Selection Sort Algorithm in DSA C# - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of the selection sort on this array?

Given the array [29, 10, 14, 37, 13], what will be the state of the array after the first full pass of selection sort?

DSA C#

int[] arr = {29, 10, 14, 37, 13};
// After first pass of selection sort

A [10, 29, 14, 37, 13]

B [10, 14, 29, 37, 13]

C [29, 10, 14, 37, 13]

D [13, 10, 14, 37, 29]

💡 Hint

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

In the first pass, selection sort finds the smallest element (10) and swaps it with the first element (29). So the array becomes [10, 29, 14, 37, 13].

❓ Predict Output

intermediate

Final sorted array after selection sort

What is the final sorted array after applying selection sort on [64, 25, 12, 22, 11]?

DSA C#

int[] arr = {64, 25, 12, 22, 11};
// After full selection sort

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

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

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

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

💡 Hint

Selection sort sorts the array in ascending order by repeatedly selecting the smallest element.

Selection sort will reorder the array to [11, 12, 22, 25, 64] after all passes.

🧠 Conceptual

advanced

Why does selection sort have O(n²) time complexity?

Selection sort has a time complexity of O(n²). Which reason best explains why?

A Because it compares each element with every other element to find the minimum in each pass.

B Because it uses extra memory to store sorted elements.

C Because it divides the array into halves and sorts recursively.

D Because it swaps elements only once per pass.

💡 Hint

Think about how many comparisons happen in each pass.

Selection sort compares each element with all others to find the minimum, leading to roughly n*(n-1)/2 comparisons, which is O(n²).

🔧 Debug

advanced

Identify the bug in this selection sort code snippet

What is the bug in the following C# selection sort code snippet?

DSA C#

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;
}

A The outer loop should run until arr.Length, not arr.Length - 1.

B The inner loop should start from 0 instead of i + 1.

C The comparison operator should be '<' to find the minimum element, not '>'.

D The swap logic is incorrect; it should swap arr[i] with arr[j].

💡 Hint

Selection sort finds the smallest element in the unsorted part.

The code uses '>' which finds the maximum element instead of the minimum. It should use '<' to find the minimum element for sorting in ascending order.

🚀 Application

expert

How many swaps does selection sort perform on a sorted array?

If selection sort is applied to an already sorted array of 5 elements, how many swaps will it perform?

A 0

B 10

C 5

D 4

💡 Hint

Selection sort swaps the minimum element with the current position in each pass, even if they are the same.

Selection sort performs one swap per pass except possibly the last. For 5 elements, it performs 4 swaps even if the array is sorted.

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