Linear Search Algorithm in DSA Go - Build from Scratch

Linear Search Algorithm

📖 Scenario: You have a list of product IDs in a store inventory. You want to find if a specific product ID is available in the list.

🎯 Goal: Build a simple program that uses linear search to find the position of a product ID in the list.

📋 What You'll Learn

Create a slice of integers called products with the exact values: 101, 203, 405, 302, 101, 509

Create an integer variable called target with the value 302

Write a for loop using the index variable i to iterate over products

Inside the loop, use an if statement to check if products[i] equals target

If found, print the message "Product found at index: " followed by the index i

If not found after the loop, print "Product not found"

💡 Why This Matters

🌍 Real World

Searching for items in a list is common in inventory management, contact lists, or any collection of data.

💼 Career

Understanding linear search helps in debugging and optimizing simple search tasks before moving to advanced algorithms.

1

Create the product list

Create a slice of integers called products with these exact values: 101, 203, 405, 302, 101, 509

DSA Go

# Your code here

Hint

Use products := []int{} to create the slice with the given numbers.

2

Set the target product ID

Create an integer variable called target and set it to 302

DSA Go

package main

func main() {
    products := []int{101, 203, 405, 302, 101, 509}
    
    // Your code here
}

Hint

Use target := 302 to create the variable.

3

Write the linear search loop

Write a for loop with index variable i to iterate over products. Inside the loop, use an if statement to check if products[i] equals target. If yes, print "Product found at index: " and the index i, then return to stop the search.

DSA Go

package main

import "fmt"

func main() {
    products := []int{101, 203, 405, 302, 101, 509}
    target := 302

    // Your code here
}

Hint

Use a for loop from 0 to len(products). Inside, check if products[i] equals target. If yes, print and return.

4

Print not found message

After the loop, print "Product not found" if the product was not found.

DSA Go

package main

import "fmt"

func main() {
    products := []int{101, 203, 405, 302, 101, 509}
    target := 302

    for i := 0; i < len(products); i++ {
        if products[i] == target {
            fmt.Println("Product found at index:", i)
            return
        }
    }

    // Your code here
}

Hint

Print "Product not found" after the loop if no match was found.

Practice

1. What is the main idea behind the Linear Search Algorithm?

easy

A. Check each element one by one until the target is found or list ends

B. Divide the list into halves repeatedly to find the target

C. Sort the list first then search using binary steps

D. Use a hash table to find the target instantly

Solution

1. Step 1: Understand linear search process

   Linear search checks each element in order, starting from the first, until it finds the target or reaches the end.

2. Step 2: Compare with other search methods

   Other methods like binary search divide the list or use extra data structures, but linear search is simple and direct.

3. Final Answer:

   Check each element one by one until the target is found or list ends -> Option A

4. Quick Check:

   Linear search = check elements one by one [OK]

Hint: Linear search checks elements in order until found [OK]

Common Mistakes:

• Confusing linear search with binary search
• Thinking linear search requires sorted list
• Assuming linear search is faster than binary search always

2. Which of the following is the correct syntax to start a linear search loop in Go for an array named arr?

easy

A. while i < len(arr) {

B. for i in arr {

C. foreach i in arr {

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

Solution

1. Step 1: Identify Go loop syntax for arrays

   In Go, the standard for loop to iterate over an array uses: for i := 0; i < len(arr); i++ { ... }

2. Step 2: Check other options for Go correctness

   Options B and C use syntax from other languages, and D is correct Go syntax.

