Group Anagrams Using Hash Map in DSA Javascript - 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.

Specifically, how does the code handle many words and their letters efficiently?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function groupAnagrams(words) {
  const map = new Map();
  for (const word of words) {
    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: The outer loop runs once per word (n times). Sorting runs once per word on word length k.

How Execution Grows With Input

As the number of words (n) grows, the code sorts each word of length k, so work grows with both n and k.

Input Size (n)	Approx. Operations
10 (words)	About 10 times sorting each word (k log k per word)
100 (words)	About 100 times sorting each word
1000 (words)	About 1000 times sorting each word

Pattern observation: The total work grows linearly with the number of words and depends on sorting each word's letters.

Final Time Complexity

Time Complexity: O(n * k log k)

This means the time grows with the number of words and the cost to sort each word's letters.

Common Mistake

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

[OK] Correct: Sorting depends on word length k, so sorting each word takes about k log k time, which adds up across all words.

Interview Connect

Understanding this time complexity helps you explain how your solution scales with input size, a key skill in interviews.

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. Group words by their length using a hash map

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

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

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

Solution

1. Step 1: Understand anagrams

   Anagrams are words with the same letters in different orders, so sorting letters gives a unique key.

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.

3. Final Answer:

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

4. Quick Check:

   Sorted word keys group anagrams = C [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 of the following JavaScript code snippets correctly initializes an empty hash map to group anagrams?

easy

A. const map = {};

B. const map = Map();

C. const map = [];

D. const map = new Set();

Solution

1. Step 1: Identify hash map initialization in JavaScript

   JavaScript objects ({}) are commonly used as hash maps for key-value pairs.

2. Step 2: Analyze the options

   {} creates a simple hash map using bracket notation like map[key]. Map() without new throws TypeError. [] creates an array unsuitable for key-value pairs. new Set() stores unique values, not mappings.

3. Final Answer:

   const map = {}; -> Option A

4. Quick Check:

   Object literal {} is simplest hash map = A [OK]

Hint: Use {} to create a simple hash map in JavaScript [OK]

Common Mistakes:

• Using array [] instead of object
• Using Set which doesn't store key-value pairs
• Confusing Map and object initialization

3. What is the output of the following code?

function groupAnagrams(words) {
  const map = {};
  for (const word of words) {
    const sorted = word.split('').sort().join('');
    if (!map[sorted]) map[sorted] = [];
    map[sorted].push(word);
  }
  return Object.values(map);
}

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

medium

A. SyntaxError

B. [['bat'], ['tab'], ['eat'], ['tea'], ['ate']]

C. [['bat', 'tab'], ['eat', 'tea', 'ate']]

D. [['bat', 'tab', 'eat', 'tea', 'ate']]

Solution

1. Step 1: Sort each word and group in map

   Words 'bat' and 'tab' both sort to 'abt', so grouped together. 'eat', 'tea', 'ate' all sort to 'aet', grouped together.

2. Step 2: Return grouped anagrams as array of arrays

   Object.values(map) returns arrays of grouped words: [['bat', 'tab'], ['eat', 'tea', 'ate']].

3. Final Answer:

   [['bat', 'tab'], ['eat', 'tea', 'ate']] -> Option C

4. Quick Check:

   Sorted keys group words = [['bat','tab'],['eat','tea','ate']] [OK]

Hint: Sort words, group by sorted key, return grouped arrays [OK]

Common Mistakes:

• Returning flat list instead of grouped arrays
• Not sorting words before grouping
• Syntax errors from missing brackets

4. Find the error in this code snippet for grouping anagrams:

function groupAnagrams(words) {
  const map = {};
  for (const word of words) {
    const sorted = word.sort().join('');
    if (!map[sorted]) map[sorted] = [];
    map[sorted].push(word);
  }
  return Object.values(map);
}

medium

A. map should be initialized as a Map object

B. word.sort() is invalid because word is a string, not an array

C. Missing return statement

D. Using Object.values on map is incorrect

Solution

1. Step 1: Identify the error in sorting

   word is a string, and strings do not have a sort() method. It must be split into an array first.

2. Step 2: Correct approach to sort letters

   Use word.split('').sort().join('') to sort letters of the string correctly.

3. Final Answer:

   word.sort() is invalid because word is a string, not an array -> Option B

4. Quick Check:

   Strings need split() before sort() = B [OK]

Hint: Split string to array before sorting letters [OK]

Common Mistakes:

• Calling sort() directly on string
• Not splitting string before sorting
• Confusing Map object with plain object

5. Given the input array ['listen', 'silent', 'enlist', 'google', 'gogole', 'abc', 'cab'], which of the following is the correct grouped anagrams output using a hash map approach?

hard

A. [['listen', 'silent', 'enlist'], ['google', 'gogole'], ['abc', 'cab']]

B. [['listen', 'silent'], ['enlist', 'google'], ['gogole', 'abc', 'cab']]

C. [['listen'], ['silent'], ['enlist'], ['google'], ['gogole'], ['abc'], ['cab']]

D. [['listen', 'silent', 'enlist', 'google', 'gogole', 'abc', 'cab']]

Solution

1. Step 1: Sort each word to find keys

   'listen', 'silent', 'enlist' all sort to 'eilnst'; 'google', 'gogole' sort to 'eggloo'; 'abc', 'cab' sort to 'abc'.

2. Step 2: Group words by sorted keys

   Group words with same sorted key together: [['listen', 'silent', 'enlist'], ['google', 'gogole'], ['abc', 'cab']].

3. Final Answer:

   [['listen', 'silent', 'enlist'], ['google', 'gogole'], ['abc', 'cab']] -> Option A

4. Quick Check:

   Sorted keys group anagrams correctly = A [OK]

Hint: Sort letters, group by key, collect arrays [OK]

Common Mistakes:

• Mixing groups with different sorted keys
• Grouping by word length instead of letters
• Returning flat list instead of grouped arrays