Count Inversions in Array in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the inversion count variable.

DSA Rust

let mut count = [1];

Drag options to blanks, or click blank then click option'

A-1

B0

C1

DNone

Attempts:

3 left

💡 Hint

Common Mistakes

Starting count from 1 or negative numbers.

We start counting inversions from zero because initially we have found none.

2fill in blank

medium

Complete the code to compare two elements for inversion.

DSA Rust

if arr[i] [1] arr[j] { count += 1; }

Drag options to blanks, or click blank then click option'

A>

B<=

C==

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using less than or equal instead of greater than.

An inversion is when a larger element appears before a smaller one, so we check if arr[i] > arr[j].

3fill in blank

hard

Fix the error in the loop to iterate over the array indices correctly.

DSA Rust

for i in 0..[1] { for j in i+1..[2] { /* count inversions */ } }

Drag options to blanks, or click blank then click option'

Aarr.len() - 1

Barr.len()

Carr.len() + 1

Darr.len() / 2

Attempts:

3 left

💡 Hint

Common Mistakes

Using arr.len() - 1 causing missed elements.

Using arr.len() + 1 causing out of bounds.

We loop from 0 to arr.len() (exclusive) to cover all indices properly.

4fill in blank

hard

Fill both blanks to complete the merge step that counts split inversions.

DSA Rust

while i < n1 && j < n2 {
    if left[i] [1] right[j] {
        arr[k] = left[i];
        i += 1;
    } else {
        arr[k] = right[j];
        count += (n1 - i) as i64;
        j += 1;
    }
    k += 1;
}

Drag options to blanks, or click blank then click option'

A<=

B>

C<

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong comparison operators causing incorrect merges or counts.

We compare if left[i] < right[j] to merge sorted arrays and count inversions when left[i] > right[j].

5fill in blank

hard

Fill all three blanks to complete the recursive merge sort function that counts inversions.

DSA Rust

fn merge_sort(arr: &mut [i32]) -> i64 {
    let len = arr.len();
    if len <= 1 {
        return 0;
    }
    let mid = len [1] 2;
    let mut count = 0;
    count += merge_sort(&mut arr[0..[2]]);
    count += merge_sort(&mut arr[[3]..len]);
    count += merge(arr, mid);
    count
}

Drag options to blanks, or click blank then click option'

A/

Bmid

D*

Attempts:

3 left

💡 Hint

Common Mistakes

Using multiplication instead of division for mid.

Using wrong slice indices.

We divide the array by length/2, then recursively sort left (0..mid) and right (mid..len) parts.

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