Selection Sort Algorithm in DSA C# - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to swap two elements in the array.

DSA C#

int temp = arr[[1]];
arr[[1]] = arr[i];
arr[i] = temp;

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

Ai

Bj

C0

DminIndex

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping with the wrong index like 'i' or 'j' instead of 'minIndex'.

Using a fixed index like 0 instead of the variable holding the smallest element.

The variable minIndex holds the index of the smallest element found. We swap the element at minIndex with the element at i.

2fill in blank

medium

Complete the code to find the index of the smallest element in the unsorted part of the array.

DSA C#

if (arr[j] [1] arr[minIndex]) {
    minIndex = j;
}

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

A>=

B<

C==

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' instead of '<' which finds the largest element instead.

Using '==' which does not help find the smallest element.

We want to find the smallest element, so we check if arr[j] is less than arr[minIndex].

3fill in blank

hard

Fix the error in the for-loop that iterates over the unsorted part of the array.

DSA C#

for (int j = i + 1; j [1] arr.Length; j++) {
    // find the smallest element
}

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

A<=

B>=

C<

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<=' causes an index out of range error.

Using '>' or '>=' causes the loop to never run.

The loop should run while j is less than arr.Length to avoid going out of bounds.

4fill in blank

hard

Fill both blanks to complete the outer loop and initialize the minimum index.

DSA C#

for (int [1] = 0; i [2] arr.Length - 1; i++) {
    int minIndex = i;
    // inner loop to find smallest element
}

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

Ai

Bj

C<

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'j' instead of 'i' for the outer loop variable.

Using '<=' which causes index out of range.

The outer loop uses i starting at 0 and runs while i < arr.Length - 1 because the last element will be sorted by then.

5fill in blank

hard

Fill all three blanks to complete the selection sort method.

DSA C#

public void SelectionSort(int[] arr) {
    for (int [1] = 0; i [2] arr.Length - 1; i++) {
        int minIndex = i;
        for (int j = i + 1; j [3] arr.Length; j++) {
            if (arr[j] < arr[minIndex]) {
                minIndex = j;
            }
        }
        int temp = arr[minIndex];
        arr[minIndex] = arr[i];
        arr[i] = temp;
    }
}

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

Ai

B<

C<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<=' in loops causing index errors.

Using wrong loop variables.

The outer loop variable is i. It runs while i < arr.Length - 1. The inner loop variable j runs while j < arr.Length.

Practice

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

easy

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

B. Use a pivot to partition the list into smaller parts

C. Swap adjacent elements if they are in the wrong order

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

Solution

1. Step 1: Understand selection sort process

   Selection sort works by finding the smallest element in the unsorted part and swapping it with the first unsorted element.

2. Step 2: Compare with other sorting methods

   Unlike divide and conquer or adjacent swaps, selection sort directly selects the minimum each time.

3. Final Answer:

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

4. Quick Check:

   Selection sort = find minimum repeatedly [OK]

Hint: Selection sort picks smallest element each pass [OK]

Common Mistakes:

• Confusing selection sort with bubble sort
• Thinking it divides the list like merge sort
• Assuming it uses a pivot like quicksort

2. Which of the following is the correct syntax to swap two elements arr[i] and arr[minIndex] in C# during selection sort?

easy

A. arr[i], arr[minIndex] = arr[minIndex], arr[i];

B. arr[i] = arr[minIndex]; arr[minIndex] = arr[i];

C. temp = arr[i]; arr[i] = arr[minIndex]; arr[minIndex] = temp;

D. swap(arr[i], arr[minIndex]);

Solution

1. Step 1: Understand swapping in C#

   C# does not support tuple unpacking or direct swap functions; we use a temporary variable to swap values.

2. Step 2: Check each option

   temp = arr[i]; arr[i] = arr[minIndex]; arr[minIndex] = temp; uses a temp variable correctly. arr[i] = arr[minIndex]; arr[minIndex] = arr[i]; overwrites arr[i] before assigning arr[minIndex]. swap(arr[i], arr[minIndex]); is invalid syntax. arr[i], arr[minIndex] = arr[minIndex], arr[i]; is Python style, not C#.

