Selection Sort Algorithm in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to start the outer loop for selection sort.

DSA Python

for i in range([1], len(arr)):
    min_idx = i

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

A0

B1

Clen(arr)

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Starting loop from 1 skips the first element.

Using len(arr) as start causes no iterations.

The outer loop starts from index 0 to go through all elements for sorting.

2fill in blank

medium

Complete the code to start the inner loop for selection sort.

DSA Python

for j in range([1], len(arr)):
    if arr[j] < arr[min_idx]:
        min_idx = j

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

A0

Bi - 1

Clen(arr)

Di + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Starting inner loop at 0 causes unnecessary comparisons.

Starting inner loop at i - 1 causes index errors.

The inner loop starts from the next element after i to find the minimum in the unsorted part.

3fill in blank

hard

Fix the error in swapping the elements after finding the minimum.

DSA Python

arr[i], arr[[1]] = arr[[1]], arr[i]

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

A0

Bi

Cmin_idx

Dj

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping with i does nothing.

Swapping with j may cause wrong swaps.

We swap the element at i with the element at min_idx, which holds the smallest value found.

4fill in blank

hard

Fill both blanks to complete the condition and the key comparison in selection sort.

DSA Python

for i in range(len(arr)):
    min_idx = i
    for j in range(i + 1, len(arr)):
        if arr[j] [1] arr[min_idx]:
            min_idx = [2]

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

A<

B>

Cj

Di

Attempts:

3 left

💡 Hint

Common Mistakes

Using > instead of < reverses sorting order.

Assigning min_idx to i instead of j misses the new minimum.

The condition checks if arr[j] is less than arr[min_idx], and if yes, updates min_idx to j.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps each number to its square if the number is less than 5.

DSA Python

squares = [1]: [2] for [3] in range(10) if [3] < 5

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

Ax

Bx**2

Di

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names inconsistently.

Not applying the condition correctly.

The dictionary comprehension uses x as the variable, maps x to x squared, and includes only x less than 5.

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 merge sorted halves

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

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

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

   Other options describe different algorithms: merge sort, bubble sort, and quicksort, not selection sort.

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 the smallest next element [OK]

Common Mistakes:

• Confusing selection sort with bubble sort
• Thinking it divides the list like merge sort
• Mixing it up with quicksort's pivot method

2. Which of the following is the correct syntax to swap two elements a[i] and a[j] in Python during selection sort?

easy

A. temp = a[i]; a[i] = a[j]; a[j] = a[i]

B. a[i], a[j] = a[j], a[i]

C. swap(a[i], a[j])

D. a[i] = a[j]; a[j] = a[i]

Solution

1. Step 1: Identify Pythonic swap syntax

   Python allows swapping two variables in one line: a[i], a[j] = a[j], a[i].

2. Step 2: Check other options for correctness

   temp = a[i]; a[i] = a[j]; a[j] = a[i] causes data loss by overwriting without properly using temp; swap(a[i], a[j]) is invalid syntax; a[i] = a[j]; a[j] = a[i] causes both to have same value.

3. Final Answer:

   a[i], a[j] = a[j], a[i] -> Option B

4. Quick Check:

   Python swap = a[i], a[j] = a[j], a[i] [OK]

Hint: Use comma swap for clean Python element exchange [OK]

Common Mistakes:

• Using a function swap() which doesn't exist
• Assigning one element twice causing data loss
• Not using tuple unpacking for swap

3. What will be the output of the following code snippet?

def selection_sort(arr):
    for i in range(len(arr)):
        min_idx = i
        for j in range(i+1, len(arr)):
            if arr[j] < arr[min_idx]:
                min_idx = j
        arr[i], arr[min_idx] = arr[min_idx], arr[i]

arr = [3, 1, 4]
selection_sort(arr)
print(arr)

medium

A. [4, 3, 1]

B. [3, 1, 4]

C. [1, 4, 3]

D. [1, 3, 4]

Solution

1. Step 1: Trace first iteration of outer loop

   i=0, min_idx=0; inner loop finds smallest element from index 1 and 2. arr[1]=1 < arr[0]=3, so min_idx=1. Swap arr[0] and arr[1]: arr becomes [1, 3, 4].

2. Step 2: Trace second iteration

   i=1, min_idx=1; inner loop checks arr[2]=4 which is not less than arr[1]=3, so no swap. Array stays [1, 3, 4].

3. Final Answer:

   [1, 3, 4] -> Option D

4. Quick Check:

   Sorted array = [1, 3, 4] [OK]

Hint: Selection sort places smallest element at front each pass [OK]

Common Mistakes:

• Not swapping when min_idx changes
• Confusing indices in inner loop
• Assuming array is unchanged

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

def selection_sort(arr):
    for i in range(len(arr)):
        min_idx = i
        for j in range(i+1, len(arr)):
            if arr[j] > arr[min_idx]:
                min_idx = j
        arr[i], arr[min_idx] = arr[min_idx], arr[i]

medium

A. Initialize min_idx to 0 instead of i

B. Swap should be done outside the outer loop

C. Change '>' to '<' in the if condition to find the smallest element

D. Use '==' instead of '>' in the if condition

Solution

1. Step 1: Analyze the if condition logic

   The code uses 'if arr[j] > arr[min_idx]' which finds the largest element, but selection sort needs the smallest.

2. Step 2: Correct the comparison operator

   Changing '>' to '<' will correctly find the smallest element to swap.

3. Final Answer:

   Change '>' to '<' in the if condition to find the smallest element -> Option C

4. Quick Check:

   Selection sort finds minimum, so use '<' [OK]

Hint: Selection sort finds minimum, so use '<' in comparison [OK]

Common Mistakes:

• Using '>' causes sorting in descending order
• Swapping outside loop breaks logic
• Wrong initialization of min_idx

5. Given a list arr = [5, 3, 8, 3, 9], how many swaps will the selection sort algorithm perform to sort this list in ascending order?

hard

A. 3

B. 4

C. 2

D. 5

Solution

1. Step 1: Trace selection sort swaps step-by-step

   Initial array: [5, 3, 8, 3, 9] - i=0: smallest is 3 at index 1, swap arr[0] and arr[1] -> [3, 5, 8, 3, 9] (1 swap) - i=1: smallest is 3 at index 3, swap arr[1] and arr[3] -> [3, 3, 8, 5, 9] (2 swaps) - i=2: smallest is 5 at index 3, swap arr[2] and arr[3] -> [3, 3, 5, 8, 9] (3 swaps) - i=3: smallest is 8 at index 3, no swap needed - i=4: last element, no swap needed

2. Step 2: Count total swaps

   Total swaps performed = 3.

3. Final Answer:

   3 -> Option A

4. Quick Check:

   Swaps counted = 3 [OK]

Hint: Count swaps only when min_idx changes and swap occurs [OK]

Common Mistakes:

• Counting swaps even when min_idx equals i
• Assuming one swap per iteration always
• Ignoring repeated elements affecting swaps