Selection Sort Algorithm in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to swap two elements in the array.

DSA Javascript

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

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

Aj

Bi

Cindex

DminIndex

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping with the wrong index like j or i itself.

Not using a temporary variable for swapping.

We swap the element at position i with the element at position minIndex to place the smallest element at the correct position.

2fill in blank

medium

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

DSA Javascript

for (let j = i + 1; j < arr.length; j++) {
  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.

Using equality or inequality operators which do not help find the smallest.

We look for the smallest element, so we check if arr[j] is less than arr[minIndex].

3fill in blank

hard

Fix the error in the loop that iterates over the array for selection sort.

DSA Javascript

for (let i = 0; i [1] arr.length - 1; i++) {
  // selection sort logic
}

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

A<=

B>

C<

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<=' causes an extra iteration leading to out-of-bounds errors.

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

We loop until i is less than arr.length - 1 because the last element will already be sorted.

4fill in blank

hard

Fill both blanks to complete the selection sort inner loop and update the smallest element index.

DSA Javascript

for (let j = [1]; j < arr.length; j++) {
  if (arr[j] [2] arr[minIndex]) {
    minIndex = j;
  }
}

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

Ai + 1

Bi

C<

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Starting from i instead of i + 1 includes the already sorted part.

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

The inner loop starts from i + 1 to check the unsorted part, and we use '<' to find the smallest element.

5fill in blank

hard

Fill the blanks to create a dictionary mapping elements to their original positions before sorting.

DSA Javascript

let positions = {};
for (let i = 0; i < arr.length; i++) {
  positions[[1]] = [2];
}
// After sorting, positions hold the index of each element
arr.sort((a, b) => a - b);
console.log(positions);

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

Aarr[i]

Bi

Carr.length

Di + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using arr.length or i + 1 as keys or values incorrectly.

Mixing keys and values in the dictionary.

We use arr[i] as the key and i as the value to map elements to their original positions. arr.length is unrelated here but included as distractor.

Practice

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

easy

A. Divide the array into halves and merge sorted halves

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

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

D. Use a pivot to partition the array 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 Merge Sort, Bubble Sort, and Quick Sort, which are different algorithms.

3. Final Answer:

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

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 uses divide and conquer
• Assuming it uses a pivot element

2. Which of the following is the correct syntax to swap two elements arr[i] and arr[minIndex] in JavaScript during Selection Sort?

easy

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

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

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

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

Solution

1. Step 1: Identify JavaScript swap syntax

   JavaScript supports swapping using destructuring assignment: [a, b] = [b, a].

2. Step 2: Check each option

   temp = arr[i]; arr[i] = arr[minIndex]; arr[minIndex] = temp; is valid but missing let or var for temp and is verbose. arr[i], arr[minIndex] = arr[minIndex], arr[i]; is invalid syntax in JavaScript. swap(arr[i], arr[minIndex]); is a function call but no such built-in function exists.

3. Final Answer:

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

4. Quick Check:

   Destructuring swap = [arr[i], arr[minIndex]] = [arr[minIndex], arr[i]]; [OK]

Hint: Use destructuring assignment for swapping in JavaScript [OK]

Common Mistakes:

• Using Python-style tuple swap syntax
• Calling undefined swap function
• Forgetting to declare temporary variable

3. What will be the output array after running Selection Sort on [29, 10, 14, 37, 13]?

medium

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

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

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

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

Solution

1. Step 1: Perform Selection Sort step-by-step

   Initial array: [29, 10, 14, 37, 13]
   Find smallest from index 0 to end: 10 at index 1, swap with 29.
   Array: [10, 29, 14, 37, 13]
   Find smallest from index 1 to end: 13 at index 4, swap with 29.
   Array: [10, 13, 14, 37, 29]
   Find smallest from index 2 to end: 14 at index 2, no swap.
   Find smallest from index 3 to end: 29 at index 4, swap with 37.
   Array: [10, 13, 14, 29, 37]

2. Step 2: Confirm final sorted array

   The array is sorted in ascending order: [10, 13, 14, 29, 37].

3. Final Answer:

   [10, 13, 14, 29, 37] -> Option D

4. Quick Check:

   Sorted array = [10, 13, 14, 29, 37] [OK]

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

Common Mistakes:

• Not swapping correctly after finding minimum
• Confusing index positions during swaps
• Assuming array is unchanged

4. Identify the error in this Selection Sort code snippet:

function selectionSort(arr) {
  for (let i = 0; i < arr.length; i++) {
    let minIndex = i;
    for (let j = i + 1; j < arr.length; j++) {
      if (arr[j] < arr[minIndex]) {
        minIndex = j;
      }
    }
    arr[i] = arr[minIndex];
    arr[minIndex] = arr[i];
  }
  return arr;
}

medium

A. Swapping lines overwrite values causing incorrect swaps

B. Loop ranges are incorrect, should be j < arr.length - 1

C. minIndex should be initialized outside the outer loop

D. Function does not return the sorted array

Solution

1. Step 1: Analyze the swap logic

   The swap is done by assigning arr[i] = arr[minIndex] first, then arr[minIndex] = arr[i]. This overwrites arr[i] before assigning to arr[minIndex], causing both to have the same value.

2. Step 2: Correct swap method

   Use a temporary variable or destructuring assignment to swap without overwriting:
   let temp = arr[i]; arr[i] = arr[minIndex]; arr[minIndex] = temp;

3. Final Answer:

   Swapping lines overwrite values causing incorrect swaps -> Option A

4. Quick Check:

   Incorrect swap overwrites values = Swapping lines overwrite values causing incorrect swaps [OK]

Hint: Use temp variable or destructuring to swap safely [OK]

Common Mistakes:

• Swapping without temp variable
• Incorrect loop boundaries
• Not returning the array

5. Given an array with duplicate values [4, 2, 4, 3, 2], how many swaps will Selection Sort perform to fully sort it?

hard

A. 3

B. 5

C. 4

D. 2

Solution

1. Step 1: Trace Selection Sort swaps on the array

   Initial array: [4, 2, 4, 3, 2]
   Pass 1: smallest from index 0-4 is 2 at index 1, swap with 4 at index 0.
   Array: [2, 4, 4, 3, 2] (1 swap)
   Pass 2: smallest from index 1-4 is 2 at index 4, swap with 4 at index 1.
   Array: [2, 2, 4, 3, 4] (2 swaps)
   Pass 3: smallest from index 2-4 is 3 at index 3, swap with 4 at index 2.
   Array: [2, 2, 3, 4, 4] (3 swaps)
   Pass 4: smallest from index 3-4 is 4 at index 3, no swap.
   Pass 5: last element, no swap.

2. Step 2: Count total swaps

   Total swaps performed = 3.

3. Final Answer:

   3 -> Option A

4. Quick Check:

   Count swaps during passes = 3 [OK]

Hint: Count swaps only when minimum index differs from current index [OK]

Common Mistakes:

• Counting all comparisons as swaps
• Ignoring swaps when minIndex equals current index
• Assuming duplicates reduce swaps to zero