Insertion Sort Algorithm in DSA Go - Time & Space Complexity

Time Complexity: Insertion Sort Algorithm

O(n²)

Understanding Time Complexity

We want to understand how the time taken by Insertion Sort 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.

func insertionSort(arr []int) {
  for i := 1; i < len(arr); i++ {
    key := arr[i]
    j := i - 1
    for 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 sorted elements to its left.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The inner loop that shifts elements to insert the key.
• How many times: The outer loop runs n-1 times, and the inner loop can run up to i times for each i from 1 to n-1.

How Execution Grows With Input

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

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: Doubling the input size roughly quadruples the work needed.

Final Time Complexity

Time Complexity: O(n²)

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

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 scales helps you understand when to use it and how it compares to faster sorting methods.

Self-Check

"What if we changed the array to be nearly 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 into their correct position

C. Swap adjacent elements if they are in 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 relative to the already sorted part.

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:

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

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 or selection sort
• Thinking insertion sort divides array recursively
• Assuming insertion sort swaps all elements repeatedly

2. Which of the following Go code snippets correctly starts the outer loop for insertion sort on a slice arr?

easy

A. for i := 1; i < len(arr); i++ {

B. for i := 0; i < len(arr); i++ {

C. for i := len(arr)-1; i >= 0; i-- {

D. for i := 1; i <= len(arr); i++ {

Solution

1. Step 1: Identify correct loop start 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

   Loop runs while i is less than length of array. for i := 1; i < len(arr); i++ { matches this exactly.

3. Final Answer:

   for i := 1; i < len(arr); i++ { -> Option A

4. Quick Check:

   Start from index 1 to end [OK]

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

Common Mistakes:

• Starting loop at 0 instead of 1
• Using <= instead of < in loop condition
• Looping backwards incorrectly

3. What will be the output slice after running insertion sort on arr := []int{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 slice

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

3. Final Answer:

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

4. Quick Check:

   Sorted ascending = [1, 2, 3, 4] [OK]

Hint: Insertion sort outputs sorted ascending array [OK]

Common Mistakes:

• Assuming no change in array
• Stopping sorting early
• Mixing up partial sorted states

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

for i := 1; i < len(arr); i++ {
    key := arr[i]
    j := i - 1
    for j >= 0 && arr[j] > key {
        arr[j+1] = arr[j]
        j--
    }
    arr[j] = key
}

medium

A. The outer loop should start from i := 0

B. The inner loop condition should be arr[j] < key

C. The variable key should be declared outside the loop

D. The final assignment should be arr[j+1] = key, not arr[j] = key

Solution

1. Step 1: Analyze inner loop and final assignment

   After shifting elements greater than key, the correct position is j+1, but code assigns key to arr[j], which is incorrect.

2. Step 2: Correct the assignment

   It should be arr[j+1] = key to place key in the right spot after shifting.

3. Final Answer:

   The final assignment should be arr[j+1] = key, not arr[j] = key -> Option D

4. Quick Check:

   Insert key at arr[j+1] after shifting [OK]

Hint: Assign key to arr[j+1], not arr[j], after shifting [OK]

Common Mistakes:

• Assigning key to wrong index after inner loop
• Reversing comparison operator in inner loop
• Starting outer loop at 0

5. Given a slice arr := []int{5, 2, 9, 1, 5, 6}, how many total element shifts occur during insertion sort?

hard

A. 7

B. 6

C. 8

D. 5

Solution

1. Step 1: Trace shifts for each insertion

   Insert 2: shift 5 -> 1 shift
   Insert 9: no shift -> 0 shifts
   Insert 1: shift 9, 5, 2 -> 3 shifts
   Insert 5: shift 9, 5 -> 2 shifts
   Insert 6: shift 9 -> 1 shift

2. Step 2: Sum all shifts

   Total shifts = 1 + 0 + 3 + 2 + 1 = 7 shifts.

3. Step 3: Verify options

   7 is present as 7.

4. Final Answer:

   7 -> Option A

5. Quick Check:

   Count shifts carefully for each insertion [OK]

Hint: Count shifts by tracing each insertion step [OK]

Common Mistakes:

• Counting comparisons instead of shifts
• Missing shifts when multiple elements move
• Confusing shifts with swaps