How to Explain Group Anagrams in a Coding Interview (Step-by-Step)

Group Anagrams is a classic Medium problem that tests whether you can identify the right key for a HashMap — the core insight that separates a clean O(n·k) solution from a messy O(n²) one.

In this post, you’ll learn exactly how to walk an interviewer through this problem: what to say, in what order, and which details to proactively highlight so you come across as confident and experienced.

Problem Statement

Examples

Constraints

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

What Is the Problem?

Before writing a single line of code, restate the problem in your own words:

“I need to partition a list of strings into groups where every string in a group is an anagram of the others. Strings that share the same characters with the same frequencies belong together.”

The Key Insight — What Makes a Good HashMap Key?

This is the pivotal moment in the interview — state this insight clearly before touching code. It shows you understand why the solution works, not just how to write it.

Step 1: Start With the Brute Force (Then Dismiss It)

“A brute force approach would compare every pair of strings to check if they’re anagrams — O(n²·k) where k is the average string length. That’s too slow for 10,000 strings.”

Then pivot:

“Instead, I’ll use a HashMap where each key is the sorted version of a string. All anagrams map to the same key, so I can group them in a single O(n·k log k) pass.”

Step 2: Explain the Algorithm in Plain English

“For each string in the input, I sort its characters to produce a key. I use that key to look up a bucket in my HashMap and append the original string to it. After processing all strings, the HashMap values are my grouped anagram lists — I just return them.”

Step 3: Talk Through the Code

from collections import defaultdict

class Solution:
    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
        groups = defaultdict(list)  # key: sorted string → value: list of anagrams

        for word in strs:
            key = "".join(sorted(word))  # Sort chars to get the canonical key
            groups[key].append(word)     # Add original word to its anagram group

        return list(groups.values())     # Return all groups as a list of lists

Point out the use of defaultdict(list) — it avoids the need to check if a key already exists before appending, keeping the code clean.

Bonus: O(n·k) Approach Using Frequency Count Key

If the interviewer asks for an approach that avoids sorting entirely:

from collections import defaultdict

class Solution:
    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
        groups = defaultdict(list)

        for word in strs:
            count = [0] * 26                    # Frequency array for a–z
            for c in word:
                count[ord(c) - ord('a')] += 1   # Increment index for each char
            key = tuple(count)                  # Tuple is hashable — use as key
            groups[key].append(word)

        return list(groups.values())

“This avoids the O(k log k) sort per word. The key is now a fixed-length tuple of 26 integers. Time drops to O(n·k) — strictly linear in total characters.”

Visual Walkthrough

Tracing through ["eat","tea","tan","ate","nat","bat"]:

Step 4: Highlight Edge Cases

Step 5: State Time and Space Complexity

Let n = number of strings, k = average length of each string.

Space is O(n·k) for both because we store all original strings in the HashMap values.

Interview Checklist for Group Anagrams

• Restate — “group strings that share the same character frequencies”
• State the key insight — sorted characters = canonical HashMap key
• Mention brute force O(n²·k) comparison, then dismiss it
• Explain the sorted-key HashMap approach in plain English
• Walk through the visual — show how eat/tea/ate → “aet”
• Use defaultdict(list) and explain why
• Handle empty string, single char, no anagrams, all anagrams edge cases
• State O(n·k log k) time and O(n·k) space for sorted approach
• Mention the frequency-tuple key as an O(n·k) follow-up optimization
• Ask before coding, don’t just dive in

Related Problems to Practice

If you can solve Group Anagrams cleanly, try these next: