Selection Sort Algorithm in DSA Go - Build from Scratch

Selection Sort Algorithm

📖 Scenario: You have a list of numbers that you want to arrange from smallest to largest, just like organizing books by height on a shelf.

🎯 Goal: You will build a program that sorts a list of numbers using the selection sort method, which finds the smallest number and places it at the start, then repeats for the rest.

📋 What You'll Learn

Create a slice of integers with exact values

Use a variable to track the current position in the slice

Implement the selection sort logic with nested loops

Print the sorted slice at the end

💡 Why This Matters

🌍 Real World

Sorting is used in many real-life tasks like organizing data, searching faster, and preparing lists for reports.

💼 Career

Understanding sorting algorithms like selection sort helps in software development, data analysis, and technical interviews.

1

Create the initial list of numbers

Create a slice called numbers with these exact integers: 64, 25, 12, 22, 11

DSA Go

# Create the slice numbers with the values 64, 25, 12, 22, 11
# Your code here

Hint

Use numbers := []int{64, 25, 12, 22, 11} to create the slice.

2

Add a variable to track the current index

Add a for loop with variable i that goes from 0 to len(numbers)-1 to track the current position in the slice

DSA Go

package main

func main() {
    numbers := []int{64, 25, 12, 22, 11}
    
    // Loop from i = 0 to len(numbers)-1
    // Your code here
}

Hint

Use a for loop with i := 0; i < len(numbers)-1; i++.

3

Implement the selection sort logic

Inside the for i loop, create a variable minIndex set to i. Then add a nested for j loop from i+1 to len(numbers). Inside the inner loop, if numbers[j] < numbers[minIndex], update minIndex = j. After the inner loop, swap numbers[i] and numbers[minIndex].

DSA Go

package main

func main() {
    numbers := []int{64, 25, 12, 22, 11}
    for i := 0; i < len(numbers)-1; i++ {
        // Add selection sort logic here
        // Your code here
    }
}

Hint

Use nested loops and swap the elements at i and minIndex.

4

Print the sorted slice

Add a fmt.Println(numbers) statement after the sorting loops to print the sorted slice. Remember to import fmt at the top.

DSA Go

package main

import "fmt"

func main() {
    numbers := []int{64, 25, 12, 22, 11}
    for i := 0; i < len(numbers)-1; i++ {
        minIndex := i
        for j := i + 1; j < len(numbers); j++ {
            if numbers[j] < numbers[minIndex] {
                minIndex = j
            }
        }
        numbers[i], numbers[minIndex] = numbers[minIndex], numbers[i]
    }
    // Print the sorted slice
    // Your code here
}

Hint

Use fmt.Println(numbers) to print the sorted slice.

Practice

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

easy

A. Repeatedly find the smallest element and move it to the front

B. Divide the list into halves and merge sorted halves

C. Swap adjacent elements if they are in wrong order repeatedly

D. Use a pivot to partition the list into smaller parts

Solution

1. Step 1: Understand selection sort process

   Selection sort finds the smallest element in the unsorted part and swaps it with the first unsorted element.

2. Step 2: Compare with other sorting methods

   Other options describe merge sort, bubble sort, and quicksort, which are different algorithms.

3. Final Answer:

   Repeatedly find the smallest element and move it to the front -> Option A

4. Quick Check:

   Selection sort = find smallest and swap [OK]

Hint: Selection sort always picks the smallest next element [OK]

Common Mistakes:

• Confusing selection sort with bubble sort
• Thinking it divides list like merge sort
• Assuming it uses a pivot like quicksort

2. Which of the following is the correct way to start the outer loop for selection sort in Go for an array arr of length n?

easy

A. for i := 1; i < n; i++ {

B. for i := 0; i < n-1; i++ {

C. for i := 0; i <= n; i++ {

D. for i := 1; i <= n; i++ {

Solution

1. Step 1: Understand loop range for selection sort

   The outer loop runs from 0 to n-2 because the last element will be sorted automatically after previous passes.

