Anagram Check Techniques in DSA Rust - Time & Space Complexity

Time Complexity: Anagram Check Techniques

O(n log n)

Understanding Time Complexity

Checking if two words are anagrams means seeing if they have the same letters in any order.

We want to know how the time needed changes as the words get longer.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn are_anagrams(s1: &str, s2: &str) -> bool {
    if s1.len() != s2.len() {
        return false;
    }
    let mut chars1: Vec = s1.chars().collect();
    let mut chars2: Vec = s2.chars().collect();
    chars1.sort_unstable();
    chars2.sort_unstable();
    chars1 == chars2
}
    

This code checks if two strings are anagrams by sorting their characters and comparing.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Sorting the characters of each string.
• How many times: Sorting runs twice, once per string, each over n characters.

How Execution Grows With Input

Sorting takes more time as the string length grows, roughly n log n steps.

Input Size (n)	Approx. Operations
10	About 10 * log(10) ≈ 33 steps per sort, done twice
100	About 100 * log(100) ≈ 664 steps per sort, done twice
1000	About 1000 * log(1000) ≈ 9966 steps per sort, done twice

Pattern observation: As input grows, operations grow a bit faster than linearly, because sorting is involved.

Final Time Complexity

Time Complexity: O(n log n)

This means the time to check anagrams grows a bit faster than the length of the strings because sorting is the main work.

Common Mistake

[X] Wrong: "Comparing characters one by one without sorting is always faster."

[OK] Correct: Without sorting or counting, you might miss differences or need to check many pairs, which can be slower or more complex.

Interview Connect

Understanding how sorting affects time helps you explain your choices clearly and shows you know how to handle string problems efficiently.

Self-Check

"What if we used a character count array instead of sorting? 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 only alphabets

Solution

1. Step 1: Understand anagram definition

   Anagrams are words or phrases made by rearranging letters of another, so character frequency must match.

2. Step 2: Compare character frequencies

   Checking if both strings have the same characters with the same counts confirms they are anagrams.

3. Final Answer:

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

4. Quick Check:

   Anagram check = same characters & frequency [OK]

Hint: Check if character counts match in both strings [OK]

Common Mistakes:

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

2. Which Rust code snippet correctly sorts the characters of a string s to help check anagrams?

easy

A. let mut chars: Vec<char> = s.chars().collect(); chars.sort();

B. let chars = s.sort();

C. let chars = s.chars().sorted();

D. let chars = s.to_vec().sort();

Solution

1. Step 1: Collect characters into a vector

   Use s.chars().collect() to get a vector of characters from the string.

2. Step 2: Sort the vector in place

   Call chars.sort() to sort characters alphabetically for comparison.

3. Final Answer:

   let mut chars: Vec<char> = s.chars().collect(); chars.sort(); -> Option A

4. Quick Check:

   Collect chars then sort vector [OK]

Hint: Collect chars into Vec then call sort() [OK]

Common Mistakes:

• Trying to sort string directly
• Using non-existent sorted() method
• Calling sort() on &str instead of Vec<char>

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

fn main() {
    let s1 = "listen";
    let s2 = "silent";
    let mut v1: Vec<char> = s1.chars().collect();
    let mut v2: Vec<char> = s2.chars().collect();
    v1.sort();
    v2.sort();
    println!("{}", v1 == v2);
}

medium

A. Runtime panic

B. true

C. Compilation error

D. false

Solution

1. Step 1: Convert strings to char vectors and sort

   Both "listen" and "silent" are converted to vectors and sorted alphabetically.

2. Step 2: Compare sorted vectors

   Sorted vectors for both strings become ['e', 'i', 'l', 'n', 's', 't'], so they are equal.

3. Final Answer:

   true -> Option B

4. Quick Check:

   Sorted chars equal means anagrams [OK]

Hint: Sort both strings and compare vectors [OK]

Common Mistakes:

• Expecting false because strings differ
• Confusing sorting with original order
• Thinking println! prints vectors directly

4. Identify the bug in this Rust function that checks if two strings are anagrams:

fn are_anagrams(s1: &str, s2: &str) -> bool {
    let mut v1: Vec<char> = s1.chars().collect();
    let mut v2: Vec<char> = s2.chars().collect();
    v1.sort();
    v2.sort();
    v1 == v2
}

fn main() {
    println!("{}", are_anagrams("abc", "cba"));
}

medium

A. Using == operator on vectors causes error

B. Missing mutable keyword for vectors

C. Should compare lengths before sorting

D. No bug, function works correctly

Solution

1. Step 1: Review vector creation and sorting

   Vectors are mutable and sorted correctly using v1.sort() and v2.sort().

2. Step 2: Check comparison and usage

   Comparing vectors with == is valid and returns true if sorted characters match.

3. Final Answer:

   No bug, function works correctly -> Option D

4. Quick Check:

   Sorting and comparing vectors works fine [OK]

Hint: Sorting vectors and comparing is valid in Rust [OK]

Common Mistakes:

• Thinking == can't compare vectors
• Assuming length check is mandatory
• Forgetting mut keyword (but it's present here)

5. You want to check if two strings are anagrams ignoring case and spaces. Which Rust approach is best?

hard

A. Sort strings without changing case or removing spaces

B. Compare strings directly after trimming spaces only

C. Convert both strings to lowercase, remove spaces, then sort and compare characters

D. Check if lengths match after removing spaces only

Solution

1. Step 1: Normalize strings by lowering case and removing spaces

   Convert both strings to lowercase and remove all spaces to ignore case and whitespace differences.

2. Step 2: Sort characters and compare

   Sort the cleaned character vectors and compare them to confirm anagram status.

3. Final Answer:

   Convert both strings to lowercase, remove spaces, then sort and compare characters -> Option C

4. Quick Check:

   Normalize case and spaces before sorting [OK]

Hint: Lowercase and remove spaces before sorting [OK]

Common Mistakes:

• Ignoring case differences
• Not removing spaces
• Comparing raw strings directly