Group Anagrams
Given an array of strings `strs`, group the anagrams together. You can return the answer in any order. Use a HashMap where the key is the sorted string.
Why Interviewers Ask This
Airbnb asks this to evaluate your ability to optimize data grouping logic efficiently. They specifically test if you can recognize that sorting strings is a reliable canonical key for anagrams, rather than relying on less efficient character counting or brute-force comparisons. This assesses your grasp of hash map trade-offs and string manipulation complexity.
How to Answer This Question
1. Clarify the problem constraints: Ask about input size, character sets (ASCII vs Unicode), and whether empty strings are possible. 2. Propose the canonical key strategy: Explain that sorting characters within each string creates a unique identifier for all anagram groups. 3. Discuss time complexity: Analyze why sorting takes O(K log K) per string, where K is string length, leading to a total complexity of O(N * K log K). 4. Walk through the algorithm: Iterate through the array, sort each string to generate the key, and append the original string to a HashMap list associated with that key. 5. Address edge cases: Mention handling nulls, single-character strings, or duplicate entries in the input array. 6. Refine the solution: Briefly mention alternative approaches like frequency arrays for strictly ASCII inputs to show depth of knowledge before finalizing the sorted-string approach.
Key Points to Cover
- Recognizing that sorting creates a deterministic canonical key for anagrams
- Explicitly stating the time complexity as O(N * K log K)
- Using a HashMap to map keys to lists of original strings
- Handling edge cases like null inputs or empty strings gracefully
- Demonstrating awareness of alternative solutions like frequency arrays
Sample Answer
To solve the Group Anagrams problem effectively, I would first ensure we understand the input constraints, such as maximum array size or character encoding, which Airbnb often values for scalable systems. My preferred approach involves using a HashMap where the key is the sorted version of each string. Since anagrams share identical character compositions, their sorted versions will be identical. For example, 'listen' and 'silent' both become 'eilnst'.
I would initialize an empty HashMap. Then, I iterate through the input array `strs`. For every string, I convert it to a character array, sort it, and join it back into a string to serve as the key. I then check if this key exists in the map; if not, I create a new list for it. Finally, I add the original unsorted string to the list corresponding to that key.
Regarding complexity, if N is the number of strings and K is the maximum length of a string, sorting each string takes O(K log K). Doing this for N strings results in O(N * K log K) time complexity. The space complexity is O(N * K) to store the result and the map keys. While a frequency count array could work for ASCII-only inputs in O(N * K), the sorting method is more robust for Unicode and generally cleaner to implement quickly during an interview. This approach balances readability with efficiency, fitting well with Airbnb's focus on clear, maintainable code.
Common Mistakes to Avoid
- Attempting to compare every pair of strings which leads to O(N^2) complexity and Time Limit Exceeded errors
- Forgetting to sort the character array correctly or modifying the original string reference incorrectly
- Ignoring the possibility of duplicate strings in the input array causing incorrect grouping logic
- Overlooking the distinction between ASCII and Unicode when discussing frequency array optimizations
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