Count Inversions in Array in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return the number of inversions in the array using a simple nested loop.

DSA C++

int countInversions(vector<int>& arr) {
    int count = 0;
    for (int i = 0; i < arr.size(); i++) {
        for (int j = i + 1; j < arr.size(); j++) {
            if (arr[i] [1] arr[j]) {
                count++;
            }
        }
    }
    return count;
}

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

A<

B==

C<=

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' instead of '>' will count non-inversions.

Using '==' will not count inversions correctly.

We count an inversion when a larger number appears before a smaller number, so the condition is arr[i] > arr[j].

2fill in blank

medium

Complete the merge function to count inversions while merging two sorted halves.

DSA C++

int merge(vector<int>& arr, int left, int mid, int right) {
    vector<int> temp;
    int i = left, j = mid + 1, inv_count = 0;
    while (i <= mid && j <= right) {
        if (arr[i] <= arr[j]) {
            temp.push_back(arr[i++]);
        } else {
            temp.push_back(arr[j++]);
            inv_count += [1];
        }
    }
    while (i <= mid) temp.push_back(arr[i++]);
    while (j <= right) temp.push_back(arr[j++]);
    for (int k = left; k <= right; k++) arr[k] = temp[k - left];
    return inv_count;
}

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

Aright - j + 1

Bj - mid

Cmid - i + 1

Di - left

Attempts:

3 left

💡 Hint

Common Mistakes

Using j - mid counts wrong elements.

Using right - j + 1 counts elements from the right half incorrectly.

When arr[i] > arr[j], all elements from i to mid are greater than arr[j], so add mid - i + 1 to inversion count.

3fill in blank

hard

Fix the error in the recursive mergeSort function to correctly count inversions.

DSA C++

int mergeSort(vector<int>& arr, int left, int right) {
    int inv_count = 0;
    if (left < right) {
        int mid = left + (right - left) / 2;
        inv_count += mergeSort(arr, left, mid);
        inv_count += mergeSort(arr, mid + 1, right);
        inv_count += [1];
    }
    return inv_count;
}

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

Amerge(arr, left, mid, right)

Bmerge(arr, mid, right, left)

Cmerge(arr, left, right, mid)

Dmerge(arr, right, mid, left)

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping parameters causes wrong merges and wrong counts.

Using mid as right or left incorrectly breaks the recursion.

The merge function must be called with left, mid, right to merge the two halves correctly and count inversions.

4fill in blank

hard

Complete the countInversions function to use the mergeSort helper for efficient inversion counting.

DSA C++

int countInversions(vector<int>& arr) {
    if (arr.size() == 0) return 0;
    return [1](arr, 0, arr.[2] - 1);
}

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

AmergeSort

Bmerge

Csize()

Dbegin()

Attempts:

3 left

💡 Hint

Common Mistakes

Calling merge directly doesn't recurse.

Using wrong range like arr.size() without -1 causes issues.

Call the recursive mergeSort function with the full array range: mergeSort(arr, 0, arr.size() - 1).

5fill in blank

hard

Fill the two blanks to complete the merge function that counts inversions and merges two halves.

DSA C++

int merge(vector<int>& arr, int left, int mid, int right) {
    vector<int> temp;
    int i = left, j = mid + 1, inv_count = 0;
    while (i <= mid && j <= right) {
        if (arr[i] [1] arr[j]) {
            temp.push_back(arr[i++]);
        } else {
            temp.push_back(arr[j++]);
            inv_count += [2];
        }
    }
    while (i <= mid) temp.push_back(arr[i++]);
    while (j <= right) temp.push_back(arr[j++]);
    for (int k = left; k <= right; k++) arr[k] = temp[k - left];
    return inv_count;
}

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

A<=

B>

Cmid - i + 1

Dright - j + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' (B) in if-condition would pick larger first, breaking sort.

Using 'right - j + 1' (D) counts wrong half.

Use '<=' (A) to check if arr[i] <= arr[j]; if not, all remaining elements in left half (mid - i + 1, C) form inversions with arr[j].

Practice

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

easy

A. The sum of all elements in the array

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

Solution

1. Step 1: Understand the definition of inversion

   An inversion is a pair (i, j) such that i < j and array[i] > array[j].

