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DSA Rustprogramming~10 mins

Comparison Based vs Non Comparison Based Sorting in DSA Rust - Interactive Comparison Practice

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Practice - 5 Tasks
Answer the questions below
1fill in blank
easy

Complete the code to perform a comparison-based sort using Rust's built-in method.

DSA Rust
let mut numbers = vec![5, 3, 8, 1];
numbers.[1]();
println!("{:?}", numbers);
Drag options to blanks, or click blank then click option'
Asort
Breverse
Cpush
Dpop
Attempts:
3 left
šŸ’” Hint
Common Mistakes
Using reverse instead of sort.
Trying to use push or pop which modify the vector size.
2fill in blank
medium

Complete the code to perform a non-comparison based sort using counting sort logic for small integers.

DSA Rust
fn counting_sort(arr: &mut [usize], max_val: usize) {
    let mut count = vec![0; max_val + 1];
    for &num in arr.iter() {
        count[[1]] += 1;
    }
    let mut idx = 0;
    for num in 0..=max_val {
        for _ in 0..count[num] {
            arr[idx] = num;
            idx += 1;
        }
    }
}

let mut data = vec![2, 5, 3, 0, 2, 3, 0, 3];
counting_sort(&mut data, 5);
println!("{:?}", data);
Drag options to blanks, or click blank then click option'
Aarr[num]
Bidx
Cnum
Dmax_val
Attempts:
3 left
šŸ’” Hint
Common Mistakes
Using idx instead of num as index.
Using arr[num] which is incorrect syntax here.
3fill in blank
hard

Fix the error in the code to correctly implement radix sort for sorting non-negative integers.

DSA Rust
fn counting_sort_for_radix(arr: &mut [u32], exp: u32) {
    let mut output = vec![0; arr.len()];
    let mut count = vec![0; 10];

    for &num in arr.iter() {
        count[((num / [1]) % 10) as usize] += 1;
    }

    for i in 1..10 {
        count[i] += count[i - 1];
    }

    for &num in arr.iter().rev() {
        let index = ((num / (exp)) % 10) as usize;
        output[count[index] - 1] = num;
        count[index] -= 1;
    }

    for i in 0..arr.len() {
        arr[i] = output[i];
    }
}

let mut data = vec![170, 45, 75, 90, 802, 24, 2, 66];
counting_sort_for_radix(&mut data, 10);
println!("{:?}", data);
Drag options to blanks, or click blank then click option'
Aarr.len()
B10
C1
Dexp
Attempts:
3 left
šŸ’” Hint
Common Mistakes
Using a fixed number like 10 instead of exp.
Using array length or 1 which do not represent the digit place.
4fill in blank
hard

Fill both blanks to create a dictionary comprehension that filters and transforms a list of numbers to their squares if they are even.

DSA Rust
let numbers = vec![1, 2, 3, 4, 5];
let squares: std::collections::HashMap<_, _> = numbers.iter()
    .filter(|&&x| x [1] 2 == 0)
    .map(|&x| (x, x [2] x))
    .collect();

println!("{:?}", squares);
Drag options to blanks, or click blank then click option'
A%
B*
C+
D-
Attempts:
3 left
šŸ’” Hint
Common Mistakes
Using addition or subtraction instead of multiplication for square.
Using wrong operator for even check.
5fill in blank
hard

Fill all three blanks to create a HashMap comprehension that stores uppercase keys and their values only if the value is positive.

DSA Rust
let data = std::collections::HashMap::from([
    ("apple", 3),
    ("banana", -1),
    ("cherry", 5),
]);

let filtered: std::collections::HashMap<_, _> = data.iter()
    .filter(|&(_, &v)| v [1] 0)
    .map(|(&k, &v)| (k.[2](), v))
    .collect();

println!("{:?}", filtered);
Drag options to blanks, or click blank then click option'
A>
Bto_uppercase
C<
Dlen
Attempts:
3 left
šŸ’” Hint
Common Mistakes
Using < instead of > for filtering.
Using len instead of to_uppercase.
Not filtering values correctly.

