Count Inversions in Array in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the inversion count variable.

DSA Java

int countInversions(int[] arr) {
    int [1] = 0;
    // rest of the code
    return 0;
}

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

Acount

BinversionCount

Ctotal

Dresult

Attempts:

3 left

💡 Hint

Common Mistakes

Using a variable name that is too generic or unclear.

Forgetting to initialize the variable.

The variable 'count' is commonly used to keep track of the number of inversions.

2fill in blank

medium

Complete the code to compare elements for inversion detection.

DSA Java

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

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

A==

B<

C>

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' instead of '>' which counts non-inversions.

Using '==' which does not detect inversions.

An inversion is when a larger number appears before a smaller number, so arr[i] > arr[j].

3fill in blank

hard

Fix the error in the merge step to count split inversions correctly.

DSA Java

while (i < left.length && j < right.length) {
    if (left[i] <= right[j]) {
        merged[k++] = left[i++];
    } else {
        merged[k++] = right[j++];
        [1] += (left.length - i);
    }
}

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

Acount

Bresult

Ctotal

DinversionCount

Attempts:

3 left

💡 Hint

Common Mistakes

Using a different variable name causing compilation errors.

Not incrementing the inversion count at all.

The variable 'count' is used to accumulate the number of inversions during merging.

4fill in blank

hard

Fill both blanks to complete the merge function that counts inversions.

DSA Java

int mergeAndCount(int[] left, int[] right, int[] merged) {
    int i = 0, j = 0, k = 0, [1] = 0;
    while (i < left.length && j < right.length) {
        if (left[i] <= right[j]) {
            merged[k++] = left[i++];
        } else {
            merged[k++] = right[j++];
            [2] += (left.length - i);
        }
    }
    while (i < left.length) merged[k++] = left[i++];
    while (j < right.length) merged[k++] = right[j++];
    return [1];
}

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

Acount

BinversionCount

Ctotal

Dresult

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names for counting and returning.

Forgetting to return the count.

The variable 'count' is used to track and return the number of inversions found during merging.

5fill in blank

hard

Fill all three blanks to complete the recursive merge sort that counts inversions.

DSA Java

int mergeSortAndCount(int[] arr) {
    if (arr.length < 2) return 0;
    int mid = arr.length / 2;
    int[] left = Arrays.copyOfRange(arr, 0, [1]);
    int[] right = Arrays.copyOfRange(arr, [2], arr.length);
    int count = mergeSortAndCount(left) + mergeSortAndCount(right);
    count += mergeAndCount(left, right, arr);
    return [3];
}

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

Amid

B0

Ccount

Darr.length

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong indices for splitting arrays.

Returning the wrong variable at the end.

The array is split at 'mid', left is from 0 to mid, right is from mid to end, and the total count is returned.

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