Selection Sort Algorithm in DSA Go

Mental Model

Selection sort finds the smallest item and puts it at the start, then repeats for the rest.

Analogy: Imagine picking the smallest apple from a basket and placing it in a new row, then picking the next smallest from the remaining apples, and so on.

[7, 3, 5, 2, 4]
 ↑
(minIndex points here)

Dry Run Walkthrough

Input: list: [7, 3, 5, 2, 4]

Goal: Sort the list in ascending order using selection sort

Step 1: Find smallest from index 0 to end and swap with index 0

[2, 3, 5, 7, 4]
 ↑

Why: Put smallest value 2 at the start

Step 2: Find smallest from index 1 to end and swap with index 1

[2, 3, 5, 7, 4]
    ↑

Why: 3 is already smallest in remaining, no change

Step 3: Find smallest from index 2 to end and swap with index 2

[2, 3, 4, 7, 5]
       ↑

Why: Swap 4 with 5 to put next smallest in place

Step 4: Find smallest from index 3 to end and swap with index 3

[2, 3, 4, 5, 7]
          ↑

Why: Swap 5 with 7 to continue sorting

Step 5: Last element is in place, sorting done

[2, 3, 4, 5, 7]

Why: All elements sorted

Result:

[2, 3, 4, 5, 7]

Annotated Code

DSA Go

package main

import "fmt"

func selectionSort(arr []int) {
    n := len(arr)
    for i := 0; i < n-1; 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]
    }
}

func main() {
    arr := []int{7, 3, 5, 2, 4}
    selectionSort(arr)
    fmt.Println(arr)
}

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

iterate over each position to place the smallest remaining element

minIndex := i

assume current position holds the smallest value

for j := i + 1; j < n; j++ {

search the rest of the array for a smaller value

if arr[j] < arr[minIndex] { minIndex = j }

update minIndex if a smaller element is found

arr[i], arr[minIndex] = arr[minIndex], arr[i]

swap smallest found with current position

Output

[2 3 4 5 7]

Complexity Analysis

Time: O(n^2) because for each element we scan the rest to find the smallest

Space: O(1) because sorting is done in place without extra storage

vs Alternative: Compared to bubble sort, selection sort does fewer swaps but still has O(n^2) comparisons

Edge Cases

empty array

function returns immediately without error

DSA Go

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

array with one element

no swaps needed, array remains same

DSA Go

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

array with all elements equal

no swaps occur, array stays unchanged

DSA Go

if arr[j] < arr[minIndex] { minIndex = j }

When to Use This Pattern

When you need a simple sorting method that repeatedly selects the smallest item and places it in order, use selection sort because it clearly separates finding minimum and swapping.

Common Mistakes

Mistake: Swapping inside the inner loop instead of after finding the minimum
Fix: Only swap once per outer loop iteration after the minimum is found

Summary

Selection sort repeatedly finds the smallest element and moves it to the front.

Use it for small or nearly sorted lists where simplicity matters more than speed.

The key insight is to separate finding the minimum from swapping to reduce unnecessary swaps.

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