Group Anagrams Using Hash Map in DSA C++

Mental Model

Group words that have the same letters by sorting them and using the sorted word as a key.

Analogy: Imagine sorting letters of each word like arranging colored beads in order. Words with the same beads in the same order belong to the same group.

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

hash map:
{
  "aet" -> ["eat", "tea", "ate"],
  "ant" -> ["tan", "nat"],
  "abt" -> ["bat"]
}

Dry Run Walkthrough

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

Goal: Group words that are anagrams together using a hash map

Step 1: Take first word "eat", sort letters to get "aet", add "eat" to hash map under key "aet"

{"aet": ["eat"]}

Why: Sorting letters creates a unique key for all anagrams

Step 2: Take second word "tea", sort letters to get "aet", add "tea" to existing list under key "aet"

{"aet": ["eat", "tea"]}

Why: Words with same sorted letters belong to same group

Step 3: Take third word "tan", sort letters to get "ant", add "tan" to hash map under key "ant"

{"aet": ["eat", "tea"], "ant": ["tan"]}

Why: New sorted key means new group

Step 4: Take fourth word "ate", sort letters to get "aet", add "ate" to list under key "aet"

{"aet": ["eat", "tea", "ate"], "ant": ["tan"]}

Why: Add to existing anagram group

Step 5: Take fifth word "nat", sort letters to get "ant", add "nat" to list under key "ant"

{"aet": ["eat", "tea", "ate"], "ant": ["tan", "nat"]}

Why: Add to existing anagram group

Step 6: Take sixth word "bat", sort letters to get "abt", add "bat" to hash map under key "abt"

{"aet": ["eat", "tea", "ate"], "ant": ["tan", "nat"], "abt": ["bat"]}

Why: New group for different letters

Result:

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

Annotated Code

DSA C++

#include <iostream>
#include <vector>
#include <string>
#include <unordered_map>
#include <algorithm>

using namespace std;

vector<vector<string>> groupAnagrams(vector<string>& strs) {
    unordered_map<string, vector<string>> hashMap;
    for (string& word : strs) {
        string sortedWord = word;
        sort(sortedWord.begin(), sortedWord.end()); // sort letters to get key
        hashMap[sortedWord].push_back(word); // group anagrams under sorted key
    }
    vector<vector<string>> result;
    for (auto& pair : hashMap) {
        result.push_back(pair.second); // collect grouped anagrams
    }
    return result;
}

int main() {
    vector<string> input = {"eat", "tea", "tan", "ate", "nat", "bat"};
    vector<vector<string>> groups = groupAnagrams(input);
    for (const auto& group : groups) {
        cout << "[";
        for (size_t i = 0; i < group.size(); ++i) {
            cout << group[i];
            if (i != group.size() - 1) cout << ", ";
        }
        cout << "]\n";
    }
    return 0;
}

sort(sortedWord.begin(), sortedWord.end()); // sort letters to get key

sort letters of current word to create unique key for anagram group

hashMap[sortedWord].push_back(word); // group anagrams under sorted key

add original word to list under sorted key in hash map

for (auto& pair : hashMap) { result.push_back(pair.second); }

collect all grouped anagrams from hash map into result vector

Output

[eat, tea, ate]

[tan, nat]

[bat]

Complexity Analysis

Time: O(n * k log k) because we sort each of the n words of length k

Space: O(n * k) because we store all words grouped in hash map

vs Alternative: Naive approach compares each word with others, O(n^2 * k), much slower than hashing with sorting

Edge Cases

empty input list

returns empty list with no groups

DSA C++

for (string& word : strs) { ... } handles empty by skipping loop

list with one word

returns list with one group containing that word

DSA C++

hashMap[sortedWord].push_back(word); adds single word group

words with same letters but different cases

treated as different groups because sorting is case sensitive

DSA C++

sort(sortedWord.begin(), sortedWord.end()); does not normalize case

When to Use This Pattern

When you need to group words that are rearrangements of each other, use a hash map with sorted word keys to quickly group anagrams.

Common Mistakes

Mistake: Using the original word as key instead of the sorted word
Fix: Sort the word first to create the key before inserting into the hash map

Mistake: Not sorting the word before using as key
Fix: Always sort letters of the word to get the correct key for grouping

Summary

Groups words that have the same letters by sorting and hashing.

Use when you want to find all anagram groups in a list of words.

Sorting each word creates a unique key that identifies its anagram group.

Practice

1. What is the main idea behind grouping anagrams using a hash map?

easy

A. Use sorted strings as keys to group words with the same characters

B. Use the length of the string as the key to group words

C. Use the first character of the string as the key

D. Use the last character of the string as the key

Solution

1. Step 1: Understand anagram grouping

   Anagrams have the same characters in different orders, so sorting the string gives a unique key for all anagrams.

2. Step 2: Use sorted string as hash map key

   By sorting each word and using it as a key, all anagrams map to the same key and can be grouped together.

