Anagram Check Techniques in DSA Go - Time & Space Complexity

Time Complexity: Anagram Check Techniques

O(n)

Understanding Time Complexity

We want to understand how the time needed to check if two words are anagrams changes as the words get longer.

How does the work grow when the input size increases?

Scenario Under Consideration

Analyze the time complexity of the following 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
}
    

This code checks if two strings are anagrams by counting characters in the first string and then subtracting counts using the second string.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Two loops each going through all characters of the strings.
• How many times: Each loop runs once over all characters, so twice over the input length.

How Execution Grows With Input

As the strings get longer, the number of steps grows roughly twice as fast because we check each character twice.

Input Size (n)	Approx. Operations
10	About 20 steps
100	About 200 steps
1000	About 2000 steps

Pattern observation: The work grows linearly with input size, doubling because of two passes.

Final Time Complexity

Time Complexity: O(n)

This means the time to check anagrams grows directly in proportion to the length of the strings.

Common Mistake

[X] Wrong: "Sorting both strings is always faster for anagram checks."

[OK] Correct: Sorting takes more time (O(n log n)) compared to counting characters (O(n)), so counting is usually faster.

Interview Connect

Understanding how to efficiently check anagrams shows you can choose the right approach for string problems, a useful skill in many coding challenges.

Self-Check

"What if we used sorting of both strings instead of counting? How would the time complexity change?"

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