Insertion Sort Algorithm in DSA Rust - Time & Space Complexity

Time Complexity: Insertion Sort Algorithm

O(n²)

Understanding Time Complexity

We want to understand how the time taken by the Insertion Sort algorithm changes as the list size grows.

How does the number of steps increase when sorting more items?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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;
    }
}
    

This code sorts an array by inserting each element into its correct place among the already sorted elements to its left.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The inner while loop that shifts elements to insert the key in the right position.
• How many times: The outer for loop runs once for each element except the first, and the inner loop can run up to i times for each i.

How Execution Grows With Input

As the list size grows, the number of comparisons and shifts grows roughly with the square of the size in the worst case.

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

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

Final Time Complexity

Time Complexity: O(n²)

This means the time taken grows roughly with the square of the number of items to sort.

Common Mistake

[X] Wrong: "Insertion Sort always takes the same time no matter the input order."

[OK] Correct: If the list is already mostly sorted, Insertion Sort runs much faster, close to linear time, because fewer shifts are needed.

Interview Connect

Understanding Insertion Sort's time complexity helps you compare simple sorting methods and recognize when they are efficient or slow in real tasks.

Self-Check

"What if we changed the array to be already sorted? How would the time complexity change?"

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