Count Inversions in Array in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the inversion count to zero.

DSA Javascript

let inversionCount = [1];

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

A1

B0

C-1

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing inversionCount to 1 or null instead of 0.

Forgetting to initialize inversionCount before counting.

The inversion count should start at zero because initially no pairs are counted.

2fill in blank

medium

Complete the code to loop through the array elements for counting inversions.

DSA Javascript

for (let i = 0; i [1] arr.length - 1; i++) {

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

A>

B<=

C<

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using <= causes index out of range in inner loop.

Using > or >= causes the loop not to run correctly.

The loop should run while i is less than arr.length - 1 to avoid out-of-bound errors.

3fill in blank

hard

Fix the error in the inner loop condition to correctly count inversions.

DSA Javascript

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

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

A<=

B>=

C>

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using <= causes index out of range error.

Using > or >= causes the loop to not run properly.

The inner loop should run while j is less than arr.length to cover all pairs after i.

4fill in blank

hard

Fill both blanks to correctly check and count an inversion pair.

DSA Javascript

if (arr[i] [1] arr[j]) {
  inversionCount [2];
}

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

A>

B<

C++

D--

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' instead of '>' reverses the inversion logic.

Using '--' decreases the count incorrectly.

An inversion is when arr[i] is greater than arr[j], so we check '>' and increment the count with '++'.

5fill in blank

hard

Fill all three blanks to complete the function that counts inversions in an array.

DSA Javascript

function countInversions(arr) {
  let inversionCount = [1];
  for (let i = 0; i [2] arr.length - 1; i++) {
    for (let j = i + 1; j [3] arr.length; j++) {
      if (arr[i] > arr[j]) {
        inversionCount++;
      }
    }
  }
  return inversionCount;
}

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

A0

B<

C<=

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using <= in loops causes index errors.

Starting count at 1 or other values causes wrong results.

Initialize count to 0, loop i while less than length-1, and loop j while less than length.

Practice

1. What does the term inversion mean in the context of an array?

easy

A. A pair of elements where the first is smaller than the second and appears before it

B. A pair of elements where the first is greater than the second and appears before it

C. The total number of elements in the array

D. The sum of all elements in the array

Solution

1. Step 1: Understand the definition of inversion

   An inversion is when a larger number appears before a smaller number in the array.

2. Step 2: Compare options with the definition

   A pair of elements where the first is greater than the second and appears before it correctly describes this as a pair where the first element is greater and comes before the second.

3. Final Answer:

   A pair of elements where the first is greater than the second and appears before it -> Option B

4. Quick Check:

   Inversion = pair (i, j) with i < j and arr[i] > arr[j] [OK]

Hint: Inversion means bigger number comes before smaller one [OK]

Common Mistakes:

• Confusing inversion with sorted order
• Thinking inversion counts pairs where first is smaller
• Mixing inversion with total elements count

2. Which of the following JavaScript code snippets correctly initializes a variable to count inversions?

easy

A. int count = 0;

B. count := 0;

C. let count = 0;

D. var count == 0;

Solution

1. Step 1: Identify correct JavaScript syntax for variable initialization

   JavaScript uses let or var with = for assignment.

2. Step 2: Check each option

   let count = 0; uses let count = 0; which is correct. count := 0; uses ':=' which is not JavaScript. int count = 0; uses 'int' which is not JavaScript. var count == 0; uses '==' which is a comparison, not assignment.

3. Final Answer:

   let count = 0; -> Option C

4. Quick Check:

   Variable assignment in JS uses =, not := or == [OK]

Hint: Use let or var with = for assignment in JavaScript [OK]

Common Mistakes:

• Using ':=' instead of '='
• Using 'int' keyword from other languages
• Using '==' instead of '=' for assignment

3. What is the number of inversions in the array [2, 4, 1, 3, 5]?

medium

A. 3

B. 2

C. 4

D. 5

Solution

1. Step 1: Identify all inversion pairs in the array

   Check pairs where first element is greater and appears before second:
   - (2,1), (4,1), (4,3)

2. Step 2: Count the total inversions

   There are 3 such pairs.

3. Final Answer:

   3 -> Option A

4. Quick Check:

   Pairs (2,1), (4,1), (4,3) = 3 inversions [OK]

Hint: Count pairs where earlier number is bigger than later [OK]

Common Mistakes:

• Counting pairs where first is smaller
• Missing some inversion pairs
• Counting total pairs instead of inversions

4. Consider this code snippet to count inversions:

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

What is the error in this code?

medium

A. The function should return arr.length instead of count

B. The inner loop should start from j = 0

C. The count variable should be initialized inside the loops

D. The comparison should be arr[i] > arr[j] to count inversions

Solution

1. Step 1: Understand the inversion condition

   An inversion is when arr[i] > arr[j] and i < j, but code checks arr[i] < arr[j].

2. Step 2: Identify the fix

   The comparison should be changed to arr[i] > arr[j] to count inversions correctly.

3. Final Answer:

   The comparison should be arr[i] > arr[j] to count inversions -> Option D

4. Quick Check:

   Inversion means first element bigger, so use > [OK]

Hint: Use > to detect inversion pairs, not < [OK]

Common Mistakes:

• Using < instead of > in condition
• Starting inner loop from 0 instead of i+1
• Resetting count inside loops

5. Given the array [8, 4, 2, 1], which approach will efficiently count the number of inversions?

hard

A. Sort the array and count inversions during merge (O(n log n) time)

B. Count inversions by summing all elements

C. Use a nested loop to compare all pairs (O(n²) time)

D. Use a hash map to store counts of elements

Solution

1. Step 1: Understand the problem size and efficiency

   For larger arrays, nested loops (O(n²)) are slow. Merge sort based counting is efficient (O(n log n)).

2. Step 2: Identify the best approach

   Counting inversions during merge sort is a classic efficient method.

3. Final Answer:

   Sort the array and count inversions during merge (O(n log n) time) -> Option A

4. Quick Check:

   Merge sort inversion count = efficient O(n log n) [OK]

Hint: Use merge sort to count inversions efficiently [OK]

Common Mistakes:

• Using nested loops for large arrays causing slow performance
• Trying to sum elements instead of counting pairs
• Using hash maps which don't help count inversions