Count Inversions in Array in DSA Typescript - 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 do we count inversions in an array?

Counting inversions helps measure how far the array is from being sorted. More inversions mean the array is less sorted.

beginner

What is the naive way 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.

intermediate

How does merge sort help count inversions efficiently?

During merge sort, when merging two sorted halves, we 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), which is much faster than the naive O(n²) approach.

What does an inversion in an array represent?

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

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

C Two equal elements next to each other

D Any two elements in the array

An inversion is when a larger element appears before a smaller element.

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

A O(n²)

B O(n)

C O(n log n)

D O(log n)

The naive method checks all pairs, leading to O(n²) time.

Which sorting algorithm is commonly used to count inversions efficiently?

A Bubble Sort

B Insertion Sort

C Merge Sort

D Selection Sort

Merge Sort can count inversions during the merge step efficiently.

During merge sort, when do we count inversions?

A After sorting completes

B When splitting the array

C Before sorting starts

D When merging two sorted halves

Inversions are counted while merging two sorted halves.

If an array is already sorted, how many inversions does it have?

A Zero

B Maximum possible

C Half the number of elements

D Depends on the array size

A sorted array has no inversions.

Explain what an inversion in an array is and why counting inversions is useful.

Think about pairs where order is reversed compared to sorted order.

Describe how merge sort can be modified to count inversions efficiently.

Focus on the merge step where two sorted halves combine.

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 of elements (i, j) such that i < j but 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 correctly describes this condition. Other options describe unrelated 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 where earlier element > later element [OK]

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

Common Mistakes:

• Confusing inversion with sorted order
• Thinking inversion counts total elements
• Mixing up greater and smaller element order

2. Which of the following TypeScript function signatures correctly represents a function to count inversions in an array of numbers?

easy

A. function countInversions(arr: string): number[]

B. function countInversions(arr: number): string

C. function countInversions(arr: number[]): number

D. function countInversions(arr: any): void

Solution

1. Step 1: Analyze the function purpose and parameters

   The function should take an array of numbers and return a number representing the count of inversions.

2. Step 2: Match the correct signature

   function countInversions(arr: number[]): number matches this: input is number array, output is number. Others mismatch input/output types.

3. Final Answer:

   function countInversions(arr: number[]): number -> Option C

4. Quick Check:

   Input array, output number count [OK]

Hint: Input array of numbers, output single number [OK]

Common Mistakes:

• Using wrong input type like string instead of array
• Returning array instead of number
• Using void return type when output is needed

3. What is the output of the following TypeScript code?

function countInversions(arr: number[]): number {
  let count = 0;
  for (let i = 0; i < arr.length; i++) {
    for (let j = i + 1; j < arr.length; j++) {
      if (arr[i] > arr[j]) count++;
    }
  }
  return count;
}

console.log(countInversions([2, 4, 1, 3, 5]));

medium

A. 3

B. 4

C. 2

D. 5

Solution

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

   Array: [2, 4, 1, 3, 5]. Inversions are (2,1), (4,1), (4,3). Total 3 inversions.

2. Step 2: Confirm the count returned by the nested loops

   The nested loops count each inversion correctly, so output is 3.

3. Final Answer:

   3 -> Option A

4. Quick Check:

   Inversions = 3 [OK]

Hint: Count pairs where earlier > later element [OK]

Common Mistakes:

• Counting pairs incorrectly by missing some
• Confusing equal elements as inversions
• Off-by-one errors in loops

4. The following code attempts to count inversions but has a bug. What is the bug?

function countInversions(arr: number[]): number {
  let count = 0;
  for (let i = 0; i <= arr.length; i++) {
    for (let j = i + 1; j < arr.length; j++) {
      if (arr[i] > arr[j]) count++;
    }
  }
  return count;
}

medium

A. The inner loop should start from j = 0 instead of i + 1

B. The outer loop condition should be < instead of <= to avoid out-of-bounds

C. The count variable should be reset inside the outer loop

D. The comparison should be arr[i] < arr[j] instead of >

Solution

1. Step 1: Analyze the outer loop condition

   The outer loop runs from i = 0 to i <= arr.length, which means i can be arr.length, causing arr[i] to be undefined (out-of-bounds).

2. Step 2: Correct the loop condition

   Changing to i < arr.length prevents out-of-bounds access and fixes the bug.

3. Final Answer:

   The outer loop condition should be < instead of <= to avoid out-of-bounds -> Option B

4. Quick Check:

   Loop bounds must prevent out-of-range access [OK]

Hint: Use < length, not <= length in loops [OK]

Common Mistakes:

• Using <= causes index out-of-bounds error
• Resetting count inside loop loses total count
• Changing comparison reverses inversion logic

5. Given the array [8, 4, 2, 1], which approach is the most efficient to count inversions and what is the inversion count?

hard

A. Use a modified merge sort to count inversions; count is 6

B. Use nested loops to count inversions; count is 3

C. Sort the array and count differences; count is 4

D. Use a hash map to count inversions; count is 5

Solution

1. Step 1: Identify inversion count manually

   Array: [8,4,2,1]. Inversions are (8,4), (8,2), (8,1), (4,2), (4,1), (2,1) totaling 6.

2. Step 2: Choose the most efficient method

   Modified merge sort counts inversions in O(n log n), better than nested loops O(n²). Hash map or sorting differences do not directly count inversions.

3. Final Answer:

   Use a modified merge sort to count inversions; count is 6 -> Option A

4. Quick Check:

   Merge sort method = 6 inversions [OK]

Hint: Use merge sort for large arrays to count inversions fast [OK]

Common Mistakes:

• Counting inversions incorrectly by missing pairs
• Using nested loops for large arrays causing slow performance
• Confusing sorting with inversion counting