Anagram Check Techniques in DSA Go - Build from Scratch

Anagram Check Techniques

📖 Scenario: You work in a word game company. You need to check if two words are anagrams. Anagrams are words made by rearranging the letters of another word, like "listen" and "silent".

🎯 Goal: Build a simple Go program that checks if two given words are anagrams using a letter counting technique.

📋 What You'll Learn

Create two string variables with exact words

Create a map to count letters

Write a loop to count letters in the first word

Write a loop to decrease counts using the second word

Check if all counts are zero to confirm anagram

Print true if anagrams, else false

💡 Why This Matters

🌍 Real World

Checking anagrams is useful in word games, puzzles, and text analysis tools.

💼 Career

Understanding maps and loops is essential for many programming jobs, especially in data processing and algorithm design.

1

Create two string variables for the words

Create two string variables called word1 and word2 with values "listen" and "silent" respectively.

DSA Go

# Your code here

Hint

Use := to create and assign string variables.

2

Create a map to count letters in the first word

Create a variable called letterCount as a map from rune to int. Initialize it with make(map[rune]int).

DSA Go

package main

func main() {
    word1 := "listen"
    word2 := "silent"
    // Create letterCount map here
    
    // Your code here
}

Hint

Use make(map[rune]int) to create an empty map for counting letters.

3

Count letters in word1 and adjust counts with word2

Use a for loop with variable ch to iterate over word1 and increase letterCount[ch] by 1. Then use another for loop with variable ch to iterate over word2 and decrease letterCount[ch] by 1.

DSA Go

package main

func main() {
    word1 := "listen"
    word2 := "silent"
    letterCount := make(map[rune]int)

    // Count letters in word1
    
    // Your code here

    // Decrease counts using word2
    
    // Your code here
}

Hint

Use for _, ch := range word to loop over each letter.

4

Check if all counts are zero and print result

Use a for loop with variables _, count to iterate over letterCount. If any count is not zero, print false and return. After the loop, print true.

DSA Go

package main

func main() {
    word1 := "listen"
    word2 := "silent"
    letterCount := make(map[rune]int)

    for _, ch := range word1 {
        letterCount[ch]++
    }

    for _, ch := range word2 {
        letterCount[ch]--
    }

    // Check counts and print result
    
    // Your code here
}

Hint

If any letter count is not zero, the words are not anagrams.

Practice

1. What is the main idea behind checking if two strings are anagrams?

easy

A. Both strings have the same characters with the same frequency

B. Both strings have the same length only

C. Both strings start with the same character

D. Both strings contain vowels only

Solution

1. Step 1: Understand anagram definition

   Anagrams are words or phrases made by rearranging letters of another, so they must have the same characters with the same counts.

2. Step 2: Compare character frequency

   Checking if two strings have the same frequency of each character confirms if they are anagrams.

3. Final Answer:

   Both strings have the same characters with the same frequency -> Option A

4. Quick Check:

   Character frequency equality = Anagram check [OK]

Hint: Check if sorted strings are equal [OK]

Common Mistakes:

• Only comparing string lengths
• Ignoring character counts
• Checking first characters only

2. Which Go code snippet correctly checks if two strings are anagrams by sorting?

easy

A. sort.Strings([]string{s1, s2}); return s1 == s2

B. s1Sorted := sort.Strings(strings.Split(s1, "")); s2Sorted := sort.Strings(strings.Split(s2, "")); return s1Sorted == s2Sorted

C. s1Sorted := strings.Split(s1, ""); sort.Strings(s1Sorted); s2Sorted := strings.Split(s2, ""); sort.Strings(s2Sorted); return strings.Join(s1Sorted, "") == strings.Join(s2Sorted, "")

D. return s1 == s2

Solution

1. Step 1: Split strings into slices of characters

   Use strings.Split(s, "") to get slice of characters for sorting.

2. Step 2: Sort slices and join back to strings

   Sort each slice with sort.Strings and join with strings.Join to compare sorted strings.

3. Final Answer:

   Splitting, sorting slices, then joining for comparison -> Option C

4. Quick Check:

   Sort slices and join = Correct anagram check [OK]

Hint: Split, sort slices, join, then compare [OK]

Common Mistakes:

• Trying to sort strings directly (not allowed)
• Comparing unsorted strings
• Using sort.Strings on whole string instead of slice

3. What is the output of this Go code snippet?

func areAnagrams(s1, s2 string) bool {
    if len(s1) != len(s2) {
        return false
    }
    count := make(map[rune]int)
    for _, ch := range s1 {
        count[ch]++
    }
    for _, ch := range s2 {
        count[ch]--
        if count[ch] < 0 {
            return false
        }
    }
    return true
}

fmt.Println(areAnagrams("listen", "silent"))
fmt.Println(areAnagrams("hello", "bello"))

medium

A. true false

B. false true

C. true true

D. false false

Solution

1. Step 1: Count characters in first string

   Map counts characters in "listen": l=1, i=1, s=1, t=1, e=1, n=1.

2. Step 2: Decrease counts for second string and check negatives

   For "silent", counts decrease to zero without negatives, so true. For "bello", 'b' not in map or count negative, so false.

3. Final Answer:

   true false -> Option A

4. Quick Check:

   Map count balance = true/false [OK]

Hint: Use map counts to verify character balance [OK]

Common Mistakes:

• Not checking length first
• Not returning false on negative count
• Confusing rune and byte iteration

4. Identify the bug in this Go function that checks anagrams:

func isAnagram(s1, s2 string) bool {
    if len(s1) != len(s2) {
        return false
    }
    count := make(map[rune]int)
    for i := 0; i < len(s1); i++ {
        count[rune(s1[i])]++
        count[rune(s2[i])]--
    }
    for _, v := range count {
        if v != 0 {
            return false
        }
    }
    return true
}

medium

A. Not checking string lengths before counting

B. Using rune conversion on byte indexing may miscount multibyte characters

C. Missing return statement at the end

D. Using map with int keys instead of rune keys

Solution

1. Step 1: Understand rune vs byte indexing in Go strings

   Indexing strings by byte (s1[i]) may split multibyte characters incorrectly, causing wrong rune counts.

2. Step 2: Identify correct iteration method

   Use range over string to get runes properly instead of byte indexing with rune conversion.

3. Final Answer:

   Using rune conversion on byte indexing may miscount multibyte characters -> Option B

4. Quick Check:

   Byte indexing ≠ rune iteration for Unicode [OK]

Hint: Use 'for _, r := range s' to handle runes correctly [OK]

Common Mistakes:

• Assuming byte indexing works for all characters
• Ignoring Unicode multibyte characters
• Not checking map values properly

5. Given a list of strings, which Go approach efficiently groups all anagrams together?

hard

A. Use string length as map key to group strings

B. Compare every pair of strings with nested loops checking character counts

C. Count vowels in each string and group by vowel count

D. Sort each string's characters and use the sorted string as a map key

Solution

1. Step 1: Understand grouping anagrams by sorted keys

   Sorting characters of each string creates a unique key for all anagrams, allowing grouping in a map efficiently.

2. Step 2: Compare with other methods

   Nested loops are slow (O(n²)), length or vowel count grouping is inaccurate for anagrams.

3. Final Answer:

   Sort each string's characters and use the sorted string as a map key -> Option D

4. Quick Check:

   Sorted string keys = efficient anagram grouping [OK]

Hint: Sort strings and group by sorted key [OK]

Common Mistakes:

• Using nested loops causing slow performance
• Grouping by length only misses anagrams
• Grouping by vowels ignores consonants