Count Inversions in Array in DSA C++ - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is an inversion in an array?

An inversion is a pair of elements where the first element is greater than the second element, but the first element appears before the second in the array.

beginner

Why is counting inversions important?

Counting inversions helps measure how far an array is from being sorted. It is useful in sorting algorithms and understanding array disorder.

beginner

What is the naive approach to count inversions in an array?

Check every pair of elements and count if the first is greater than the second. This takes O(n²) time for an array of size n.

intermediate

How does the merge sort technique help count inversions efficiently?

During merge sort, while merging two sorted halves, count how many elements from the right half are placed before elements from the left half. This counts inversions in O(n log n) time.

intermediate

What is the time complexity of counting inversions using merge sort?

The time complexity is O(n log n), where n is the size of the array.

What does an inversion in an array represent?

A Any two elements in the array

B A pair where the first element is greater and appears before the second

C Two equal elements next to each other

D A pair where the first element is smaller and appears before the second

An inversion is when a larger element appears before a smaller one in the array.

What is the time complexity of the naive method to count inversions?

A O(n log n)

B O(n)

C O(log n)

D O(n²)

The naive method checks all pairs, resulting in O(n²) time.

Which algorithm is commonly used to count inversions efficiently?

A Merge Sort

B Bubble Sort

C Linear Search

D Binary Search

Merge Sort can count inversions during the merge step efficiently.

During merge sort, when do we count inversions?

A When merging two sorted halves

B When dividing the array

C Before sorting starts

D After sorting completes

Inversions are counted while merging two sorted halves.

What is the main advantage of using merge sort for counting inversions?

A It sorts the array in O(n²) time

B It uses less memory

C It counts inversions in O(n log n) time

D It does not require recursion

Merge sort counts inversions efficiently in O(n log n) time.

Explain how merge sort can be used to count inversions in an array.

Think about how merging two sorted lists can reveal inversions.

Describe the difference between the naive and efficient methods for counting inversions.

Consider time complexity and approach.

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