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

Time Complexity: Count Inversions in Array

O(n²)

Understanding Time Complexity

Counting inversions means finding how many pairs in an array are out of order.

We want to know how the time to count these pairs grows as the array gets bigger.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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;
}
    

This code counts how many pairs (i, j) have arr[i] > arr[j] with i < j.

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 times, and for each, the inner loop runs up to n times, so roughly n x n = n² times.

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: When input size doubles, operations roughly quadruple, showing a square growth.

Final Time Complexity

Time Complexity: O(n²)

This means the time to count inversions grows roughly with the square of the array size.

Common Mistake

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

[OK] Correct: Because you must compare pairs, and many pairs exist, so one pass cannot see all pairs needed.

Interview Connect

Understanding this helps you see why some problems need careful approaches to avoid slow solutions.

Self-Check

"What if we used a divide-and-conquer method like merge sort 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. 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