Selection Sort Algorithm in DSA Rust - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Selection Sort on a small array

What is the output of the following Rust code after performing selection sort on the array?

DSA Rust

fn selection_sort(arr: &mut [i32]) {
    let n = arr.len();
    for i in 0..n {
        let mut min_index = i;
        for j in (i + 1)..n {
            if arr[j] < arr[min_index] {
                min_index = j;
            }
        }
        arr.swap(i, min_index);
    }
}

fn main() {
    let mut data = [64, 25, 12, 22, 11];
    selection_sort(&mut data);
    println!("{:?}", data);
}

A [25, 11, 12, 22, 64]

B [64, 25, 12, 22, 11]

C [11, 22, 12, 25, 64]

D [11, 12, 22, 25, 64]

💡 Hint

Selection sort finds the smallest element and places it at the start, then repeats for the rest.

The selection sort algorithm repeatedly selects the smallest remaining element and swaps it to the front. After sorting, the array is in ascending order: [11, 12, 22, 25, 64].

🧠 Conceptual

intermediate

Number of swaps in selection sort

For an array of 5 elements, how many swaps does the selection sort algorithm perform in the worst case?

A 4

B 5

C 10

D 0

💡 Hint

Selection sort swaps only once per outer loop iteration if needed.

Selection sort performs exactly n-1 swaps in the worst case, where n is the number of elements. For 5 elements, it performs 4 swaps.

🔧 Debug

advanced

Identify the bug in this selection sort implementation

What error does the following Rust code produce when run?

DSA Rust

fn selection_sort(arr: &mut [i32]) {
    let n = arr.len();
    for i in 0..n {
        let mut min_index = i;
        for j in i..n {
            if arr[j] < arr[min_index] {
                min_index = j;
            }
        }
        arr.swap(i, min_index);
    }
}

fn main() {
    let mut data = [3, 1, 2];
    selection_sort(&mut data);
    println!("{:?}", data);
}

A No error, output: [1, 2, 3]

B Panic at runtime due to out of bounds indexing

C Infinite loop

D Compilation error: mismatched types

💡 Hint

Check the inner loop range carefully.

The inner loop starts at i, which includes the current element. This is valid and does not cause errors. The code runs correctly and outputs the sorted array.

❓ Predict Output

advanced

Selection sort on a reversed array

What is the output of the following Rust code after sorting the reversed array?

DSA Rust

fn selection_sort(arr: &mut [i32]) {
    let n = arr.len();
    for i in 0..n {
        let mut min_index = i;
        for j in (i + 1)..n {
            if arr[j] < arr[min_index] {
                min_index = j;
            }
        }
        arr.swap(i, min_index);
    }
}

fn main() {
    let mut data = [5, 4, 3, 2, 1];
    selection_sort(&mut data);
    println!("{:?}", data);
}

A [1, 3, 2, 4, 5]

B [1, 2, 3, 4, 5]

C [5, 4, 3, 2, 1]

D [2, 1, 3, 4, 5]

💡 Hint

Selection sort always sorts in ascending order.

The reversed array is sorted into ascending order by selection sort, resulting in [1, 2, 3, 4, 5].

🧠 Conceptual

expert

Time complexity of selection sort

What is the time complexity of selection sort in all cases (best, average, worst)?

A O(n) best, O(n^2) worst

B O(n log n) best, O(n^2) worst

C O(n^2) for all cases

D O(n) for all cases

💡 Hint

Selection sort always scans the entire unsorted part to find the minimum.

Selection sort always performs n*(n-1)/2 comparisons regardless of input order, so time complexity is O(n^2) in all cases.

Practice

1. What is the main idea behind the Selection Sort Algorithm?

easy

A. Repeatedly find the smallest element and move it to the front

B. Divide the list into halves and sort each half separately

C. Swap adjacent elements if they are in wrong order

D. Insert each element into its correct position in a sorted sublist

Solution

1. Step 1: Understand selection sort process

   Selection sort finds the smallest element in the unsorted part and swaps it with the first unsorted element.

2. Step 2: Compare with options

   Repeatedly find the smallest element and move it to the front matches this description exactly, others describe different sorting methods.

