a canonical form so that all anagrams share the same representation. For example, sorting the string or using a count of characters. -
Hashing & Grouping:
Once you have a canonical representation, you can use a hash map (dictionary) to group the original strings. Each unique key (canonical form) will map to a list of strings that are anagrams of one another.
Approach 1: Sorting Each String
Idea
-
Sort Each String:
For each string, sort its characters. All anagrams will result in the same sorted string. -
Use Hash Map:
Use the sorted string as a key in a hash map and append the original string to the list associated with that key.
Step-by-Step Walkthrough (Visual Example)
Imagine processing the list: ["eat", "tea", "tan", "ate", "nat", "bat"].
- "eat" → "aet"
- Map becomes:
{ "aet": ["eat"] }
- Map becomes:
- "tea" → "aet"
- Map updates to:
{ "aet": ["eat", "tea"] }
- Map updates to:
- "tan" → "ant"
- Map becomes:
{ "aet": ["eat", "tea"], "ant": ["tan"] }
- Map becomes:
- "ate" → "aet"
- Map updates to:
{ "aet": ["eat", "tea", "ate"], "ant": ["tan"] }
- Map updates to:
- "nat" → "ant"
- Map updates to:
{ "aet": ["eat", "tea", "ate"], "ant": ["tan", "nat"] }
- Map updates to:
- "bat" → "abt"
- Map becomes:
{ "aet": ["eat", "tea", "ate"], "ant": ["tan", "nat"], "abt": ["bat"] }
- Map becomes:
Finally, return the lists of grouped anagrams from the map.
Complexity Analysis
-
Time Complexity: O(n * k log k)
Where n is the number of strings and k is the maximum length of a string (due to sorting each string). -
Space Complexity: O(n * k)
For storing the hash map keys and lists of strings.
Python Code (Sorting Approach)
Java Code (Sorting Approach)
Approach 2: Character Count as Key
Idea
-
Character Frequency Count:
Instead of sorting, count the frequency of each letter in the string. Since there are 26 lowercase letters, you can represent the count as an array of length 26. -
Convert Count to a String Key:
Use the frequency array to form a unique key (for example, by concatenating counts with a delimiter) so that anagrams produce the same key.
Step-by-Step Walkthrough
For the string "tea":
- Count array might be:
[1, 0, 0, 0, 1, 0, ... 1, ...0](with1for 't', 'e', and 'a'). - Convert this array into a key string like:
"1#0#0#0#1#0#...#1#...#0". - Group all strings with the same key.
Complexity Analysis
-
Time Complexity: O(n * k)
Where n is the number of strings and k is the length of each string (since counting is O(k)). -
Space Complexity: O(n * k)
For the hash map storage.
Python Code (Character Count Approach)
Java Code (Character Count Approach)
Common Mistakes & Edge Cases
-
Not Handling Empty Strings:
- Ensure your solution can handle an input like
[""]. Even an empty string should be grouped correctly.
- Ensure your solution can handle an input like
-
Improper Key Generation:
- When using the character count approach, forgetting to include delimiters or not converting the count into a hashable type (like a tuple in Python) may lead to incorrect grouping.
-
Duplicate Strings:
- Ensure that duplicate strings (which are valid anagrams) are included in the correct group.
-
Large Input Sizes:
- Consider time and space optimizations when handling up to 10⁴ strings. The character count method may be more efficient for long strings with many characters.
Alternative Variations & Related Problems
-
Related Problems:
- Valid Anagram: Check if two strings are anagrams.
- Longest Anagram Chain: While not exactly anagrams, it also involves string manipulation and grouping based on transformation rules.
- Group Shifted Strings: Another grouping problem based on a different transformation rule.
-
Alternative Approaches:
- Bit Masking:
For small alphabets, one might explore bit masking techniques; however, since anagrams require counts (not just existence), this is less applicable. - Counting Sort Approach:
In languages where counting sort is efficient, you might use a counting sort algorithm to generate the canonical key.
- Bit Masking:
Final Thoughts
Understanding the key idea behind grouping anagrams—mapping strings to a canonical representation—is critical.
- The sorting approach is intuitive and simple to implement.
- The character count approach optimizes performance by avoiding the O(k log k) cost of sorting, reducing it to O(k) per string.
Both methods showcase essential techniques for solving problems that involve hashing and grouping. Practice these approaches, and try modifying the solutions for similar problems to build a strong foundation for coding interviews.
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