Linear Search Algorithm in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return the index of the target element in the slice.

DSA Go

func linearSearch(arr []int, target int) int {
    for i := 0; i < len(arr); i++ {
        if arr[i] == [1] {
            return i
        }
    }
    return -1
}

Drag options to blanks, or click blank then click option'

Atarget

Blen(arr)

Carr

Di

Attempts:

3 left

💡 Hint

Common Mistakes

Using the loop index 'i' instead of the target value for comparison.

Comparing with the entire array instead of a single element.

The code compares each element arr[i] with the target value to find a match.

2fill in blank

medium

Complete the code to return -1 if the target is not found after the loop.

DSA Go

func linearSearch(arr []int, target int) int {
    for i := 0; i < len(arr); i++ {
        if arr[i] == target {
            return i
        }
    }
    return [1]
}

Drag options to blanks, or click blank then click option'

A-1

B0

Clen(arr)

Di

Attempts:

3 left

💡 Hint

Common Mistakes

Returning 0 which could be a valid index.

Returning the length of the array which is out of bounds.

Returning -1 indicates the target was not found in the slice.

3fill in blank

hard

Fix the error in the loop condition to avoid out-of-range errors.

DSA Go

func linearSearch(arr []int, target int) int {
    for i := 0; i <= [1]; i++ {
        if arr[i] == target {
            return i
        }
    }
    return -1
}

Drag options to blanks, or click blank then click option'

Alen(arr)

Bi

Clen(arr) - 1

Dtarget

Attempts:

3 left

💡 Hint

Common Mistakes

Using <= len(arr) causes index out of range error.

Using 'i' or 'target' as loop boundary is incorrect.

The loop should run while i is less than the last valid index, which is len(arr) - 1.

4fill in blank

hard

Fill both blanks to create a linear search that returns true if target is found, false otherwise.

DSA Go

func contains(arr []int, target int) bool {
    for [1] := 0; [1] < len(arr); [1]++ {
        if arr[[1]] == [2] {
            return true
        }
    }
    return false
}

Drag options to blanks, or click blank then click option'

Ai

Btarget

Cj

Dlen(arr)

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'j' instead of 'i' inconsistently.

Comparing with 'len(arr)' instead of 'target'.

The loop variable 'i' is used to check each element against 'target'.

5fill in blank

hard

Fill in the blanks to create a linear search that returns the index of the first occurrence of target or -1 if not found.

DSA Go

func findIndex(arr []int, [1] int) int {
    for [2] := 0; [2] < len(arr); [2]++ {
        if arr[[2]] == [1] {
            return [2]
        }
    }
    return -1
}

Drag options to blanks, or click blank then click option'

Atarget

Bi

Cindex

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names inconsistently.

Returning the wrong variable as index.

The function parameter is 'target', the loop variable is 'i', and the index returned is 'i'.

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