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DSA Goprogramming~3 mins

Why Group Anagrams Using Hash Map in DSA Go?

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The Big Idea

What if you could find all matching word puzzles instantly without endless checking?

The Scenario

Imagine you have a big list of words and you want to find which words are anagrams of each other, like "listen" and "silent". Doing this by checking every pair of words one by one is like trying to find matching socks in a huge messy drawer by pulling out two socks at a time and comparing them.

The Problem

Checking every pair of words manually takes a very long time as the list grows. It's easy to make mistakes, miss pairs, or waste time repeating checks. This slow and tiring process makes it hard to organize or find anagrams quickly.

The Solution

Using a hash map, we can group words by a key that represents their letters sorted in order. This way, all anagrams share the same key and get grouped together instantly. It's like sorting socks by color first, so matching pairs are easy to spot without checking each one individually.

Before vs After
✗ Before
for i := 0; i < len(words); i++ {
  for j := i + 1; j < len(words); j++ {
    if areAnagrams(words[i], words[j]) {
      // group them
    }
  }
}
✓ After
groups := make(map[string][]string)
for _, word := range words {
  key := sortLetters(word)
  groups[key] = append(groups[key], word)
}
What It Enables

This method lets you quickly and easily find all groups of anagrams in a list, no matter how big, saving time and effort.

Real Life Example

Spell-checkers and word games use this to suggest words that are anagrams or to find all possible word combinations from a set of letters.

Key Takeaways

Manual pairwise checking is slow and error-prone.

Hash maps group anagrams by a sorted key efficiently.

This approach scales well for large word lists.

Practice

(1/5)
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
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