3. Final Answer:

   Use sorted strings as keys to group words with the same characters -> Option A

4. Quick Check:

   Sorted string key = group anagrams [OK]

Hint: Sort each word to find its anagram group key [OK]

Common Mistakes:

• Using string length as key misses different anagrams
• Using first or last character groups unrelated words
• Not sorting strings before using as keys

2. Which of the following C++ code snippets correctly declares a hash map to group anagrams where the key is a string and the value is a vector of strings?

easy

A. std::unordered_map anagramGroups;

B. std::map> anagramGroups;

C. std::unordered_map> anagramGroups;

D. std::unordered_map anagramGroups;

Solution

1. Step 1: Identify key and value types

   The key should be a string (sorted word), and the value should be a vector of strings (group of anagrams).

2. Step 2: Match correct declaration

   std::unordered_map> anagramGroups; uses unordered_map with key as string and value as vector<string>, which is correct for grouping anagrams.

3. Final Answer:

   std::unordered_map> anagramGroups; -> Option C

4. Quick Check:

   Key=string, Value=vector<string> [OK]

Hint: Use unordered_map for grouping [OK]

Common Mistakes:

• Using int as key instead of string
• Using vector<char> as key (not hashable)
• Using string as value instead of vector<string>

3. What is the output of the following C++ code snippet?

std::vector words = {"eat", "tea", "tan", "ate", "nat", "bat"};
std::unordered_map> groups;
for (auto &word : words) {
    std::string sortedWord = word;
    std::sort(sortedWord.begin(), sortedWord.end());
    groups[sortedWord].push_back(word);
}
for (auto &pair : groups) {
    std::cout << "Group: ";
    for (auto &w : pair.second) std::cout << w << " ";
    std::cout << std::endl;
}

medium

A. Group: eat tea
Group: tan bat
Group: ate nat

B. Group: eat tea ate
Group: tan nat
Group: bat

C. Group: ate tea
Group: bat tan
Group: eat nat

D. Group: eat tan bat
Group: tea nat ate

Solution

1. Step 1: Sort each word and group by sorted key

   "eat", "tea", "ate" all sort to "aet" so grouped together; "tan", "nat" sort to "ant"; "bat" sorts to "abt".

2. Step 2: Print groups

   Groups printed are: "eat tea ate", "tan nat", and "bat".

3. Final Answer:

   Group: eat tea ate Group: tan nat Group: bat -> Option B

4. Quick Check:

   Sorted keys group anagrams correctly [OK]

Hint: Sort words, group by sorted key, print groups [OK]

Common Mistakes:

• Not sorting words before grouping
• Mixing groups by original word order
• Printing keys instead of grouped words

4. Identify the error in the following code snippet that tries to group anagrams:

std::vector words = {"abc", "bca", "cab"};
std::unordered_map> groups;
for (auto word : words) {
    std::string key = word;
    std::sort(word.begin(), word.end());
    groups[key].push_back(word);
}

medium

A. Sorting is done on 'word' but 'key' is used as map key, causing wrong grouping

B. The vector 'words' is not initialized correctly

C. The map should use vector<char> as key instead of string

D. The loop variable should be a reference to avoid copying

Solution

1. Step 1: Analyze sorting and key usage

   The code sorts 'word' but uses 'key' (unsorted) as the map key, so grouping is incorrect.

2. Step 2: Correct usage

   Sort 'key' and use it as the map key to group anagrams properly.

3. Final Answer:

   Sorting is done on 'word' but 'key' is used as map key, causing wrong grouping -> Option A

4. Quick Check:

   Sort key before using in map [OK]

Hint: Sort the key string, not the original word [OK]

Common Mistakes:

• Sorting wrong variable before using as key
• Using unsorted string as map key
• Ignoring reference vs copy in loop

5. Given a list of strings, how would you modify the grouping anagrams approach to handle case-insensitive grouping (e.g., "Eat" and "tea" should be grouped together)?

hard

A. Sort strings without changing case

B. Use the original string as key without any changes

C. Group by string length only

D. Convert each string to lowercase before sorting and using as key

Solution

1. Step 1: Normalize case for consistent keys

   Convert each string to lowercase so that case differences do not affect grouping.

2. Step 2: Sort the lowercase string and use as key

   Sorting the lowercase string ensures all case variants of anagrams map to the same key.

3. Step 3: Group anagrams using normalized keys

   Use the lowercase sorted string as the hash map key to group anagrams case-insensitively.

4. Final Answer:

   Convert each string to lowercase before sorting and using as key -> Option D

5. Quick Check:

   Lowercase + sort = case-insensitive grouping [OK]

Hint: Lowercase strings before sorting to ignore case [OK]

Common Mistakes:

• Not normalizing case before grouping
• Using original case strings as keys
• Grouping by length or unsorted strings