Anagram Check Techniques in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if two strings have the same length.

DSA Rust

fn are_anagrams(s1: &str, s2: &str) -> bool {
    if s1.len() [1] s2.len() {
        return false;
    }
    true
}

Drag options to blanks, or click blank then click option'

A==

B!=

C>

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using '==' instead of '!=' causes wrong logic.

We return false if the lengths are not equal, so the condition should check for inequality.

2fill in blank

medium

Complete the code to sort the characters of a string.

DSA Rust

fn sorted_chars(s: &str) -> Vec<char> {
    let mut chars: Vec<char> = s.chars().collect();
    chars.[1]();
    chars
}

Drag options to blanks, or click blank then click option'

Areverse

Bpop

Csort

Dshuffle

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'reverse' only reverses order, doesn't sort.

We use the 'sort' method to arrange characters in ascending order.

3fill in blank

hard

Fix the error in comparing sorted character vectors for anagram check.

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();
    v2.sort();
    v1 [1] v2
}

Drag options to blanks, or click blank then click option'

A>

B!=

C<

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using '!=' causes wrong result.

To check if two strings are anagrams, their sorted character vectors must be equal.

4fill in blank

hard

Fill both blanks to count character frequencies using a HashMap.

DSA Rust

use std::collections::HashMap;

fn char_count(s: &str) -> HashMap<char, usize> {
    let mut map = HashMap::new();
    for c in s.chars() {
        *map.entry(c).[1](0) += 1;
    }
    map
}

fn are_anagrams(s1: &str, s2: &str) -> bool {
    char_count(s1) [2] char_count(s2)
}

Drag options to blanks, or click blank then click option'

Aor_insert

B==

Cinsert

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'insert' instead of 'or_insert' causes panic.

Use 'or_insert(0)' to initialize count, and '==' to compare frequency maps.

5fill in blank

hard

Fill all three blanks to implement an anagram check using character counts.

DSA Rust

use std::collections::HashMap;

fn are_anagrams(s1: &str, s2: &str) -> bool {
    if s1.len() != s2.len() {
        return false;
    }
    let mut count = HashMap::new();
    for c in s1.chars() {
        *count.entry(c).[1](0) += 1;
    }
    for c in s2.chars() {
        let counter = count.entry(c).[2](0);
        if *counter == 0 {
            return false;
        }
        *counter -= [3];
    }
    true
}

Drag options to blanks, or click blank then click option'

Aor_insert

Bor_default

C1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'or_default' instead of 'or_insert' may cause confusion.

Use 'or_insert(0)' to initialize counts and subtract 1 for each character in s2.

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