3. Final Answer:

   Repeatedly find the smallest element and move it to the front -> Option A

4. Quick Check:

   Selection sort = find smallest and swap [OK]

Hint: Selection sort always picks smallest next element [OK]

Common Mistakes:

• Confusing selection sort with bubble sort
• Thinking it divides list into halves
• Mixing it up with insertion sort

2. Which of the following is the correct Rust syntax to swap two elements a[i] and a[min_index] in a vector a?

easy

A. a.swap_elements(i, min_index);

B. swap(a[i], a[min_index]);

C. a[i], a[min_index] = a[min_index], a[i];

D. a.swap(i, min_index);

Solution

1. Step 1: Recall Rust vector swap method

   Rust vectors have a method called swap that takes two indices and swaps their elements.

2. Step 2: Match syntax with options

   a.swap(i, min_index); uses a.swap(i, min_index); which is correct. Others are invalid Rust syntax or methods.

3. Final Answer:

   a.swap(i, min_index); -> Option D

4. Quick Check:

   Rust vector swap method = a.swap(i, j) [OK]

Hint: Use vector.swap(index1, index2) to swap elements [OK]

Common Mistakes:

• Trying to swap elements directly with assignment
• Using non-existent swap_elements method
• Using Python-style tuple swap syntax

3. What will be the output vector after running selection sort on vec![3, 1, 4, 2]?

medium

A. [1, 3, 2, 4]

B. [4, 3, 2, 1]

C. [1, 2, 3, 4]

D. [2, 1, 3, 4]

Solution

1. Step 1: Perform selection sort step-by-step

   Initial vector: [3, 1, 4, 2] - Find smallest (1), swap with first element: [1, 3, 4, 2] - Find smallest in remaining (3,4,2) is 2, swap with second element: [1, 2, 4, 3] - Find smallest in last two (4,3) is 3, swap with third element: [1, 2, 3, 4]

2. Step 2: Confirm final sorted vector

   The vector is sorted ascending: [1, 2, 3, 4]

3. Final Answer:

   [1, 2, 3, 4] -> Option C

4. Quick Check:

   Selection sort output = sorted ascending [OK]

Hint: Selection sort outputs sorted ascending list [OK]

Common Mistakes:

• Confusing selection sort with reverse sorting
• Stopping after first swap
• Not swapping correctly

4. Consider this Rust code snippet for selection sort. What is the bug?

fn selection_sort(a: &mut Vec<i32>) {
    let n = a.len();
    for i in 0..=n {
        let mut min_index = i;
        for j in i+1..n {
            if a[j] < a[min_index] {
                min_index = j;
            }
        }
        a.swap(i, min_index);
    }
}

medium

A. min_index should be initialized to 0, not i

B. The outer loop should run until n-1, not n

C. The swap method does not exist for vectors

D. The inner loop should start from 0, not i+1

Solution

1. Step 1: Analyze loop ranges

   The outer loop runs from 0 to n inclusive, meaning it runs n+1 times. But the last element does not need to be checked because it will already be sorted.

2. Step 2: Correct loop range for outer loop

   The outer loop should run from 0 to n-1 to avoid unnecessary iteration and potential out-of-bound errors.

3. Final Answer:

   The outer loop should run until n-1, not n -> Option B

4. Quick Check:

   Outer loop range = 0..n-1 [OK]

Hint: Outer loop runs to n-1, not n [OK]

Common Mistakes:

• Running outer loop one step too far
• Starting inner loop incorrectly
• Misunderstanding swap method existence

5. You want to sort a vector of integers but only partially: you want the first 3 smallest elements sorted at the front, and the rest can remain unsorted. How can you modify selection sort to do this efficiently?

hard

A. Run the outer loop only 3 times instead of full length

B. Run the inner loop only 3 times for each outer iteration

C. Sort the entire vector then shuffle last elements randomly

D. Use bubble sort for first 3 elements and selection sort for rest

Solution

1. Step 1: Understand partial sorting requirement

   We only want the first 3 smallest elements sorted at the front, so we only need to find and place the smallest element 3 times.

2. Step 2: Modify selection sort outer loop

   By running the outer loop 3 times, selection sort will place the 3 smallest elements in order at the front, leaving the rest unsorted.

3. Final Answer:

   Run the outer loop only 3 times instead of full length -> Option A

4. Quick Check:

   Partial sort = limit outer loop iterations [OK]

Hint: Limit outer loop to number of elements to sort [OK]

Common Mistakes:

• Modifying inner loop instead of outer loop
• Sorting entire vector unnecessarily
• Mixing sorting algorithms incorrectly