Linear Search Algorithm in DSA Go - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Linear Search on an Array

What is the output of this Go code that performs a linear search for the value 7 in the array?

DSA Go

package main
import "fmt"
func linearSearch(arr []int, target int) int {
    for i, v := range arr {
        if v == target {
            return i
        }
    }
    return -1
}
func main() {
    arr := []int{3, 5, 7, 9, 11}
    index := linearSearch(arr, 7)
    fmt.Println(index)
}

A 1

B 3

C 2

D -1

💡 Hint

Remember that array indices start at 0 in Go.

The value 7 is at index 2 in the array [3, 5, 7, 9, 11]. The linear search returns the index where the target is found.

🧠 Conceptual

intermediate

Number of Comparisons in Worst Case

If you perform a linear search for a value that is not in an array of length 10, how many comparisons will the algorithm make in the worst case?

A 10

B 9

C 11

D 1

💡 Hint

Think about how many elements the algorithm checks when the value is not found.

In the worst case, linear search checks every element once, so it makes 10 comparisons for an array of length 10.

🔧 Debug

advanced

Find the Error in Linear Search Implementation

What error will this Go code cause when running the linear search function?

DSA Go

package main
import "fmt"
func linearSearch(arr []int, target int) int {
    for i := 0; i < len(arr); i++ {
        if arr[i] == target {
            return i
        }
    }
    return -1
}
func main() {
    arr := []int{1, 2, 3}
    fmt.Println(linearSearch(arr, 2))
}

A returns -1 always

B runtime error: index out of range

C compilation error: missing colon

D returns 0 always

💡 Hint

Check the loop condition carefully.

The loop uses i <= len(arr), which tries to access arr[len(arr)] causing an index out of range runtime error.

🚀 Application

advanced

Linear Search on a Linked List

Given a singly linked list with nodes containing values [4, 8, 15, 16, 23], what is the output of searching for value 15 using linear search?

DSA Go

type Node struct {
    value int
    next  *Node
}

func linearSearch(head *Node, target int) int {
    index := 0
    current := head
    for current != nil {
        if current.value == target {
            return index
        }
        current = current.next
        index++
    }
    return -1
}

// List: 4 -> 8 -> 15 -> 16 -> 23 -> nil
// Search for 15

A -1

B 1

C 3

D 2

💡 Hint

Count nodes starting from zero at the head.

The value 15 is at the third node, which is index 2 starting from 0.

🧠 Conceptual

expert

Time Complexity of Linear Search in Different Data Structures

Which statement correctly describes the time complexity of linear search for an element in an unsorted array and an unsorted linked list, both of size n?

A Both have O(n) time complexity because each element must be checked sequentially.

B Array search is O(log n), linked list search is O(n).

C Array search is O(1), linked list search is O(n).

D Both have O(1) time complexity because elements are accessed directly.

💡 Hint

Think about how linear search works on unsorted data.

Linear search checks each element one by one in both arrays and linked lists, so both have O(n) time complexity.

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