Group Anagrams Using Hash Map in DSA Typescript - Time & Space Complexity

Time Complexity: Group Anagrams Using Hash Map

O(n * k log k)

Understanding Time Complexity

We want to understand how the time needed to group anagrams grows as the list of words gets bigger.

How does the program's work increase when we add more words or longer words?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

This code groups words that are anagrams by sorting each word and using the sorted word as a key in a map.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Looping over each word and sorting its letters.
• How many times: Once for each word in the list.

How Execution Grows With Input

As the number of words grows, the program sorts each word, so work grows with both number of words and word length.

Input Size (n)	Approx. Operations
10 words, length 5	About 10 x sorting 5 letters
100 words, length 5	About 100 x sorting 5 letters
1000 words, length 5	About 1000 x sorting 5 letters

Pattern observation: The total work grows roughly linearly with the number of words, but sorting each word depends on its length.

Final Time Complexity

Time Complexity: O(n * k log k)

This means the time grows with the number of words (n) times the cost to sort each word of length k.

Common Mistake

[X] Wrong: "Sorting each word is constant time, so total time is just O(n)."

[OK] Correct: Sorting depends on word length k, so sorting each word takes about k log k time, not constant.

Interview Connect

Understanding this helps you explain how hashing and sorting combine to solve grouping problems efficiently.

Self-Check

"What if we used character count arrays as keys instead of sorting? How would the time complexity change?"

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