3. Final Answer:

   for i := 0; i < len(arr); i++ { -> Option D

4. Quick Check:

   Go for loop syntax = for i := 0; i < len(arr); i++ { [OK]

Hint: Use Go's for loop with index and length [OK]

Common Mistakes:

• Using Python or JavaScript loop syntax in Go
• Forgetting to initialize loop variable
• Using while loop which is not common in Go

3. What will be the output of the following Go code?

package main
import "fmt"
func main() {
  arr := []int{4, 2, 7, 1, 9}
  target := 7
  index := -1
  for i := 0; i < len(arr); i++ {
    if arr[i] == target {
      index = i
      break
    }
  }
  fmt.Println(index)
}

medium

A. 3

B. -1

C. 2

D. 7

Solution

1. Step 1: Trace the loop to find target 7 in array

   Array is [4, 2, 7, 1, 9]. Loop checks elements: index 0=4, 1=2, 2=7 matches target.

2. Step 2: Assign index and break loop

   When arr[2] == 7, index is set to 2 and loop breaks, so index remains 2.

3. Final Answer:

   2 -> Option C

4. Quick Check:

   Target 7 found at index 2 [OK]

Hint: Stop loop when target found, print index [OK]

Common Mistakes:

• Returning element value instead of index
• Not breaking loop and returning last index
• Confusing index with value

4. Find the error in this Go linear search code snippet:

arr := []int{3, 5, 8, 6}
target := 6
index := -1
for i := 0; i <= len(arr); i++ {
  if arr[i] == target {
    index = i
    break
  }
}
fmt.Println(index)

medium

A. The target value should be declared as a pointer

B. Loop condition should be i < len(arr), not i <= len(arr)

C. Variable 'index' should be initialized to 0 instead of -1

D. The array should be sorted before searching

Solution

1. Step 1: Check loop boundary condition

   The loop uses i <= len(arr), but valid indices are 0 to len(arr)-1. Using <= causes out-of-range error.

2. Step 2: Correct loop condition

   Change loop to i < len(arr) to avoid accessing invalid index.

3. Final Answer:

   Loop condition should be i < len(arr), not i <= len(arr) -> Option B

4. Quick Check:

   Loop index must be less than length [OK]

Hint: Use i < len(arr) to avoid index out of range [OK]

Common Mistakes:

• Using <= instead of < in loop condition
• Initializing index incorrectly
• Thinking array must be sorted for linear search

5. You have a list of student names: ["Anna", "Bob", "Clara", "David", "Eva"]. You want to find the position of "David" using linear search in Go. Which code snippet correctly implements this and prints the index or -1 if not found?

hard

A. names := []string{"Anna", "Bob", "Clara", "David", "Eva"}
target := "David"
index := -1
for i := 0; i < len(names); i++ {
if names[i] == target {
index = i
break
}
}
fmt.Println(index)

B. names := []string{"Anna", "Bob", "Clara", "David", "Eva"}
target := "David"
index := 0
for i := 1; i <= len(names); i++ {
if names[i] == target {
index = i
break
}
}
fmt.Println(index)

C. names := []string{"Anna", "Bob", "Clara", "David", "Eva"}
target := "David"
index := -1
for i := 0; i < len(names); i++ {
if names[i] != target {
index = i
break
}
}
fmt.Println(index)

D. names := []string{"Anna", "Bob", "Clara", "David", "Eva"}
target := "David"
index := -1
for i := 0; i < len(names); i++ {
if names[i] == target {
index = i + 1
break
}
}
fmt.Println(index)

Solution

1. Step 1: Check loop and index initialization

   names := []string{"Anna", "Bob", "Clara", "David", "Eva"} target := "David" index := -1 for i := 0; i < len(names); i++ { if names[i] == target { index = i break } } fmt.Println(index) initializes index to -1 and loops from 0 to len(names)-1 correctly.

2. Step 2: Verify condition and output

   It compares names[i] == target and breaks on match, printing correct zero-based index or -1 if not found.

3. Final Answer:

   Option A code correctly implements linear search and prints index or -1 -> Option A

4. Quick Check:

   Linear search with correct loop and condition [OK]

Hint: Loop from 0 to len-1, compare equal, break and print index [OK]

Common Mistakes:

• Using wrong loop bounds causing runtime error
• Setting index incorrectly on no match
• Adding 1 to index causing off-by-one error