Insertion Sort Algorithm in DSA Go - Build from Scratch

Insertion Sort Algorithm

📖 Scenario: You have a list of unsorted numbers representing daily temperatures. You want to arrange them from lowest to highest to easily see the trend.

🎯 Goal: Build a program that sorts a list of integers using the insertion sort algorithm step-by-step.

📋 What You'll Learn

Create a slice of integers with exact values

Use a variable to track the current index in the sorting process

Implement the insertion sort logic using loops

Print the sorted slice at the end

💡 Why This Matters

🌍 Real World

Sorting data like temperatures helps in analyzing trends and making decisions, such as weather forecasting or climate studies.

💼 Career

Understanding sorting algorithms is fundamental for software development roles, especially when optimizing data processing and improving performance.

1

Create the initial slice of temperatures

Create a slice called temps with these exact integer values: 29, 15, 42, 8, 33

DSA Go

# Create the slice temps with values 29, 15, 42, 8, 33
# Your code here

Hint

Use temps := []int{29, 15, 42, 8, 33} to create the slice.

2

Add a variable to track the current index

Add a variable called i of type int to track the current index in the sorting process, starting from 1

DSA Go

package main

import "fmt"

func main() {
    temps := []int{29, 15, 42, 8, 33}
    // Declare variable i starting at 1
    // Your code here
}

Hint

Write i := 1 to start from the second element.

3

Implement the insertion sort logic

Use a for loop with variable i from 1 to the length of temps. Inside it, use a variable key to hold temps[i] and another variable j starting at i - 1. Use a for loop to move elements greater than key one position ahead. Insert key at the correct position.

DSA Go

package main

import "fmt"

func main() {
    temps := []int{29, 15, 42, 8, 33}
    i := 1
    // Implement insertion sort logic here
    // Your code here
}

Hint

Use nested loops to shift elements and insert the key in the right place.

4

Print the sorted slice

Print the sorted temps slice using fmt.Println(temps)

DSA Go

package main

import "fmt"

func main() {
    temps := []int{29, 15, 42, 8, 33}
    for i := 1; i = 0 && temps[j] > key {
            temps[j+1] = temps[j]
            j--
        }
        temps[j+1] = key
    }
    // Print the sorted temps slice
    // Your code here
}

Hint

Use fmt.Println(temps) to display the sorted slice.

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