How to Solve Group Anagrams

To solve the Group Anagrams problem, use array scanning combined with hash lookups to group strings that share the same set of characters. Sorting each string or utilizing frequency counts helps identify anagrams. The approach should focus on leveraging efficient data structures like hash tables.

Problem Statement

Given a list of strings, the task is to group the anagrams together. An anagram is a word formed by rearranging the letters of another word. For instance, 'eat', 'tea', and 'ate' are anagrams. You can return the answer in any order, but all anagrams should be grouped together.

For example, the input ['eat', 'tea', 'tan', 'ate', 'nat', 'bat'] should result in the output [['bat'], ['nat', 'tan'], ['ate', 'eat', 'tea']]. You need to efficiently group these strings using appropriate data structures and algorithms while adhering to the given constraints.

Examples

Example 1

Input: strs = ["eat","tea","tan","ate","nat","bat"]

Output: [["bat"],["nat","tan"],["ate","eat","tea"]]

Example 2

Input: strs = [""]

Output: [[""]]

Example details omitted.

Example 3

Input: strs = ["a"]

Output: [["a"]]

Example details omitted.

Constraints

• 1 <= strs.length <= 104
• 0 <= strs[i].length <= 100
• strs[i] consists of lowercase English letters.

Solution Approach

Array Scanning with Hash Lookup

The key approach for solving this problem is using hash tables to group the anagrams. For each string, we can either sort it or count the frequency of characters, then use the sorted version or character count as the key in the hash table. Strings that produce the same key will be grouped together as anagrams. This allows efficient grouping in linear time per string after sorting, leveraging hash lookups for quick grouping.

Optimized Sorting Method

An alternate approach to solve the problem involves sorting each string individually. Sorting will allow anagrams to have the same sequence of characters. By using the sorted version of the string as the key in the hash table, you can group anagrams efficiently. This method runs in O(n * k log k) time complexity, where n is the number of strings and k is the length of the longest string.

Frequency Count Hashing

Instead of sorting strings, a frequency count approach can be used. Count the occurrences of each character in the string and use this frequency count as a key. Strings with identical character frequencies will be grouped as anagrams. This method provides an optimized time complexity of O(n * k) where n is the number of strings and k is the length of each string.

Complexity Analysis

The time complexity depends on the approach used. Using sorting, the time complexity is O(n * k log k), where n is the number of strings and k is the length of the longest string. With frequency counting, the time complexity is O(n * k), where n is the number of strings and k is the length of each string. The space complexity is O(n) for storing the grouped anagrams in the hash table.

What Interviewers Usually Probe

• Do you understand how hash tables can be used to group strings efficiently?
• Can you explain the trade-off between sorting strings and using frequency counts?
• Will you consider edge cases such as empty strings and strings with single characters?

Common Pitfalls or Variants

Common pitfalls

• Forgetting to handle empty strings, which should be grouped correctly.
• Not efficiently handling large inputs, especially when the number of strings is high (up to 10^4).
• Assuming the problem only requires sorting the input, rather than considering the hashing approach.

Follow-up variants

• Group Anagrams with case sensitivity
• Group Anagrams with longer strings up to length 100
• Group Anagrams with additional constraints like handling non-English characters

How GhostInterview Helps

• Screenshot or capture input for a hands-on approach to hashing anagrams.
• Walkthroughs include the entire answer path and complexity breakdown.
• Provide live screen-share workflows to walk through the hashing solution.

Topic Pages

• Array
• Hash Table
• String
• Sorting

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• How GhostInterview Works - Review the screenshot, reasoning, and answer flow before using the solver in a live interview.

FAQ

What is the best approach for grouping anagrams in terms of performance?

Using hash tables with either a sorted key or a frequency count of characters provides the most efficient grouping. The frequency count approach is usually better for performance when avoiding unnecessary sorting.

How do I handle edge cases like empty strings in this problem?

Empty strings should be treated like any other string and grouped in their own group. Make sure to include logic that correctly handles this case.

Is sorting the strings the only way to group anagrams?

No, another approach is to use a frequency count of characters as a key for the hash table. Sorting is not necessary and can be more expensive in terms of time complexity.

How does the time complexity differ between sorting and frequency counting for grouping anagrams?

Sorting each string takes O(k log k) time per string, where k is the length of the string. Frequency counting is more efficient with a time complexity of O(k) per string, making it faster for large inputs.

What are the space complexities of the sorting and frequency count methods?

Both methods use O(n) space for the hash table to store the grouped anagrams, where n is the number of strings. The sorting method may also use additional space for sorting the strings.