2. Step 2: Compare options with definition

   A pair of elements where the first is greater than the second and appears before it matches the definition exactly, others describe different concepts.

3. Final Answer:

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

4. Quick Check:

   Inversion = pair with first > second [OK]

Hint: Inversion means earlier element is bigger than a later one [OK]

Common Mistakes:

• Confusing inversion with sorted pairs
• Thinking inversion counts total elements
• Mixing up greater and smaller conditions

2. Which of the following C++ code snippets correctly declares a function to count inversions in an array?

easy

A. int countInversions(vector<int>& arr);

B. void countInversions(vector<int> arr);

C. int countInversions(int arr[]);

D. int countInversions(vector<int> arr);

Solution

1. Step 1: Check function return type and parameter passing

   The function should return an integer count and take array by reference for efficiency.

2. Step 2: Analyze each option

   int countInversions(vector<int>& arr); returns int and takes vector by reference, correct. void countInversions(vector<int> arr); returns void, wrong. int countInversions(int arr[]); uses raw array without size info, less preferred. int countInversions(vector<int> arr); passes vector by value, inefficient.

3. Final Answer:

   int countInversions(vector<int>& arr); -> Option A

4. Quick Check:

   Return int, pass vector by reference [OK]

Hint: Count returns int; pass vector by reference to avoid copying [OK]

Common Mistakes:

• Using void return type
• Passing vector by value causing copies
• Using raw arrays without size info

3. What is the output of the following C++ code snippet?

std::vector<int> arr = {2, 4, 1, 3, 5};
int count = 0;
for (int i = 0; i < arr.size(); i++) {
  for (int j = i + 1; j < arr.size(); j++) {
    if (arr[i] > arr[j]) count++;
  }
}
std::cout << count << std::endl;

medium

A. 2

B. 4

C. 3

D. 5

Solution

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

   Pairs where first > second and first index < second index are: (2,1), (4,1), (4,3).

2. Step 2: Count these pairs

   There are exactly 3 such pairs.

3. Final Answer:

   3 -> Option C

4. Quick Check:

   Inversions = 3 [OK]

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

Common Mistakes:

• Counting pairs where first is smaller
• Missing some inversion pairs
• Confusing index order

4. Find the error in this C++ function to count inversions using a brute force approach:

int countInversions(std::vector<int> arr) {
  int count = 0;
  for (int i = 0; i < arr.size(); i++) {
    for (int j = 0; j < arr.size(); j++) {
      if (arr[i] > arr[j]) count++;
    }
  }
  return count;
}

medium

A. Function should return void, not int

B. Inner loop should start from j = i + 1, not 0

C. Comparison should be arr[i] < arr[j]

D. Count variable should be initialized outside the function

Solution

1. Step 1: Analyze loop ranges for counting inversions

   Inversion pairs require i < j, so inner loop must start from j = i + 1.

2. Step 2: Identify the error in the given code

   Inner loop starts from 0, counting pairs twice and including invalid pairs where j < i.

3. Final Answer:

   Inner loop should start from j = i + 1, not 0 -> Option B

4. Quick Check:

   Inner loop from i+1 to end [OK]

Hint: Inner loop must start after outer index to avoid double count [OK]

Common Mistakes:

• Starting inner loop from 0 causing double counting
• Wrong comparison direction
• Incorrect return type

5. Given the array {8, 4, 2, 1}, what is the number of inversions using an efficient merge sort based approach?

hard

A. 6

B. 4

C. 5

D. 3

Solution

1. Step 1: Understand inversions in the array

   All pairs where first > second and first index < second index are: (8,4), (8,2), (8,1), (4,2), (4,1), (2,1).

2. Step 2: Count inversions carefully

   There are 6 pairs total, but merge sort counts inversions during merging. The efficient approach counts 6 inversions.

3. Step 3: Verify merge sort inversion count

   Merge sort counts 6 inversions for this array.

4. Final Answer:

   6 -> Option A

5. Quick Check:

   Inversions = 6 [OK]

Hint: Count pairs where left element is bigger during merge [OK]

Common Mistakes:

• Under-counting inversions by missing pairs
• Confusing brute force and merge sort counts
• Miscounting pairs with equal elements