Array Access and Update at Index in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to access the element at index 2 in the array.

DSA Go

arr := [5]int{10, 20, 30, 40, 50}
value := arr[[1]]
fmt.Println(value)

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

A4

B1

C3

D2

Attempts:

3 left

💡 Hint

Common Mistakes

Using index 3 or 4 which accesses wrong elements.

Forgetting that indices start at 0.

The index 2 accesses the third element in the array, which is 30.

2fill in blank

medium

Complete the code to update the element at index 1 to 100.

DSA Go

arr := [3]int{5, 10, 15}
arr[[1]] = 100
fmt.Println(arr)

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

A0

B1

C2

D3

Attempts:

3 left

💡 Hint

Common Mistakes

Using index 3 which is out of range.

Updating the wrong index.

Index 1 is the second element, which we update to 100.

3fill in blank

hard

Fix the error in the code to correctly update the last element of the array.

DSA Go

arr := [4]int{1, 2, 3, 4}
arr[[1]] = 10
fmt.Println(arr)

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

A3

B0

C2

D4

Attempts:

3 left

💡 Hint

Common Mistakes

Using index 4 which is out of range.

Confusing length with last index.

The last element of an array with length 4 is at index 3.

4fill in blank

hard

Fill both blanks to access and update the middle element of the array.

DSA Go

arr := [5]int{10, 20, 30, 40, 50}
arr[[1]] = arr[[2]] + 5
fmt.Println(arr)

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

A2

B1

C3

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using different indices for access and update.

Choosing wrong middle index.

The middle element in an array of length 5 is at index 2. We update it by adding 5 to its current value.

5fill in blank

hard

Fill all three blanks to double the first element, set the last element to 0, and print the array.

DSA Go

arr := [4]int{2, 4, 6, 8}
arr[[1]] = arr[[2]] * 2
arr[[3]] = 0
fmt.Println(arr)

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

A0

B3

D2

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up indices for first and last elements.

Using out of range indices.

We double the first element at index 0, then set the last element at index 3 to 0.

Practice

1. What does the expression arr[2] do in an array arr?

easy

A. Accesses the element at index 3 of the array

B. Accesses the element at index 2 of the array

C. Updates the element at index 2 of the array

D. Deletes the element at index 2 of the array

Solution

1. Step 1: Understand array indexing

   Array indexing starts at 0, so index 2 means the third element.

2. Step 2: Identify the operation

   Using arr[2] accesses (reads) the element at index 2 without changing it.

3. Final Answer:

   Accesses the element at index 2 of the array -> Option B

4. Quick Check:

   arr[2] reads element at index 2 [OK]

Hint: Remember arrays start at 0 index [OK]

Common Mistakes:

• Thinking index 2 is the second element
• Confusing access with update
• Assuming indexing starts at 1

2. Which of the following is the correct syntax to update the element at index 1 in a Go array named nums to 10?

easy

A. nums[1] = 10

B. nums(1) = 10

C. nums[1] := 10

D. nums.update(1, 10)

Solution

1. Step 1: Recall Go array update syntax

   In Go, to update an element, use array[index] = value.

2. Step 2: Check each option

   nums[1] = 10 uses correct syntax. nums(1) = 10 uses parentheses which is invalid. nums[1] := 10 uses := which is for declaration, not update. nums.update(1, 10) uses a method that does not exist.

3. Final Answer:

   nums[1] = 10 -> Option A

4. Quick Check:

   Use square brackets and = for update [OK]

Hint: Use square brackets and = to update array element [OK]

Common Mistakes:

• Using parentheses instead of square brackets
• Using := instead of = for update
• Assuming arrays have update methods

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

package main
import "fmt"
func main() {
  arr := [4]int{5, 10, 15, 20}
  arr[2] = 25
  fmt.Println(arr)
}

medium

A. [5 10 25 20]

B. [5 10 15 20]

C. [25 10 15 20]

D. [5 25 15 20]

Solution

1. Step 1: Understand initial array and update

   Initial array is [5, 10, 15, 20]. The code updates element at index 2 (third element) to 25.

2. Step 2: Apply the update and print

   After update, array becomes [5, 10, 25, 20]. Printing shows this updated array.

3. Final Answer:

   [5 10 25 20] -> Option A

4. Quick Check:

   Index 2 updated to 25 [OK]

Hint: Index 2 means third element update [OK]

Common Mistakes:

• Confusing index 2 with index 1 or 3
• Not updating the array before print
• Expecting commas in printed array

4. Identify the error in the following Go code snippet:

package main
func main() {
  arr := [3]int{1, 2, 3}
  arr[3] = 4
}

medium

A. Cannot assign to arr[3] because arrays are immutable

B. Syntax error due to missing import

C. Index out of range error at arr[3]

D. No error, code runs fine

Solution

1. Step 1: Check array size and index used

   Array has size 3 with valid indices 0, 1, 2. Accessing arr[3] is invalid.

2. Step 2: Identify error type

   Accessing arr[3] causes runtime panic: index out of range error.

3. Final Answer:

   Index out of range error at arr[3] -> Option C

4. Quick Check:

   Index must be less than array length [OK]

Hint: Array indices go from 0 to length-1 [OK]

Common Mistakes:

• Assuming arr[3] is valid for size 3 array
• Thinking arrays are immutable in Go
• Ignoring runtime errors

5. Given an array arr := [5]int{1, 2, 3, 4, 5}, which code snippet correctly doubles the value at each index using array access and update?

hard

A. for i := 1; i <= len(arr); i++ { arr[i] = arr[i] * 2 }

B. for i := 0; i < len(arr); i++ { arr[i] *= 2; arr[i+1] = arr[i] }

C. for i := 0; i < len(arr); i++ { arr[i+1] = arr[i] * 2 }

D. for i := 0; i < len(arr); i++ { arr[i] = arr[i] * 2 }

Solution

1. Step 1: Understand array length and valid indices

   Array length is 5, valid indices are 0 to 4. Loop must run from 0 to 4.

2. Step 2: Analyze each loop snippet

   for i := 0; i < len(arr); i++ { arr[i] = arr[i] * 2 } loops correctly and updates each element by doubling. for i := 1; i <= len(arr); i++ { arr[i] = arr[i] * 2 } uses i from 1 to 5, arr[5] is invalid index. for i := 0; i < len(arr); i++ { arr[i+1] = arr[i] * 2 } updates arr[i+1], which causes out of range at last iteration. for i := 0; i < len(arr); i++ { arr[i] *= 2; arr[i+1] = arr[i] } updates arr[i+1] inside loop, causing out of range and incorrect logic.

3. Final Answer:

   for i := 0; i < len(arr); i++ { arr[i] = arr[i] * 2 } -> Option D

4. Quick Check:

   Loop from 0 to len-1 and update arr[i] [OK]

Hint: Loop indices 0 to len-1 for safe access [OK]

Common Mistakes:

• Looping beyond array length causing index error
• Updating wrong indices inside loop
• Using <= instead of < in loop condition