Insertion Sort Algorithm in DSA Rust - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Insertion Sort on a small array

What is the output of the following Rust code after sorting the array using insertion sort?

DSA Rust

fn insertion_sort(arr: &mut [i32]) {
    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;
    }
}

fn main() {
    let mut data = [5, 2, 9, 1, 5];
    insertion_sort(&mut data);
    println!("{:?}", data);
}

A [5, 2, 9, 1, 5]

B [1, 5, 2, 5, 9]

C [9, 5, 5, 2, 1]

D [1, 2, 5, 5, 9]

💡 Hint

Insertion sort places each element in its correct position by shifting larger elements to the right.

Insertion sort iterates through the array, inserting each element into the sorted portion on the left. The final sorted array is [1, 2, 5, 5, 9].

🧠 Conceptual

intermediate

Number of comparisons in insertion sort

For an array of size 5 sorted in descending order, how many comparisons does insertion sort perform in the worst case?

A 5

B 15

C 10

D 20

💡 Hint

Worst case is when each new element is smaller than all sorted elements.

Insertion sort compares each element with all previously sorted elements in the worst case. For size 5, comparisons = 1+2+3+4 = 10.

🔧 Debug

advanced

Identify the bug in this insertion sort implementation

What error will this Rust code produce when run?

DSA Rust

fn insertion_sort(arr: &mut [i32]) {
    for i in 1..arr.len() {
        let key = arr[i];
        let mut j = i + 1;
        while j > 0 && arr[j] > key {
            arr[j] = arr[j - 1];
            j -= 1;
        }
        arr[j] = key;
    }
}

fn main() {
    let mut data = [3, 1, 4];
    insertion_sort(&mut data);
    println!("{:?}", data);
}

A Index out of bounds panic

B No error, prints [1, 3, 4]

C Type mismatch error

D Infinite loop

💡 Hint

Check the condition inside the while loop carefully.

The condition arr[j] > key accesses arr[j] when j can be equal to arr.len(), causing out of bounds panic.

📝 Syntax

advanced

Which option will cause a syntax error in Rust?

Which of the following insertion sort code snippets will cause a syntax error?

A

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;
}

B

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;
}

C

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D

or 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;
}

💡 Hint

Look for missing semicolons in Rust code.

Option A is missing a semicolon after arr[j] = arr[j - 1], causing a syntax error.

🚀 Application

expert

Insertion sort on linked list output

Given a singly linked list with values 4 -> 3 -> 1 -> 2 -> null, what is the state of the list after applying insertion sort?

A 2 -> 1 -> 3 -> 4 -> null

B 1 -> 2 -> 3 -> 4 -> null

C 1 -> 3 -> 2 -> 4 -> null

D 4 -> 3 -> 2 -> 1 -> null

💡 Hint

Insertion sort on linked list rearranges nodes to ascending order.

Insertion sort on linked list sorts nodes by inserting each node into the sorted part. Final sorted list is 1 -> 2 -> 3 -> 4 -> null.

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