Count Inversions in Array in DSA Java - 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 the correct ascending order.

🎯 Goal: Build a Java program that counts the number of inversions in an array. An inversion is when a larger number appears before a smaller number in the list.

📋 What You'll Learn

Create an integer array called arr with the exact values {2, 4, 1, 3, 5}.

Create an integer variable called inversionCount and initialize it to 0.

Write a nested for loop using variables i and j to compare elements in arr and count inversions by increasing inversionCount.

Print the value of inversionCount.

💡 Why This Matters

🌍 Real World

Counting inversions helps in understanding how far a list is from being sorted, useful in sorting algorithms and measuring disorder in data.

💼 Career

This concept is important in software engineering roles that involve algorithm optimization, data analysis, and performance tuning.

1

Create the array

Create an integer array called arr with the exact values {2, 4, 1, 3, 5}.

DSA Java

// Create the array arr with values {2, 4, 1, 3, 5}
// Your code here

Hint

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

2

Initialize inversion count

Create an integer variable called inversionCount and initialize it to 0 inside the main method after the array.

DSA Java

public class Main {
    public static void main(String[] args) {
        int[] arr = {2, 4, 1, 3, 5};
        // Initialize inversionCount to 0
        // Your code here
    }
}

Hint

Use int inversionCount = 0; to start counting from zero.

3

Count inversions with nested loops

Use nested for loops with variables i and j to compare elements in arr. For each pair where arr[i] > arr[j] and i < j, increase inversionCount by 1.

DSA Java

public class Main {
    public static void main(String[] args) {
        int[] arr = {2, 4, 1, 3, 5};
        int inversionCount = 0;
        // Use nested for loops to count inversions
        // Your code here
    }
}

Hint

Use two loops: outer loop with i from 0 to length-1, inner loop with j from i+1 to length-1. Increase inversionCount when arr[i] > arr[j].

4

Print the inversion count

Print the value of inversionCount using System.out.println.

DSA Java

public class Main {
    public static void main(String[] args) {
        int[] arr = {2, 4, 1, 3, 5};
        int inversionCount = 0;
        for (int i = 0; i  arr[j]) {
                    inversionCount++;
                }
            }
        }
        // Print the inversionCount
        // Your code here
    }
}

Hint

Use System.out.println(inversionCount); to show the result.

Practice

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

easy

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

B. A pair of elements where the first is smaller 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 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 -> Option A

4. Quick Check:

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

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

Common Mistakes:

• Confusing inversion with sorted order
• Thinking inversion counts equal elements
• Mixing up element positions

2. Which of the following Java code snippets correctly initializes an integer array named arr with values 3, 1, 2?

easy

A. int arr = {3, 1, 2};

B. int[] arr = new int(3, 1, 2);

C. int[] arr = {3, 1, 2};

D. int arr[] = new int[3];

Solution

1. Step 1: Check array declaration syntax

   In Java, arrays are declared with square brackets and initialized with curly braces or using new keyword with curly braces.

2. Step 2: Validate each option

   int[] arr = {3, 1, 2}; uses correct syntax: int[] arr = {3, 1, 2}; int arr = {3, 1, 2}; misses brackets, int[] arr = new int(3, 1, 2); uses wrong constructor syntax, int arr[] = new int[3]; does not initialize to the given values (zeros instead).

3. Final Answer:

   B -> Option C

4. Quick Check:

   Correct array init = int[] arr = {3, 1, 2}; [OK]

Hint: Use int[] arr = {values}; to initialize arrays [OK]

Common Mistakes:

• Missing square brackets in declaration
• Using parentheses instead of braces
• Incorrect use of new keyword

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

medium

A. 2

B. 5

C. 4

D. 3

Solution

1. Step 1: Identify all inversion pairs

   Pairs where first element is greater and appears before second: (2,1), (4,1), (4,3), (4,5) is not inversion, (3,5) no inversion. Actually, count carefully: (2,1), (4,1), (4,3) are 3 inversions. But check if (2,3) or (2,5) are inversions? No. So total 3 inversions.

2. Step 2: Count the pairs

   Counting these pairs gives 3 inversions.

3. Final Answer:

   3 -> Option D

4. Quick Check:

   Inversions in [2,4,1,3,5] = 3 [OK]

Hint: Count pairs where left element > right element [OK]

Common Mistakes:

• Counting pairs with equal elements
• Missing some inversion pairs
• Counting sorted pairs as inversions

4. Consider the following Java method to count inversions. What is the error?

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

medium

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

B. The inner loop should start from j = 0

C. The count variable should be initialized inside the loops

D. The method should return void instead of int

Solution

1. Step 1: Analyze the inversion condition

   Inversion means arr[i] > arr[j] where i < j, but code checks arr[i] < arr[j], which is incorrect.

2. Step 2: Identify the fix

   Change condition to arr[i] > arr[j] to correctly count inversions.

3. Final Answer:

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

4. Quick Check:

   Correct condition = arr[i] > arr[j] [OK]

Hint: Inversion means left element greater than right [OK]

Common Mistakes:

• Using wrong comparison operator
• Starting inner loop incorrectly
• Resetting count inside loops

5. Given the array [8, 4, 2, 1], what is the total number of inversions? Also, explain why this number is significant in sorting algorithms.

hard

A. 6; It shows the array is nearly sorted

B. 6; It shows the array is sorted in descending order

C. 4; It shows the array is sorted in ascending order

D. 3; It shows the array has few inversions

Solution

1. Step 1: Count inversions in the array

   Pairs: (8,4), (8,2), (8,1), (4,2), (4,1), (2,1) total 6 inversions.

2. Step 2: Understand significance in sorting

   Maximum inversions mean array is sorted in descending order, which is worst case for some sorting algorithms.

3. Final Answer:

   6; It shows the array is sorted in descending order -> Option B

4. Quick Check:

   Inversions = 6; descending order = max inversions [OK]

Hint: Max inversions = descending order array [OK]

Common Mistakes:

• Miscounting inversion pairs
• Confusing max inversions with sorted array
• Ignoring significance in sorting