Group Anagrams Using Hash Map in DSA Typescript - Build from Scratch

Group Anagrams Using Hash Map

📖 Scenario: You work in a word game company. You want to group words that are anagrams of each other to help players find word matches quickly.

🎯 Goal: Build a program that groups a list of words into groups where each group contains words that are anagrams of each other.

📋 What You'll Learn

Create an array of words

Create a hash map (object) to group anagrams

Use a loop to process each word and group them by sorted characters

Print the grouped anagrams as arrays

💡 Why This Matters

🌍 Real World

Grouping anagrams helps in word games, search engines, and spell checkers to find similar words quickly.

💼 Career

Understanding hash maps and grouping techniques is essential for software developers working with data processing and optimization.

1

Create the list of words

Create an array called words with these exact strings: "eat", "tea", "tan", "ate", "nat", "bat".

DSA Typescript

// Create the array called words with the given strings
// Your code here

Hint

Use const words = ["eat", "tea", "tan", "ate", "nat", "bat"] to create the array.

2

Create the hash map to group anagrams

Create an empty object called anagramGroups to store groups of anagrams.

DSA Typescript

const words = ["eat", "tea", "tan", "ate", "nat", "bat"]
// Create an empty object called anagramGroups
// Your code here

Hint

Use const anagramGroups: { [key: string]: string[] } = {} to create an empty object.

3

Group words by sorted characters

Use a for loop with variable word to iterate over words. Inside the loop, create a variable sortedWord that sorts the characters of word. Then, add word to the array in anagramGroups[sortedWord], creating the array if it does not exist.

DSA Typescript

const words = ["eat", "tea", "tan", "ate", "nat", "bat"]
const anagramGroups: { [key: string]: string[] } = {}

// Use a for loop to group words by sorted characters
// Your code here

Hint

Split the word into characters, sort them, join back to get sortedWord. Use if (!anagramGroups[sortedWord]) to create array if missing.

4

Print the grouped anagrams

Print the values of anagramGroups using console.log(Object.values(anagramGroups)).

DSA Typescript

const words = ["eat", "tea", "tan", "ate", "nat", "bat"]
const anagramGroups: { [key: string]: string[] } = {}

for (const word of words) {
    const sortedWord = word.split("").sort().join("")
    if (!anagramGroups[sortedWord]) {
        anagramGroups[sortedWord] = []
    }
    anagramGroups[sortedWord].push(word)
}

// Print the grouped anagrams
// Your code here

Hint

Use console.log(Object.values(anagramGroups)) to print the grouped anagrams as arrays.

Practice

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

easy

A. Count the number of vowels in each word and group by count

B. Use sorted strings as keys to group words with the same letters

C. Group words by their length using a hash map

D. Use the first letter of each word as the key in the hash map

Solution

1. Step 1: Understand anagrams

   Anagrams are words with the same letters in different orders.

2. Step 2: Use sorted string as key

   Sorting letters of each word gives a unique key for all its anagrams.

3. Final Answer:

   Use sorted strings as keys to group words with the same letters -> Option B

4. Quick Check:

   Sorted word key groups anagrams = A [OK]

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

Common Mistakes:

• Grouping by word length only
• Using first letter as key
• Counting vowels instead of letters

2. Which TypeScript syntax correctly initializes a hash map to group anagrams?

easy

A. const map = new Map();

B. const map = new Map();

C. const map = new Map();

D. const map = new Map();

Solution

1. Step 1: Identify key and value types

   Keys are sorted strings, values are arrays of strings (anagrams).

2. Step 2: Match types in Map initialization

   Map matches key as string and value as string array.

3. Final Answer:

   const map = new Map(); -> Option A

4. Quick Check:

   Map key=string, value=string array = D [OK]

Hint: Map keys are strings; values are arrays of strings [OK]

Common Mistakes:

• Using number as key instead of string
• Using number array as value
• Confusing value type with number

3. What is the output of this TypeScript code?

function groupAnagrams(strs: string[]): string[][] {
  const map = new Map();
  for (const str of strs) {
    const key = str.split('').sort().join('');
    if (!map.has(key)) map.set(key, []);
    map.get(key)!.push(str);
  }
  return Array.from(map.values());
}

console.log(groupAnagrams(['eat', 'tea', 'tan', 'ate', 'nat', 'bat']));

medium

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

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

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

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

Solution

1. Step 1: Sort each word to create keys

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

2. Step 2: Group words by keys in map

   Map groups: "aet" -> ["eat", "tea", "ate"], "ant" -> ["tan", "nat"], "abt" -> ["bat"].

3. Final Answer:

   [["eat", "tea", "ate"], ["tan", "nat"], ["bat"]] -> Option C

4. Quick Check:

   Sorted keys group anagrams = B [OK]

Hint: Sort letters to find groups, then collect values [OK]

Common Mistakes:

• Mixing groups by length or first letter
• Returning individual words instead of groups
• Incorrect grouping order

4. Identify the bug in this TypeScript code for grouping anagrams:

function groupAnagrams(strs: string[]): string[][] {
  const map = new Map();
  for (const str of strs) {
    const key = str.split('').sort();
    if (!map.has(key)) map.set(key, []);
    map.get(key)!.push(str);
  }
  return Array.from(map.values());
}

medium

A. The key is an array, should be a string after join()

B. map.get(key) returns undefined, should check with has() first

C. Using Map instead of plain object causes error

D. Missing return type annotation on function

Solution

1. Step 1: Check key creation

   str.split('').sort() returns an array, not a string key.

2. Step 2: Fix key by joining sorted array

   Use str.split('').sort().join('') to create string key for map.

3. Final Answer:

   The key is an array, should be a string after join() -> Option A

4. Quick Check:

   Key must be string, not array = A [OK]

Hint: Join sorted array to form string key [OK]

Common Mistakes:

• Using array as map key without joining
• Ignoring map.has() check
• Confusing Map with plain object

5. Given the input array ['abc', 'bca', 'cab', 'xyz', 'zyx', 'zxy', 'aabb', 'baba'], which code snippet correctly groups anagrams using a hash map and filters out groups with less than 3 words?

hard

A. const map = new Map();
for (const str of strs) {
const key = str[0];
if (!map.has(key)) map.set(key, []);
map.get(key)!.push(str);
}
return Array.from(map.values()).filter(group => group.length >= 3);

B. const map = new Map();
for (const str of strs) {
const key = str.length.toString();
if (!map.has(key)) map.set(key, []);
map.get(key)!.push(str);
}
return Array.from(map.values()).filter(group => group.length >= 3);

C. const map = new Map();
for (const str of strs) {
const key = str.split('').reverse().join('');
if (!map.has(key)) map.set(key, []);
map.get(key)!.push(str);
}
return Array.from(map.values()).filter(group => group.length >= 3);

D. const map = new Map();
for (const str of strs) {
const key = str.split('').sort().join('');
if (!map.has(key)) map.set(key, []);
map.get(key)!.push(str);
}
return Array.from(map.values()).filter(group => group.length >= 3);

Solution

1. Step 1: Create keys by sorting letters

   Sorting letters groups anagrams correctly, e.g., "abc", "bca", "cab" all become "abc".

2. Step 2: Filter groups with 3 or more words

   Use filter to keep only groups where length >= 3.

3. Final Answer:

   Code snippet using sorted keys and filtering groups by length >= 3 -> Option D

4. Quick Check:

   Sort keys + filter groups >= 3 = C [OK]

Hint: Sort letters for keys, then filter groups by size [OK]

Common Mistakes:

• Using word length or first letter as key
• Reversing letters instead of sorting
• Not filtering groups by size