Practice

(1/5)
1. Which of the following sorting algorithms is non-comparison based?
easy
A. Heap Sort
B. Merge Sort
C. Quick Sort
D. Counting Sort

Solution

  1. Step 1: Understand comparison based sorting

    Comparison based sorting algorithms decide order by comparing elements, like Merge Sort, Quick Sort, and Heap Sort.
  2. Step 2: Identify non-comparison based sorting

    Counting Sort uses element values directly to sort without comparing elements.
  3. Final Answer:

    Counting Sort -> Option D
  4. Quick Check:

    Non-comparison based = Counting Sort [OK]
Hint: Non-comparison sorts use element values, not comparisons [OK]
Common Mistakes:
  • Confusing Merge Sort as non-comparison
  • Thinking Quick Sort is non-comparison
  • Assuming Heap Sort is non-comparison
2. Which of the following is the correct Rust syntax to call a non-comparison based sorting function counting_sort on a mutable vector arr?
easy
A. counting_sort(&mut arr);
B. counting_sort(arr);
C. counting_sort(&arr);
D. counting_sort(mut arr);

Solution

  1. Step 1: Understand Rust borrowing rules for mutable vectors

    To modify a vector inside a function, we pass a mutable reference using &mut.
  2. Step 2: Identify correct function call syntax

    Calling counting_sort(&mut arr); passes a mutable reference correctly.
  3. Final Answer:

    counting_sort(&mut arr); -> Option A
  4. Quick Check:

    Mutable reference call = counting_sort(&mut arr); [OK]
Hint: Use &mut to pass mutable vector to sorting function [OK]
Common Mistakes:
  • Passing vector without reference
  • Passing immutable reference &arr
  • Using invalid syntax like mut arr
3. What will be the output of the following Rust code snippet?
let mut arr = vec![4, 2, 2, 8, 3, 3, 1];
counting_sort(&mut arr);
println!("{:?}", arr);

Assuming counting_sort correctly sorts the vector in ascending order.
medium
A. [1, 2, 3, 4, 8]
B. [1, 2, 2, 3, 3, 4, 8]
C. [8, 4, 3, 3, 2, 2, 1]
D. [4, 2, 2, 8, 3, 3, 1]

Solution

  1. Step 1: Understand counting sort behavior

    Counting sort sorts elements in ascending order, preserving duplicates.
  2. Step 2: Sort the given vector

    Sorted vector is [1, 2, 2, 3, 3, 4, 8].
  3. Final Answer:

    [1, 2, 2, 3, 3, 4, 8] -> Option B
  4. Quick Check:

    Sorted ascending = [1, 2, 2, 3, 3, 4, 8] [OK]
Hint: Counting sort outputs sorted ascending with duplicates intact [OK]
Common Mistakes:
  • Reversing order to descending
  • Removing duplicates incorrectly
  • Printing original unsorted array
4. The following Rust code attempts to sort a vector using counting sort but causes a compilation error. What is the error?
fn counting_sort(arr: Vec<i32>) {
    // sorting logic
}

fn main() {
    let mut data = vec![3, 1, 4];
    counting_sort(data);
    println!("{:?}", data);
}
medium
A. Missing mutable keyword for data
B. counting_sort function missing return type
C. Cannot print data after move error
D. Syntax error in println! macro

Solution

  1. Step 1: Identify ownership move in function call

    Passing data by value moves ownership to counting_sort.
  2. Step 2: Understand why println! causes error

    After move, data is invalid in main, so printing causes compilation error.
  3. Final Answer:

    Cannot print data after move error -> Option C
  4. Quick Check:

    Ownership moved, data invalid after call [OK]
Hint: Passing vector by value moves ownership, invalid after call [OK]
Common Mistakes:
  • Assuming mutable keyword fixes ownership move
  • Thinking missing return type causes error
  • Blaming println! syntax
5. You have a large dataset of integers ranging from 0 to 1000. Which sorting approach is more efficient and why?
hard
A. Counting Sort is best due to O(n + k) time with small range
B. Quick Sort is best due to average O(n log n) time
C. Bubble Sort is best for simplicity
D. Merge Sort is best for stable sorting

Solution

  1. Step 1: Analyze dataset characteristics

    Dataset has integers in range 0 to 1000, so k=1001 is small compared to n.
  2. Step 2: Compare sorting algorithm complexities

    Counting Sort runs in O(n + k), which is efficient for small k. Quick, Merge are O(n log n). Bubble Sort is slow O(n²).
  3. Step 3: Choose best approach

    Counting Sort is more efficient here due to linear time with small range.
  4. Final Answer:

    Counting Sort is best due to O(n + k) time with small range -> Option A
  5. Quick Check:

    Small range + large data = Counting Sort best [OK]
Hint: Small range data? Use Counting Sort for linear time [OK]
Common Mistakes:
  • Choosing Quick Sort ignoring range advantage
  • Picking Bubble Sort for large data
  • Ignoring Counting Sort's linear time benefit
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10 questions~5 mineasy hard