3. Final Answer:

   temp = arr[i]; arr[i] = arr[minIndex]; arr[minIndex] = temp; -> Option C

4. Quick Check:

   Use temp variable to swap in C# [OK]

Hint: Use temp variable to swap in C# [OK]

Common Mistakes:

• Trying to swap without a temp variable
• Using Python or other language syntax
• Overwriting one element before saving it

3. What will be the output of the following C# code snippet after applying selection sort?

int[] arr = {64, 25, 12, 22, 11};
// Apply selection sort here
for (int i = 0; i < arr.Length - 1; i++) {
    int minIndex = i;
    for (int j = i + 1; j < arr.Length; j++) {
        if (arr[j] < arr[minIndex]) {
            minIndex = j;
        }
    }
    int temp = arr[minIndex];
    arr[minIndex] = arr[i];
    arr[i] = temp;
}
Console.WriteLine(string.Join(", ", arr));

medium

A. 11, 12, 22, 25, 64

B. 64, 25, 12, 22, 11

C. 25, 12, 22, 11, 64

D. 11, 22, 12, 25, 64

Solution

1. Step 1: Trace the selection sort iterations

   First pass finds 11 (smallest) and swaps with 64. Array: [11, 25, 12, 22, 64]. Second pass finds 12 and swaps with 25. Array: [11, 12, 25, 22, 64]. Third pass finds 22 and swaps with 25. Array: [11, 12, 22, 25, 64]. Fourth pass finds 25 (already in place). Final sorted array is [11, 12, 22, 25, 64].

2. Step 2: Confirm final output

   The printed output is the sorted array joined by commas.

3. Final Answer:

   11, 12, 22, 25, 64 -> Option A

4. Quick Check:

   Selection sort output = sorted array [OK]

Hint: Selection sort outputs sorted array ascending [OK]

Common Mistakes:

• Not swapping correctly during iterations
• Confusing indexes and swapping wrong elements
• Assuming original array stays unchanged

4. Identify the error in the following selection sort code snippet and choose the correct fix:

int[] arr = {3, 1, 4, 2};
for (int i = 0; i < arr.Length; i++) {
    int minIndex = i;
    for (int j = i + 1; j < arr.Length; j++) {
        if (arr[j] > arr[minIndex]) {
            minIndex = j;
        }
    }
    int temp = arr[minIndex];
    arr[minIndex] = arr[i];
    arr[i] = temp;
}

medium

A. Change 'if (arr[j] > arr[minIndex])' to 'if (arr[j] < arr[minIndex])'

B. Change outer loop to 'for (int i = 0; i < arr.Length - 1; i++)'

C. Remove the swap code inside the loop

D. Initialize minIndex to 0 instead of i

Solution

1. Step 1: Analyze the comparison condition

   The code uses 'arr[j] > arr[minIndex]' which finds the maximum instead of minimum, so it sorts in descending order, not ascending.

2. Step 2: Correct the comparison for ascending order

   Changing the condition to 'arr[j] < arr[minIndex]' finds the minimum element as required by selection sort.

3. Final Answer:

   Change 'if (arr[j] > arr[minIndex])' to 'if (arr[j] < arr[minIndex])' -> Option A

4. Quick Check:

   Comparison must find minimum element [OK]

Hint: Use '<' to find minimum in selection sort [OK]

Common Mistakes:

• Using '>' instead of '<' in inner loop
• Looping one extra time causing index errors
• Not swapping elements properly

5. Given an array int[] arr = {5, 3, 8, 6, 2, 7, 4, 1}, how many swaps will the selection sort algorithm perform to sort it in ascending order?

hard

A. 6

B. 5

C. 7

D. 8

Solution

1. Step 1: Understand swap count in selection sort

   Selection sort swaps once per outer loop iteration if the minimum is not already at the current position.

2. Step 2: Count swaps for given array

   Array length is 8, so outer loop runs 7 times (0 to 6). Each iteration finds minimum and swaps if needed. Here, minimum is at current index in one iteration, so total swaps are 5.

3. Final Answer:

   5 -> Option B

4. Quick Check:

   Swaps = number of times minimum is not at current index [OK]

Hint: Selection sort swaps at most n-1 times, but only when needed [OK]

Common Mistakes:

• Counting swaps as number of elements
• Assuming swaps happen every iteration
• Confusing swaps with comparisons