Why Anagram Check Techniques in DSA Go?

The Big Idea

What if you could instantly tell if two words are made of the same letters without checking each one?

The Scenario

Imagine you want to check if two words are made of the same letters, like "listen" and "silent". Doing this by hand means comparing each letter one by one, which is slow and confusing.

The Problem

Manually checking each letter is slow and easy to make mistakes. You might miss letters or count wrong. It gets worse with longer words or many pairs to check.

The Solution

Anagram check techniques use simple rules and tools to quickly compare words. They count letters or sort them, making the check fast and error-free.

Before vs After

✗ Before

func areAnagramsManual(s1, s2 string) bool {
  // Compare letters one by one manually
  for i := 0; i < len(s1); i++ {
    // manual checks here
  }
  return false
}

✓ After

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
}

What It Enables

This lets you quickly and correctly find if two words are anagrams, even for long words or many checks.

Real Life Example

Spell checkers and word games use anagram checks to find matching words or suggest corrections fast.

Key Takeaways

Manual letter-by-letter checks are slow and error-prone.

Anagram techniques use counting or sorting for fast, reliable checks.

This helps in games, spell checkers, and text analysis.

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