What if you could find all matching word puzzles instantly without endless checking?
Why Group Anagrams Using Hash Map in DSA Go?
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.
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.
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.
for i := 0; i < len(words); i++ { for j := i + 1; j < len(words); j++ { if areAnagrams(words[i], words[j]) { // group them } } }
groups := make(map[string][]string)
for _, word := range words {
key := sortLetters(word)
groups[key] = append(groups[key], word)
}This method lets you quickly and easily find all groups of anagrams in a list, no matter how big, saving time and effort.
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.
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
Solution
Step 1: Understand anagrams
Anagrams are words with the same letters in different orders.Step 2: Use sorted string as key
Sorting letters of each word gives a unique key for all its anagrams.Final Answer:
Use sorted strings as keys to group words with the same letters -> Option CQuick Check:
Sorted string key = group anagrams [OK]
- Using word length as key misses different anagrams
- Using first or last letter groups unrelated words
- Not sorting letters before using as key
Solution
Step 1: Identify map key and value types
Key should be string (sorted word), value should be slice of strings (anagrams).Step 2: Check Go map initialization syntax
make(map[string][]string) creates a map with string keys and string slice values.Final Answer:
groups := make(map[string][]string) -> Option AQuick Check:
Correct map type and initialization [OK]
- Using map[string]string instead of map[string][]string
- Using slice instead of map
- Using wrong key or value types
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)
}Solution
Step 1: Sort letters of each word
"bat" and "tab" sorted to "abt"; "eat" and "tea" sorted to "aet".Step 2: Group words by sorted key
Words with key "abt" grouped as ["bat", "tab"], with key "aet" grouped as ["eat", "tea"].Final Answer:
map[abt:[bat tab] aet:[eat tea]] -> Option DQuick Check:
Sorted keys group anagrams correctly [OK]
- Not sorting letters before grouping
- Using original words as keys
- Incorrect map output format
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
}Solution
Step 1: Check key creation logic
Key is set as letters[0], only first letter of sorted word, not full sorted string.Step 2: Understand impact on grouping
Using only first letter groups unrelated words, failing to group anagrams correctly.Final Answer:
Using letters[0] as key instead of full sorted string -> Option BQuick Check:
Full sorted string needed as key [OK]
- Using partial key instead of full sorted string
- Forgetting to join letters into string key
- Not sorting letters before key creation
["listen", "silent", "enlist", "google", "gogole", "abc"], which Go code snippet correctly returns the grouped anagrams as a slice of slices of strings (ignoring order)?Solution
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].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.Final Answer:
Option A code correctly groups anagrams and returns slice of slices -> Option AQuick Check:
Sorted key + map to slices + collect groups = correct [OK]
- Using word length as key groups wrong words
- Using map[string]string loses multiple anagrams per key
- Using original word as key misses grouping
Ready to prove it?
Pass the test and this pattern moves unknown
