Count Inversions in Array in DSA Rust - 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 used in algorithms like sorting and in problems related to order and ranking.

intermediate

Which algorithm efficiently counts inversions in O(n log n) time?

A modified merge sort algorithm can count inversions efficiently by counting how many elements from the right half are placed before elements from the left half during merging.

intermediate

In the merge step of merge sort, how do we count inversions?

When an element from the right half is placed before an element from the left half, it means all remaining elements in the left half form inversions with that element. We add the count of these remaining elements to the inversion count.

beginner

What is the output of counting inversions for the array [2, 4, 1, 3, 5]?

The number of inversions is 3. The pairs are (2,1), (4,1), and (4,3).

What does an inversion in an array represent?

A Two equal elements next to each other

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

C A pair where the first element is smaller and appears after the second

D Any two elements in the array

An inversion is specifically a pair where the first element is greater and appears before the second.

Which sorting algorithm is commonly modified to count inversions efficiently?

A Merge Sort

B Bubble Sort

C Insertion Sort

D Selection Sort

Merge Sort can be modified to count inversions during the merge step efficiently.

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

A O(log n)

B O(n^2)

C O(n log n)

D O(n)

The modified merge sort counts inversions in O(n log n) time.

In the merge step, when do we add to the inversion count?

A When the array is already sorted

B When an element from the left half is smaller

C When elements are equal

D When an element from the right half is placed before an element from the left half

Inversions are counted when an element from the right half is placed before elements from the left half.

How many inversions are in the array [3, 1, 2]?

A 2

B 0

C 3

D 1

The inversions are (3,1) and (3,2), so total 2.

Explain how the modified merge sort counts inversions in an array.

Think about how merge sort splits and merges arrays.

Describe what an inversion means and why counting them is useful.

Consider pairs where order is reversed.

Practice

1. What does the term inversion mean in the context of an array?

easy

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

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. The sum of all elements in the array

Solution

1. Step 1: Understand the definition of inversion

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

2. Step 2: Compare options with definition

   A pair of elements where the first is greater than the second and appears before it matches this definition exactly, while others describe different concepts.

3. Final Answer:

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

4. Quick Check:

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

Hint: Inversion means earlier bigger number than a later one [OK]

Common Mistakes:

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

2. Which Rust code snippet correctly declares a function to count inversions in an array slice?

easy

A. fn count_inversions(arr: &[i32]) -> i32 { 0 }

B. fn count_inversions(arr: Vec i32 { 0 }

C. fn count_inversions(arr: &[i32]) -> usize { 0 }

D. fn count_inversions(arr: &mut [i32]) -> usize { 0 }

Solution

1. Step 1: Identify correct function signature for counting inversions

   The function should take an immutable slice of integers and return the count as usize (non-negative integer).

2. Step 2: Compare options

   fn count_inversions(arr: &[i32]) -> usize { 0 } uses &[i32] for input and usize for output, which is correct. fn count_inversions(arr: Vec i32 { 0 } takes ownership (Vec) and returns i32 (can be negative). fn count_inversions(arr: &mut [i32]) -> usize { 0 } uses mutable slice unnecessarily. fn count_inversions(arr: &[i32]) -> i32 { 0 } returns i32 which can be negative, not ideal for count.

3. Final Answer:

   fn count_inversions(arr: &[i32]) -> usize { 0 } -> Option C

4. Quick Check:

   Immutable slice input + usize output = correct signature [OK]

Hint: Use &[i32] input and usize output for count functions [OK]

Common Mistakes:

• Using Vec instead of slice for input
• Returning signed integer for count
• Using mutable slice when not needed

3. What is the output of the following Rust code snippet counting inversions?

fn main() {
    let arr = [2, 4, 1, 3, 5];
    let mut count = 0;
    for i in 0..arr.len() {
        for j in i+1..arr.len() {
            if arr[i] > arr[j] {
                count += 1;
            }
        }
    }
    println!("{}", count);
}

medium

A. 4

B. 3

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 count from nested loops

   Loops check all pairs i < j. Each inversion increments count by 1. Final count is 3.

3. Final Answer:

   3 -> Option B

4. Quick Check:

   Count inversions by pairs = 3 [OK]

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

Common Mistakes:

• Counting non-inversion pairs
• Missing some inversion pairs
• Confusing indices order

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

fn count_inversions(arr: &[i32]) -> usize {
    let mut count = 0;
    for i in 0..arr.len() {
        for j in 0..arr.len() {
            if arr[i] > arr[j] {
                count += 1;
            }
        }
    }
    count
}

medium

A. Inner loop should start from i+1, not 0

B. Count variable should be i32, not usize

C. Function should take mutable slice

D. Comparison operator should be < instead of >

Solution

1. Step 1: Analyze loop ranges for counting inversions

   Inversion counts pairs where i < j and arr[i] > arr[j]. Inner loop starting at 0 compares all pairs including i >= j, causing double counting and invalid pairs.

2. Step 2: Correct inner loop start index

   Inner loop should start at i+1 to ensure only pairs with i < j are counted once.

3. Final Answer:

   Inner loop should start from i+1, not 0 -> Option A

4. Quick Check:

   Loop indices must ensure i < j for correct inversion count [OK]

Hint: Inner loop must start after outer index to avoid double count [OK]

Common Mistakes:

• Starting inner loop at 0 causing double counting
• Using wrong comparison operator
• Changing input slice to mutable unnecessarily

5. Given the array [8, 4, 2, 1], what is the number of inversions? Use the merge sort based approach to count inversions efficiently.

hard

A. 5

B. 3

C. 4

D. 6

Solution

1. Step 1: Understand merge sort inversion counting

   Merge sort counts inversions by counting how many elements from right subarray are placed before left subarray elements during merge.

2. Step 2: Count inversions in [8,4,2,1]

   Pairs: (8,4), (8,2), (8,1), (4,2), (4,1), (2,1) total 6 inversions.

3. Final Answer:

   6 -> Option D

4. Quick Check:

   All pairs where left > right count to 6 [OK]

Hint: Count pairs where left number bigger than right during merge [OK]

Common Mistakes:

• Counting only adjacent pairs
• Missing some inversion pairs
• Confusing merge sort steps