Insertion Sort Algorithm in DSA C# - 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 bigger lists?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

for (int i = 1; i < arr.Length; i++) {
    int key = arr[i];
    int j = i - 1;
    while (j >= 0 && arr[j] > key) {
        arr[j + 1] = arr[j];
        j--;
    }
    arr[j + 1] = key;
}
    

This code sorts an array by inserting each element into its correct place among the already sorted elements on the 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.
• How many times: The outer for loop runs once for each element (n-1 times), and the inner loop can run up to i times for each outer step.

How Execution Grows With Input

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

Input Size (n)	Approx. Operations
10	About 45 comparisons and shifts
100	About 5,000 comparisons and shifts
1000	About 500,000 comparisons and shifts

Pattern observation: When input size doubles, operations roughly quadruple, showing a squared growth.

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 faster because the inner loop does fewer shifts.

Interview Connect

Knowing how Insertion Sort behaves helps you understand sorting basics and compare with faster methods, which is a useful skill in interviews.

Self-Check

"What if we used a binary search to find the insertion point? How would the time complexity change?"

Practice

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

easy

A. It builds the sorted array one element at a time by inserting elements in their correct position.

B. It divides the array into halves and sorts each half recursively.

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

D. It selects the smallest element and swaps it with the first unsorted element.

Solution

1. Step 1: Understand insertion sort process

   Insertion sort takes one element at a time and inserts it into the correct position in the already sorted part of the array.

2. Step 2: Compare with other sorting methods

   Other options describe merge sort (B), bubble sort (C), and selection sort (D), which are different algorithms.

3. Final Answer:

   It builds the sorted array one element at a time by inserting elements in their correct position. -> Option A

4. Quick Check:

   Insertion sort = Insert elements step-by-step [OK]

Hint: Insertion sort inserts elements one by one in order [OK]

Common Mistakes:

• Confusing insertion sort with bubble sort
• Thinking insertion sort divides array recursively
• Mixing up selection sort with insertion sort

2. Which of the following is the correct syntax to start the outer loop for insertion sort in C# for an array arr of length n?

easy

A. for (int i = 0; i < n; i++)

B. for (int i = 1; i <= n; i++)

C. for (int i = n - 1; i >= 0; i--)

D. for (int i = 1; i < n; i++)

Solution

1. Step 1: Identify starting index for insertion sort

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

2. Step 2: Check loop conditions

   The loop runs while i < n to cover all elements. for (int i = 1; i < n; i++) matches this.

3. Final Answer:

   for (int i = 1; i < n; i++) -> Option D

4. Quick Check:

   Start from 1 to n-1 for insertion sort loop [OK]

Hint: Start outer loop at 1, not 0, for insertion sort [OK]

Common Mistakes:

• Starting loop at 0 instead of 1
• Using <= n instead of < n causing out-of-bounds
• Counting backwards incorrectly

3. What will be the output array after applying insertion sort on int[] arr = {4, 3, 2, 1}?

medium

A. [1, 3, 2, 4]

B. [4, 3, 2, 1]

C. [1, 2, 3, 4]

D. [2, 3, 4, 1]

Solution

1. Step 1: Trace insertion sort on given array

   Start with 4 (sorted). Insert 3 before 4 -> [3,4,2,1]. Insert 2 before 3 -> [2,3,4,1]. Insert 1 before 2 -> [1,2,3,4].

2. Step 2: Confirm final sorted array

   The array is sorted in ascending order: [1, 2, 3, 4].

3. Final Answer:

   [1, 2, 3, 4] -> Option C

4. Quick Check:

   Insertion sort sorts ascending -> [1, 2, 3, 4] [OK]

Hint: Insertion sort sorts array ascending step-by-step [OK]

Common Mistakes:

• Assuming array stays unchanged
• Mixing up positions during insertion
• Stopping before fully sorting

4. Identify the error in this insertion sort code snippet:

for (int i = 1; i < arr.Length; i++) {
    int key = arr[i];
    int j = i + 1;
    while (j >= 0 && arr[j] > key) {
        arr[j + 1] = arr[j];
        j--;
    }
    arr[j + 1] = key;
}

medium

A. The variable j should start at i - 1 instead of i + 1.

B. The while loop condition should use j < arr.Length instead of j >= 0.

C. The key should be assigned after the while loop, not before.

D. The array index arr[j + 1] causes out-of-bounds error.

Solution

1. Step 1: Check initialization of j

   In insertion sort, j starts at the element before i, so it should be i - 1, not i + 1.

2. Step 2: Verify while loop condition and shifting

   The condition j >= 0 is correct to move backwards. The shifting logic is correct if j starts properly.

3. Final Answer:

   The variable j should start at i - 1 instead of i + 1. -> Option A

4. Quick Check:

   j starts at i-1 to compare previous elements [OK]

Hint: j must start left of i to compare previous elements [OK]

Common Mistakes:

• Starting j at i+1 causing wrong comparisons
• Incorrect while loop boundary causing errors
• Misplacing key assignment inside loop

5. Given a nearly sorted array int[] arr = {1, 2, 4, 3, 5, 6}, how many element shifts will insertion sort perform to fully sort it?

hard

A. 2

B. 1

C. 3

D. 4

Solution

1. Step 1: Identify where shifts occur

   Only the element 3 is out of order. It needs to be inserted before 4.

2. Step 2: Count shifts to insert 3

   To insert 3, j=2 (arr[2]=4 > 3), shift arr[3]=arr[2] (1 shift), j=1 (arr[1]=2 < 3), stop. Place arr[2]=3 (not a shift).

3. Step 3: Check if other shifts occur

   No other elements need shifting. Total shifts = 1.

4. Final Answer:

   1 -> Option B

5. Quick Check:

   One shift when inserting 3 [OK]

Hint: Count only elements moved to insert out-of-place items [OK]

Common Mistakes:

• Counting swaps instead of shifts
• Ignoring that placing key is not a shift
• Assuming all elements shift