2. Step 2: Check each option's correctness

   for i := 0; i < n-1; i++ { uses i < n-1, which is correct. Others either go out of bounds or start at wrong index.

3. Final Answer:

   for i := 0; i < n-1; i++ { -> Option B

4. Quick Check:

   Outer loop runs 0 to n-2 [OK]

Hint: Outer loop stops at n-2 because last element is sorted [OK]

Common Mistakes:

• Using <= n causing index out of range
• Starting loop at 1 instead of 0
• Loop running too many times

3. What will be the output of the following Go code snippet after applying one full pass of selection sort on the array?

arr := []int{29, 10, 14, 37, 13}
n := len(arr)
minIndex := 0
for j := 1; j < n; j++ {
    if arr[j] < arr[minIndex] {
        minIndex = j
    }
}
arr[0], arr[minIndex] = arr[minIndex], arr[0]
fmt.Println(arr)

medium

A. [10 13 14 37 29]

B. [13 10 14 37 29]

C. [29 10 14 37 13]

D. [10 29 14 37 13]

Solution

1. Step 1: Find minimum element index in full array

   Minimum element is 10 at index 1, so minIndex = 1.

2. Step 2: Swap arr[0] and arr[minIndex]

   Swap arr[0] (29) with arr[1] (10), resulting in [10 29 14 37 13].

3. Final Answer:

   [10 29 14 37 13] -> Option D

4. Quick Check:

   First pass places smallest element at front [OK]

Hint: First pass swaps smallest element to front [OK]

Common Mistakes:

• Not swapping elements after finding minimum
• Confusing minIndex with current index
• Printing array before swap

4. Identify the error in the following Go code implementing selection sort:

func selectionSort(arr []int) {
    n := len(arr)
    for i := 0; i < n; i++ {
        minIndex := i
        for j := i + 1; j <= n; j++ {
            if arr[j] < arr[minIndex] {
                minIndex = j
            }
        }
        arr[i], arr[minIndex] = arr[minIndex], arr[i]
    }
}

medium

A. Swap should be done only if minIndex != i

B. minIndex should be initialized outside the outer loop

C. Inner loop condition should be j < n, not j <= n

D. Outer loop should run from i := 1 to n-1

Solution

1. Step 1: Check inner loop boundary

   Inner loop uses j <= n, which causes index out of range error because valid indices are 0 to n-1.

2. Step 2: Correct inner loop condition

   It should be j < n to stay within array bounds.

3. Final Answer:

   Inner loop condition should be j < n, not j <= n -> Option C

4. Quick Check:

   Array indices go from 0 to n-1 [OK]

Hint: Array indices end at n-1, not n [OK]

Common Mistakes:

• Using <= n causing runtime panic
• Initializing minIndex outside outer loop
• Unnecessary swap checks

5. Given the array arr := []int{64, 25, 12, 22, 11}, after how many swaps will the array be fully sorted using selection sort, and what will be the final sorted array?

hard

A. 4 swaps; [11 12 22 25 64]

B. 5 swaps; [11 12 22 25 64]

C. 4 swaps; [64 25 12 22 11]

D. 3 swaps; [11 12 22 25 64]

Solution

1. Step 1: Count swaps in each selection sort pass

   Pass 1: swap 64 and 11 -> [11 25 12 22 64]
   Pass 2: swap 25 and 12 -> [11 12 25 22 64]
   Pass 3: swap 25 and 22 -> [11 12 22 25 64]
   Pass 4: swap 25 and 25 (no change, but swap counted as 1)
   Total swaps = 4 (last swap is with itself, usually counted as swap)

2. Step 2: Final sorted array

   The array after sorting is [11 12 22 25 64].

3. Final Answer:

   4 swaps; [11 12 22 25 64] -> Option A

4. Quick Check:

   Selection sort swaps once per pass, total n-1 swaps [OK]

Hint: Selection sort swaps once per pass, total n-1 swaps [OK]

Common Mistakes:

• Counting swaps incorrectly when swapping element with itself
• Assuming more swaps than passes
• Confusing sorted array with original