Count Inversions in Array in DSA Javascript - Build from Scratch

Count Inversions in Array

📖 Scenario: Imagine you have a list of numbers representing the order of books on a shelf. You want to find out how many pairs of books are out of order compared to their correct sorted order. This helps you understand how messy the shelf is.

🎯 Goal: You will write a program to count how many pairs of numbers in an array are inverted, meaning the first number is bigger than the second but appears before it in the list.

📋 What You'll Learn

Create an array called arr with the exact values [2, 4, 1, 3, 5].

Create a variable called count and set it to 0 to keep track of inversions.

Use a nested for loop with variables i and j to compare elements and count inversions.

Print the value of count to show the total number of inversions.

💡 Why This Matters

🌍 Real World

Counting inversions helps in understanding how disordered a list is, which is useful in sorting algorithms and analyzing data sequences.

💼 Career

This concept is important for software engineers working on optimization, data analysis, and algorithm design.

1

Create the array

Create an array called arr with these exact values: [2, 4, 1, 3, 5].

DSA Javascript

// Your code here

Hint

Use const arr = [2, 4, 1, 3, 5]; to create the array.

2

Create the inversion counter

Create a variable called count and set it to 0 to count the number of inversions.

DSA Javascript

const arr = [2, 4, 1, 3, 5];
// Your code here

Hint

Use let count = 0; to start counting inversions.

3

Count inversions using nested loops

Use a nested for loop with variables i and j to compare elements in arr. Increase count by 1 whenever arr[i] is greater than arr[j] and i < j.

DSA Javascript

const arr = [2, 4, 1, 3, 5];
let count = 0;
// Your code here

Hint

Use two loops: the outer loop with i from 0 to arr.length - 1, and the inner loop with j from i + 1 to arr.length - 1. Inside, check if arr[i] > arr[j] and increase count.

4

Print the total inversion count

Print the value of count to display the total number of inversions in the array.

DSA Javascript

const arr = [2, 4, 1, 3, 5];
let count = 0;
for (let i = 0; i  arr[j]) {
            count++;
        }
    }
}
// Your code here

Hint

Use console.log(count); to print the total inversions.

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