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

Time Complexity: Count Inversions in Array

O(n²)

Understanding Time Complexity

We want to understand how the time needed to count inversions in an array changes as the array grows.

Specifically, how does the number of steps increase when the array gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function countInversions(arr) {
  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 of elements are out of order in the array by checking every pair.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Comparing pairs of elements inside two nested loops.
• 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 because every new element is compared with many others.

Input Size (n)	Approx. Operations
10	About 45 comparisons
100	About 4,950 comparisons
1000	About 499,500 comparisons

Pattern observation: The operations grow roughly by the square of the input size, so doubling the array size makes the work about four times bigger.

Final Time Complexity

Time Complexity: O(n²)

This means the time needed grows quickly as the array gets bigger, because every element is compared with many others.

Common Mistake

[X] Wrong: "Since we only have one loop inside another, the time is just twice the input size, so O(2n)."

[OK] Correct: The loops multiply their work, not add. So the inner loop runs many times for each outer loop step, making it grow like n times n, not just 2 times n.

Interview Connect

Understanding this helps you explain why simple pair checking is slow for big data and why smarter methods are needed.

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

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

C. The total number of elements in the array

D. The sum of all elements in the array

Solution

1. Step 1: Understand the definition of inversion

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

2. Step 2: Compare options with the definition

   A pair of elements where the first is greater than the second and appears before it correctly describes this as a pair where the first element is greater and comes before the second.

3. Final Answer:

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

4. Quick Check:

   Inversion = pair (i, j) with i < j and arr[i] > arr[j] [OK]

Hint: Inversion means bigger number comes before smaller one [OK]

Common Mistakes:

• Confusing inversion with sorted order
• Thinking inversion counts pairs where first is smaller
• Mixing inversion with total elements count

2. Which of the following JavaScript code snippets correctly initializes a variable to count inversions?

easy

A. int count = 0;

B. count := 0;

C. let count = 0;

D. var count == 0;

Solution

1. Step 1: Identify correct JavaScript syntax for variable initialization

   JavaScript uses let or var with = for assignment.

2. Step 2: Check each option

   let count = 0; uses let count = 0; which is correct. count := 0; uses ':=' which is not JavaScript. int count = 0; uses 'int' which is not JavaScript. var count == 0; uses '==' which is a comparison, not assignment.

3. Final Answer:

   let count = 0; -> Option C

4. Quick Check:

   Variable assignment in JS uses =, not := or == [OK]

Hint: Use let or var with = for assignment in JavaScript [OK]

Common Mistakes:

• Using ':=' instead of '='
• Using 'int' keyword from other languages
• Using '==' instead of '=' for assignment

3. What is the number of inversions in the array [2, 4, 1, 3, 5]?

medium

A. 3

B. 2

C. 4

D. 5

Solution

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

   Check pairs where first element is greater and appears before second:
   - (2,1), (4,1), (4,3)

2. Step 2: Count the total inversions

   There are 3 such pairs.

3. Final Answer:

   3 -> Option A

4. Quick Check:

   Pairs (2,1), (4,1), (4,3) = 3 inversions [OK]

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

Common Mistakes:

• Counting pairs where first is smaller
• Missing some inversion pairs
• Counting total pairs instead of inversions

4. Consider this code snippet to count inversions:

function countInversions(arr) {
  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;
}

What is the error in this code?

medium

A. The function should return arr.length instead of count

B. The inner loop should start from j = 0

C. The count variable should be initialized inside the loops

D. The comparison should be arr[i] > arr[j] to count inversions

Solution

1. Step 1: Understand the inversion condition

   An inversion is when arr[i] > arr[j] and i < j, but code checks arr[i] < arr[j].

2. Step 2: Identify the fix

   The comparison should be changed to arr[i] > arr[j] to count inversions correctly.

3. Final Answer:

   The comparison should be arr[i] > arr[j] to count inversions -> Option D

4. Quick Check:

   Inversion means first element bigger, so use > [OK]

Hint: Use > to detect inversion pairs, not < [OK]

Common Mistakes:

• Using < instead of > in condition
• Starting inner loop from 0 instead of i+1
• Resetting count inside loops

5. Given the array [8, 4, 2, 1], which approach will efficiently count the number of inversions?

hard

A. Sort the array and count inversions during merge (O(n log n) time)

B. Count inversions by summing all elements

C. Use a nested loop to compare all pairs (O(n²) time)

D. Use a hash map to store counts of elements

Solution

1. Step 1: Understand the problem size and efficiency

   For larger arrays, nested loops (O(n²)) are slow. Merge sort based counting is efficient (O(n log n)).

2. Step 2: Identify the best approach

   Counting inversions during merge sort is a classic efficient method.

3. Final Answer:

   Sort the array and count inversions during merge (O(n log n) time) -> Option A

4. Quick Check:

   Merge sort inversion count = efficient O(n log n) [OK]

Hint: Use merge sort to count inversions efficiently [OK]

Common Mistakes:

• Using nested loops for large arrays causing slow performance
• Trying to sum elements instead of counting pairs
• Using hash maps which don't help count inversions