What is the time complexity of Group Anagrams?

The optimal solution has a time complexity of O(nk).

What data structures are used to solve Group Anagrams?

This problem uses Array, Hash Table, String, Sorting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Group Anagrams?

Insight 1: Anagrams have the same characters with the same frequencies, regardless of order. Insight 2: A hash table (or dictionary) is a suitable data structure to group anagrams. The key is a representation of the character frequencies, and the value is a list of strings that share those frequencies.

Which companies ask LeetCode 49. Group Anagrams in interviews?

This problem is frequently asked at Accenture, Adobe, Affirm, Agoda, Amazon and 75 more companies.

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Group Anagrams - LeetCode 49 Solution

Group Anagrams - Complete Solution Guide

Group Anagrams is LeetCode problem 49, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given an array of strings strs , group the anagrams together. You can return the answer in any order . Example 1: Input: strs = ["eat","tea","tan","ate","nat","bat"] Output: [["bat"],["nat","tan"],["ate","eat","tea"]] Explanation: There is no string in strs that can be rearranged to form "bat" . The strings "nat" and "tan" are anagrams as they can be rearranged to form each other. The strings "ate" , "eat" , and "tea" are anagrams as they can be rearranged to form each other. Example 2: Input: s

Detailed Explanation

The problem asks us to group anagrams from a given list of strings. Anagrams are words or phrases formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once. The input is an array of strings, and the output should be a list of lists, where each inner list contains strings that are anagrams of each other. The order of the output lists and the order of the strings within each list doesn't matter.

Solution Approach

The solution uses a hash table (implemented as a dictionary/map) to group anagrams. For each string in the input list, it calculates a character count. This count, represented as a tuple, array or string, serves as the key in the hash table. If a string with the same character count has already been encountered, the current string is added to the list of anagrams associated with that key. Otherwise, a new entry is created in the hash table with the new key and a list containing the current string.

Step-by-Step Algorithm

Step 1: Initialize an empty hash table (dictionary/map) called `anagram_map`.
Step 2: Iterate through each string `s` in the input list `strs`.
Step 3: For each string `s`, calculate the character count of each letter. This can be done by initializing an array of size 26 (for each lowercase English letter) and incrementing the corresponding index for each character in the string.
Step 4: Convert the character count array to a hashable key. In Python, this is done by converting the list to a tuple. In Java and C++, the count array is converted into a String representation. In C a custom key generation to convert the count to string is implemented.
Step 5: Check if the key exists in the `anagram_map`.
Step 6: If the key exists, append the string `s` to the list of anagrams associated with that key.
Step 7: If the key does not exist, create a new entry in the `anagram_map` with the key and a list containing the string `s`.
Step 8: After iterating through all the strings, collect all the values (lists of anagrams) from the `anagram_map` and return them as a list of lists.

Key Insights

Insight 1: Anagrams have the same characters with the same frequencies, regardless of order.
Insight 2: A hash table (or dictionary) is a suitable data structure to group anagrams. The key is a representation of the character frequencies, and the value is a list of strings that share those frequencies.
Insight 3: Representing the character counts as a string, tuple or array is a common technique for identifying anagrams. The character counts can serve as a unique identifier for each group of anagrams.

Complexity Analysis

Time Complexity: O(nk)

Space Complexity: O(nk)

Topics

This problem involves: Array, Hash Table, String, Sorting.

Companies

Asked at: Accenture, Adobe, Affirm, Agoda, Amazon, Anduril, Apple, Atlassian, Autodesk, BNY Mellon, BP, BlackRock, Bloomberg, Cisco, Citadel, Compass, ConsultAdd, Coupang, Dell, Disney, DoorDash, EPAM Systems, Expedia, Faire, Goldman Sachs, Google, HCL, HashedIn, IBM, Infosys, Intuit, J.P. Morgan, MSCI, Meta, Microsoft, Morgan Stanley, Myntra, NetApp, Niantic, Nielsen, Nutanix, Nvidia, Oracle, Palo Alto Networks, PayPal, Paycom, PhonePe, Publicis Sapient, Qualtrics, Ripple, SAP, Salesforce, ServiceNow, Siemens, Sigmoid, Smartsheet, Snap, Splunk, Tesla, Texas Instruments, TikTok, Turing, Twilio, Uber, Veeva Systems, Verily, Visa, Walmart Labs, Wayfair, Whatnot, Wipro, Workday, Yahoo, Yandex, Yelp, Zoho, Zopsmart, athenahealth, eBay, persistent systems.

