Count Inversions in Array in DSA Typescript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Count inversions in a small array

What is the output of the following TypeScript code that counts inversions in the array?

DSA Typescript

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]));

A 2

B 4

C 5

D 3

💡 Hint

Count pairs (i, j) where i < j and arr[i] > arr[j].

The inversions are (2,1), (4,1), and (4,3), totaling 3.

❓ Predict Output

intermediate

Count inversions using merge sort approach

What is the output of this TypeScript code that counts inversions using a merge sort method?

DSA Typescript

function mergeSortCount(arr: number[]): [number[], number] {
  if (arr.length <= 1) return [arr, 0];
  const mid = Math.floor(arr.length / 2);
  const [left, leftCount] = mergeSortCount(arr.slice(0, mid));
  const [right, rightCount] = mergeSortCount(arr.slice(mid));
  let i = 0, j = 0, count = leftCount + rightCount;
  const merged: number[] = [];
  while (i < left.length && j < right.length) {
    if (left[i] <= right[j]) {
      merged.push(left[i++]);
    } else {
      merged.push(right[j++]);
      count += left.length - i;
    }
  }
  return [merged.concat(left.slice(i)).concat(right.slice(j)), count];
}

const [sorted, inversions] = mergeSortCount([8, 4, 2, 1]);
console.log(inversions);

A 6

B 5

C 4

D 7

💡 Hint

Count how many elements from left are greater than elements from right during merge.

The inversions are (8,4), (8,2), (8,1), (4,2), (4,1), and (2,1), totaling 6.

🔧 Debug

advanced

Identify the error in inversion count code

What error does this TypeScript code produce when counting inversions?

DSA Typescript

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([1, 3, 2]));

A SyntaxError: Unexpected token

B TypeError: Cannot read property 'undefined' of undefined

C Output: 1

D Output: 0

💡 Hint

Check the loop boundary conditions carefully.

The inner loop uses j <= arr.length, which accesses arr[arr.length], undefined, causing a TypeError when comparing.

🧠 Conceptual

advanced

Time complexity of counting inversions using merge sort

What is the time complexity of counting inversions in an array of size n using the merge sort approach?

A O(n^2)

B O(log n)

C O(n log n)

D O(n)

💡 Hint

Think about how merge sort divides and merges the array.

Merge sort divides the array into halves recursively and merges them in O(n) time each level, resulting in O(n log n) overall.

🚀 Application

expert

Number of inversions after swapping two elements

Given the array [3, 1, 2, 5, 4], what is the number of inversions after swapping elements at indices 1 and 4?

A 6

B 3

C 5

D 2

💡 Hint

First count inversions in original array, then count after swap.

Original inversions: (3,1), (3,2), (5,4) = 3. After swapping elements at indices 1 and 4, array becomes [3,4,2,5,1]. New inversions: (3,2), (3,1), (4,2), (4,1), (2,1), (5,1) = 6.

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