Anagram Check Techniques in DSA Rust - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of sorting-based anagram check

What is the output of this Rust code that checks if two strings are anagrams by sorting their characters?

DSA Rust

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_unstable();
    v2.sort_unstable();
    v1 == v2
}

fn main() {
    let result = are_anagrams("listen", "silent");
    println!("{}", result);
}

A true

B false

C Compilation error

D Runtime panic

💡 Hint

Sorting characters and comparing vectors checks if strings have same letters.

The function sorts both strings' characters and compares them. Since "listen" and "silent" have the same letters, the result is true.

❓ Predict Output

intermediate

Output of frequency count anagram check

What does this Rust code print when checking if two strings are anagrams using character frequency counts?

DSA Rust

use std::collections::HashMap;

fn are_anagrams(s1: &str, s2: &str) -> bool {
    if s1.len() != s2.len() {
        return false;
    }
    let mut freq = HashMap::new();
    for c in s1.chars() {
        *freq.entry(c).or_insert(0) += 1;
    }
    for c in s2.chars() {
        let count = freq.entry(c).or_insert(0);
        if *count == 0 {
            return false;
        }
        *count -= 1;
    }
    true
}

fn main() {
    let result = are_anagrams("triangle", "integral");
    println!("{}", result);
}

A true

B Compilation error

C false

D Runtime panic

💡 Hint

Counting characters and comparing counts confirms anagrams.

The code counts characters in first string, then decreases counts for second string. If all counts match, strings are anagrams.

❓ Predict Output

advanced

Output of anagram check ignoring case and spaces

What is the output of this Rust code that checks if two strings are anagrams ignoring case and spaces?

DSA Rust

fn normalize(s: &str) -> Vec<char> {
    let mut chars: Vec<char> = s.chars()
        .filter(|c| !c.is_whitespace())
        .map(|c| c.to_ascii_lowercase())
        .collect();
    chars.sort_unstable();
    chars
}

fn are_anagrams(s1: &str, s2: &str) -> bool {
    normalize(s1) == normalize(s2)
}

fn main() {
    let result = are_anagrams("Dormitory", "Dirty room");
    println!("{}", result);
}

A Runtime panic

B false

C true

D Compilation error

💡 Hint

Removing spaces and ignoring case helps detect anagrams with different formatting.

The code normalizes both strings by removing spaces and converting to lowercase, then sorts characters. Both normalized vectors match.

❓ Predict Output

advanced

Output of anagram check with Unicode characters

What does this Rust code print when checking if two Unicode strings are anagrams?

DSA Rust

use std::collections::HashMap;

fn are_anagrams(s1: &str, s2: &str) -> bool {
    if s1.chars().count() != s2.chars().count() {
        return false;
    }
    let mut freq = HashMap::new();
    for c in s1.chars() {
        *freq.entry(c).or_insert(0) += 1;
    }
    for c in s2.chars() {
        let count = freq.entry(c).or_insert(0);
        if *count == 0 {
            return false;
        }
        *count -= 1;
    }
    true
}

fn main() {
    let result = are_anagrams("résumé", "sérumé");
    println!("{}", result);
}

A false

B true

C Compilation error

D Runtime panic

💡 Hint

Counting Unicode characters correctly is key for anagram checks with accents.

The code counts Unicode characters properly and compares frequencies. Both strings have same characters with accents.

🧠 Conceptual

expert

Choosing the best anagram check method for large inputs

Which anagram check technique is most efficient for very large strings with only ASCII letters?

A Converting strings to sets and comparing sets for equality

B Using nested loops to check each character presence in the other string

C Sorting both strings and comparing sorted results

D Counting character frequencies using a fixed-size array and comparing counts

💡 Hint

Consider time complexity and fixed character set size.

Counting frequencies with a fixed-size array (size 26 for ASCII letters) is O(n) and uses constant space, faster than sorting O(n log n). Nested loops are O(n²). Sets ignore duplicates and fail for anagrams with repeated letters.

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