Insertion Sort Algorithm in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to start the outer loop from the second element in the array.

DSA Rust

for i in [1]..arr.len() {
    // insertion sort logic
}

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

A0

B1

Carr.len()

D2

Attempts:

3 left

💡 Hint

Common Mistakes

Starting the loop from 0 causes unnecessary comparisons.

Starting from arr.len() causes the loop to never run.

The insertion sort starts from the second element (index 1) because the first element is considered sorted.

2fill in blank

medium

Complete the code to store the current element to be inserted into the sorted part.

DSA Rust

let key = arr[[1]];

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A0

Barr.len() - 1

Cj

Di

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 or j instead of i causes wrong element selection.

Using arr.len() - 1 accesses the last element always, which is incorrect.

The variable key holds the element at index i which we want to insert into the sorted part.

3fill in blank

hard

Fix the error in the while loop condition to correctly compare elements for shifting.

DSA Rust

while j >= 0 && arr[j] [1] key {
    arr[j + 1] = arr[j];
    j -= 1;
}

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

A>

B<

C==

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using less than operator causes incorrect shifting.

Using equality operator causes infinite loops.

The condition should check if arr[j] is greater than key to shift elements right.

4fill in blank

hard

Fill both blanks to correctly insert the key into its sorted position.

DSA Rust

arr[j [1] 1] = [2];

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A+

B-

Ckey

Dj

Attempts:

3 left

💡 Hint

Common Mistakes

Using '-' instead of '+' causes wrong insertion index.

Assigning j instead of key inserts wrong value.

We insert the key at position j + 1 after shifting elements.

5fill in blank

hard

Fill all three blanks to complete the insertion sort inner loop and update index.

DSA Rust

let mut j = i [1] 1;
while j >= 0 && arr[j] [2] key {
    arr[j + 1] = arr[j];
    j [3] 1;
}

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

A-

B>

C-=

D+=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+' instead of '-' for j initialization.

Using '<' instead of '>' in condition.

Using '+=' instead of '-=' to update j.

We start j at i - 1, shift elements greater than key, and decrement j by 1 each loop.

Practice

1. What is the main idea behind the Insertion Sort algorithm?

easy

A. Divide the array into halves and sort each half recursively

B. Build the sorted array one element at a time by inserting elements in their correct position

C. Swap adjacent elements if they are in the wrong order repeatedly

D. Select the smallest element and swap it with the first unsorted element

Solution

1. Step 1: Understand insertion sort process

   Insertion sort builds the sorted list by taking one element at a time and placing it in the correct position among the already sorted elements.

2. Step 2: Compare with other sorting methods

   Unlike divide and conquer (merge sort) or swapping adjacent elements (bubble sort), insertion sort inserts elements directly into their correct place.

3. Final Answer:

   Build the sorted array one element at a time by inserting elements in their correct position -> Option B

4. Quick Check:

   Insertion sort = Insert elements correctly one by one [OK]

Hint: Insertion sort inserts elements into sorted part step-by-step [OK]

Common Mistakes:

• Confusing insertion sort with bubble sort
• Thinking it divides array like merge sort
• Assuming it swaps only smallest elements

2. Which of the following Rust code snippets correctly starts the insertion sort outer loop?

easy

A. for i in 1..arr.len() {

B. for i in 0..arr.len() {

C. for i in (1..arr.len()).rev() {

D. for i in 0..arr.len()-1 {

Solution

1. Step 1: Identify insertion sort loop start

   Insertion sort starts from the second element (index 1) because the first element is considered sorted.

2. Step 2: Check Rust range syntax

   The range 1..arr.len() correctly iterates from index 1 to the last index (exclusive of arr.len()).

3. Final Answer:

   for i in 1..arr.len() { -> Option A

4. Quick Check:

   Start loop at 1 for insertion sort [OK]

Hint: Insertion sort outer loop starts at index 1 [OK]

Common Mistakes:

• Starting loop at 0 instead of 1
• Using reversed range incorrectly
• Off-by-one errors in range

3. What will be the output of the following Rust code after running insertion sort on arr = [4, 3, 1, 5]?

let mut arr = vec![4, 3, 1, 5];
for i in 1..arr.len() {
    let key = arr[i];
    let mut j = i;
    while j > 0 && arr[j - 1] > key {
        arr[j] = arr[j - 1];
        j -= 1;
    }
    arr[j] = key;
}
println!("{:?}", arr);

medium

A. [3, 4, 1, 5]

B. [4, 3, 1, 5]

C. [1, 3, 4, 5]

D. [1, 4, 3, 5]

Solution

1. Step 1: Trace first iteration (i=1)

   key=3, compare with 4; 4 > 3 so shift 4 right, insert 3 at index 0. Array: [3,4,1,5]

2. Step 2: Trace second iteration (i=2)

   key=1, compare with 4 and 3; both >1 so shift them right, insert 1 at index 0. Array: [1,3,4,5]

3. Step 3: Trace third iteration (i=3)

   key=5, compare with 4; 4 < 5 so no shift, insert 5 at index 3. Array unchanged: [1,3,4,5]

4. Final Answer:

   [1, 3, 4, 5] -> Option C

5. Quick Check:

   Sorted array after insertion sort = [1, 3, 4, 5] [OK]

Hint: Trace each key insertion step-by-step [OK]

Common Mistakes:

• Not shifting elements correctly
• Off-by-one errors in while loop
• Printing original array without sorting

4. Identify the bug in this Rust insertion sort snippet:

let mut arr = vec![2, 5, 3];
for i in 1..arr.len() {
    let key = arr[i];
    let mut j = i - 1;
    while j >= 0 && arr[j] > key {
        arr[j + 1] = arr[j];
        j -= 1;
    }
    arr[j + 1] = key;
}

medium

A. The key is not assigned correctly

B. The array is not mutable

C. The for loop range is incorrect

D. The variable j is unsigned and j >= 0 check causes infinite loop

Solution

1. Step 1: Analyze variable j type and condition

   In Rust, usize is unsigned, so j >= 0 is always true, causing infinite loop when j underflows.

2. Step 2: Understand while loop behavior

   When j reaches 0 and decrements, it wraps to a large number, so loop never ends.

3. Final Answer:

   The variable j is unsigned and j >= 0 check causes infinite loop -> Option D

4. Quick Check:

   Unsigned j with j >= 0 causes infinite loop [OK]

Hint: Unsigned index can't be checked with j >= 0 safely [OK]

Common Mistakes:

• Ignoring unsigned integer wrap-around
• Assuming j can be negative
• Not using a different loop condition

5. You want to modify insertion sort in Rust to sort a vector of tuples Vec<(i32, String)> by the integer value. Which change is needed in the comparison inside the while loop?

hard

A. Change arr[j - 1] > key to arr[j - 1].0 > key.0

B. Change arr[j - 1] > key to arr[j - 1].1 > key.1

C. No change needed, tuples compare automatically by second element

D. Change arr[j - 1] > key to arr[j - 1].1.len() > key.1.len()

Solution

1. Step 1: Understand tuple sorting by first element

   To sort by the integer part of the tuple, compare the first element (.0) of each tuple.

2. Step 2: Modify comparison in while loop

   Replace arr[j - 1] > key with arr[j - 1].0 > key.0 to compare integers correctly.

3. Final Answer:

   Change arr[j - 1] > key to arr[j - 1].0 > key.0 -> Option A

4. Quick Check:

   Compare tuple first elements for sorting [OK]

Hint: Compare tuple.0 to sort by integer part [OK]

Common Mistakes:

• Comparing string part instead of integer
• Assuming tuples compare by second element automatically
• Using length of string for comparison