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

Time Complexity: Group Anagrams Using Hash Map

O(n x 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.

func groupAnagrams(strs []string) [][]string {
    m := make(map[string][]string)
    for _, s := range strs {
        chars := []rune(s)
        sort.Slice(chars, func(i, j int) bool { return chars[i] < chars[j] })
        key := string(chars)
        m[key] = append(m[key], s)
    }
    result := [][]string{}
    for _, v := range m {
        result = append(result, v)
    }
    return result
}
    

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

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Loop over all words and sorting each word's 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) inside a loop.

Input Size (n)	Approx. Operations
10 words, avg length 5	10 x 5 log 5 ≈ 10 x 11.6 = 116
100 words, avg length 5	100 x 5 log 5 ≈ 100 x 11.6 = 1160
1000 words, avg length 5	1000 x 5 log 5 ≈ 1000 x 11.6 = 11600

Pattern observation: The operations grow roughly linearly with the number of words, multiplied by sorting cost per word.

Final Time Complexity

Time Complexity: O(n x k log k)

This means the time grows linearly with the number of words and depends on sorting each word's letters.

Common Mistake

[X] Wrong: "The time complexity is just O(n) because we only loop once over the words."

[OK] Correct: Sorting each word takes extra time depending on word length, so the total time depends on both number of words and word length.

Interview Connect

Understanding this complexity 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. Use the last letter of each word as the key

B. Use the length of words as keys to group anagrams

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

D. Use the first letter of each word as the key

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 C

4. Quick Check:

   Sorted string key = group anagrams [OK]

Hint: Sort letters to find anagram groups quickly [OK]

Common Mistakes:

• Using word length as key misses different anagrams
• Using first or last letter groups unrelated words
• Not sorting letters before using as key

2. Which Go code snippet correctly initializes a hash map to group anagrams?

easy

A. groups := make(map[string][]string)

B. groups := map[string]string{}

C. groups := make([]string, 0)

D. groups := map[int][]string{}

Solution

1. Step 1: Identify map key and value types

   Key should be string (sorted word), value should be slice of strings (anagrams).

2. Step 2: Check Go map initialization syntax

   make(map[string][]string) creates a map with string keys and string slice values.

3. Final Answer:

   groups := make(map[string][]string) -> Option A

4. Quick Check:

   Correct map type and initialization [OK]

Hint: Use map[string][]string for grouping anagrams [OK]

Common Mistakes:

• Using map[string]string instead of map[string][]string
• Using slice instead of map
• Using wrong key or value types

3. What is the output of the following Go code?

package main
import (
  "fmt"
  "sort"
  "strings"
)
func groupAnagrams(strs []string) map[string][]string {
  groups := make(map[string][]string)
  for _, word := range strs {
    letters := strings.Split(word, "")
    sort.Strings(letters)
    key := strings.Join(letters, "")
    groups[key] = append(groups[key], word)
  }
  return groups
}
func main() {
  input := []string{"bat", "tab", "eat", "tea"}
  result := groupAnagrams(input)
  fmt.Println(result)
}

medium

A. map[bat:[bat] tab:[tab] eat:[eat] tea:[tea]]

B. map[abt:[bat] tab:[tab] aet:[eat] tea:[tea]]

C. map[bat:[bat tab] eat:[eat tea]]

D. map[abt:[bat tab] aet:[eat tea]]

Solution

1. Step 1: Sort letters of each word

   "bat" and "tab" sorted to "abt"; "eat" and "tea" sorted to "aet".

2. Step 2: Group words by sorted key

   Words with key "abt" grouped as ["bat", "tab"], with key "aet" grouped as ["eat", "tea"].

3. Final Answer:

   map[abt:[bat tab] aet:[eat tea]] -> Option D

4. Quick Check:

   Sorted keys group anagrams correctly [OK]

Hint: Sorted keys map to anagram groups [OK]

Common Mistakes:

• Not sorting letters before grouping
• Using original words as keys
• Incorrect map output format

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

func groupAnagrams(strs []string) map[string][]string {
  groups := make(map[string][]string)
  for _, word := range strs {
    letters := strings.Split(word, "")
    sort.Strings(letters)
    key := letters[0]
    groups[key] = append(groups[key], word)
  }
  return groups
}

medium

A. Appending word to wrong map key

B. Using letters[0] as key instead of full sorted string

C. Not initializing the map before use

D. Missing import for strings package

Solution

1. Step 1: Check key creation logic

   Key is set as letters[0], only first letter of sorted word, not full sorted string.

2. Step 2: Understand impact on grouping

   Using only first letter groups unrelated words, failing to group anagrams correctly.

3. Final Answer:

   Using letters[0] as key instead of full sorted string -> Option B

4. Quick Check:

   Full sorted string needed as key [OK]

Hint: Use full sorted string as key, not just first letter [OK]

Common Mistakes:

• Using partial key instead of full sorted string
• Forgetting to join letters into string key
• Not sorting letters before key creation

5. Given the input list ["listen", "silent", "enlist", "google", "gogole", "abc"], which Go code snippet correctly returns the grouped anagrams as a slice of slices of strings (ignoring order)?

hard

A. func groupAnagrams(strs []string) [][]string {
groups := make(map[string][]string)
for _, word := range strs {
letters := strings.Split(word, "")
sort.Strings(letters)
key := strings.Join(letters, "")
groups[key] = append(groups[key], word)
}
result := [][]string{}
for _, group := range groups {
result = append(result, group)
}
return result
}

B. func groupAnagrams(strs []string) [][]string {
groups := make(map[int][]string)
for _, word := range strs {
key := len(word)
groups[key] = append(groups[key], word)
}
result := [][]string{}
for _, group := range groups {
result = append(result, group)
}
return result
}

C. func groupAnagrams(strs []string) [][]string {
groups := make(map[string]string)
for _, word := range strs {
letters := strings.Split(word, "")
sort.Strings(letters)
key := strings.Join(letters, "")
groups[key] = word
}
result := [][]string{}
for _, group := range groups {
result = append(result, []string{group})
}
return result
}

D. func groupAnagrams(strs []string) [][]string {
groups := make(map[string][]string)
for _, word := range strs {
key := word
groups[key] = append(groups[key], word)
}
result := [][]string{}
for _, group := range groups {
result = append(result, group)
}
return result
}

Solution

1. Step 1: Use sorted string as key and map to slice of strings

   func groupAnagrams(strs []string) [][]string { groups := make(map[string][]string) for _, word := range strs { letters := strings.Split(word, "") sort.Strings(letters) key := strings.Join(letters, "") groups[key] = append(groups[key], word) } result := [][]string{} for _, group := range groups { result = append(result, group) } return result } correctly sorts letters, joins them as key, and appends words to groups[key].

2. Step 2: Convert map values to slice of slices

   func groupAnagrams(strs []string) [][]string { groups := make(map[string][]string) for _, word := range strs { letters := strings.Split(word, "") sort.Strings(letters) key := strings.Join(letters, "") groups[key] = append(groups[key], word) } result := [][]string{} for _, group := range groups { result = append(result, group) } return result } collects all groups into a slice of slices and returns it.

3. Final Answer:

   Option A code correctly groups anagrams and returns slice of slices -> Option A

4. Quick Check:

   Sorted key + map to slices + collect groups = correct [OK]

Hint: Sort letters, map to slices, then collect groups [OK]

Common Mistakes:

• Using word length as key groups wrong words
• Using map[string]string loses multiple anagrams per key
• Using original word as key misses grouping