Group Anagrams | LeetCode 49

Medium

Array Hash Table String Sorting

Time: O(nk)Space: O(nk)

Solution Code

Available in:

import java.util.*;

class Solution {
    public List<List<String>> groupAnagrams(String[] strs) {
        Map<String, List<String>> anagramMap = new HashMap<>();
        for (String s : strs) {
            int[] count = new int[26];
            for (char c : s.toCharArray()) {
                count[c - 'a']++;
            }
            
            StringBuilder sb = new StringBuilder();
            for(int i = 0; i < 26; i++){
                sb.append(count[i]);
                sb.append("#");
            }

            String key = sb.toString();
            if (!anagramMap.containsKey(key)) {
                anagramMap.put(key, new ArrayList<>());
            }
            anagramMap.get(key).add(s);
        }
        return new ArrayList<>(anagramMap.values());
    }
}

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Problem Info

Description

Problem 49: Group Anagrams - Detailed explanation and solution approaches.

Topics

Array Hash Table String Sorting

Companies

AccentureAdobeAffirmAgodaAmazonAndurilAppleAtlassian

Stats

Difficulty:Medium

Languages:4

Companies:80

Problem #:49

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Initialize an empty hash table (dictionary/map) called `anagram_map`.
2
Step 2: Iterate through each string `s` in the input list `strs`.
3
Step 3: For each string `s`, calculate the character count of each letter. This can be done by initializing an array of size 26 (for each lowercase English letter) and incrementing the corresponding index for each character in the string.
4
Step 4: Convert the character count array to a hashable key. In Python, this is done by converting the list to a tuple. In Java and C++, the count array is converted into a String representation. In C a custom key generation to convert the count to string is implemented.
5
Step 5: Check if the key exists in the `anagram_map`.
6
Step 6: If the key exists, append the string `s` to the list of anagrams associated with that key.
7
Step 7: If the key does not exist, create a new entry in the `anagram_map` with the key and a list containing the string `s`.
8
Step 8: After iterating through all the strings, collect all the values (lists of anagrams) from the `anagram_map` and return them as a list of lists.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(nk)

💾

Space Complexity

O(nk)

📝

Why?

The time complexity is determined by the nested loops: the outer loop iterates through the input strings, and the inner loop iterates through the characters of each string. The space complexity is determined by the number of lists created within the hashmap to store anagrams, which in the worst case, is proportional to the total number of characters across all strings.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Sorting the strings

Another approach is to sort each string and use the sorted string as the key in the hash table. Anagrams will have the same sorted string representation.

Time: O(nk log k), where n is the number of strings and k is the average length of a string. Sorting each string takes O(k log k) time.Space: O(nk), as we are storing the strings in the hashmap.

Trade-offs: This approach is slightly slower than the character counting approach due to the sorting operation. However, it might be more intuitive for some developers. Character counting is generally preferred because of its lower time complexity of O(nk) compared to O(nk log k) of sorting.

✨

Key Insights

1Insight 1: Anagrams have the same characters with the same frequencies, regardless of order.
2Insight 2: A hash table (or dictionary) is a suitable data structure to group anagrams. The key is a representation of the character frequencies, and the value is a list of strings that share those frequencies.
3Insight 3: Representing the character counts as a string, tuple or array is a common technique for identifying anagrams. The character counts can serve as a unique identifier for each group of anagrams.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

String

A sequence of characters. Like a word or sentence in programming!

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Empty string. The solution correctly handles empty strings. An empty string will have a character count of all zeros, and it will be grouped with other empty strings.
Edge case 2: Input list containing only one string. The solution still works correctly. The single string will be added to a list, and the output will be a list containing that list.
Edge case 3: Strings with different lengths. The character count correctly distinguishes strings of different lengths that are not anagrams. For instance, 'abc' and 'abcd' have different character counts.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the character count. For example, not handling different cases of letters (uppercase vs. lowercase) or using incorrect indexing for the character count array.

✗Common mistake 2: Not using a hashable data structure as the key for the hash table. For example, using a list directly as a key in Python will result in a `TypeError`. The list needs to be converted to a tuple or a string.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Spell checking and suggestion. Anagram grouping can be used to identify possible corrections for misspelled words. If a word is not found in a dictionary, its anagrams can be generated and checked for validity.

🏦Real-world scenario 2: Data analysis and pattern recognition. Finding anagrams can be useful for finding patterns or relationships in textual data. For example, identifying common character combinations in names or product descriptions.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 20, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Group Anagrams"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?