Count Inversions in Array in DSA Java - Time & Space Complexity

Time Complexity: Count Inversions in Array

O(n²)

Understanding Time Complexity

We want to understand how long it takes to count inversions in an array as the array grows.

The question is: how does the work increase when the array size gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

This code counts how many pairs of elements are out of order in the array.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Comparing pairs of elements to check if one is greater than the other.
• How many times: The outer loop runs n-1 times, and for each outer loop, the inner loop runs fewer times, roughly n times in total.

How Execution Grows With Input

As the array size grows, the number of comparisons grows much faster.

Input Size (n)	Approx. Operations
10	About 45 comparisons
100	About 4,950 comparisons
1000	About 499,500 comparisons

Pattern observation: The number of operations grows roughly with the square of the input size.

Final Time Complexity

Time Complexity: O(n²)

This means if the array doubles in size, the work to count inversions roughly quadruples.

Common Mistake

[X] Wrong: "Counting inversions can be done in linear time by just one pass through the array."

[OK] Correct: Because each element must be compared with many others to find all inversions, one pass is not enough.

Interview Connect

Understanding this helps you explain how nested loops affect performance and why more efficient methods exist.

Self-Check

"What if we used a merge sort based approach to count inversions? How would the time complexity change?"

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