Array Access and Update at Index in DSA Go - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Array Update at Specific Index

What is the output of the following Go code after updating the array element at index 2?

DSA Go

package main
import "fmt"
func main() {
  arr := [5]int{10, 20, 30, 40, 50}
  arr[2] = 99
  fmt.Println(arr)
}

A [10 20 99 40 50]

B [10 99 30 40 50]

C [10 20 30 40 50]

D [99 20 30 40 50]

💡 Hint

Remember that array indices start at 0 in Go.

The code updates the element at index 2 (third element) to 99. The rest remain unchanged.

❓ Predict Output

intermediate

Accessing Array Element by Index

What will be printed by this Go program?

DSA Go

package main
import "fmt"
func main() {
  arr := [4]string{"a", "b", "c", "d"}
  fmt.Println(arr[3])
}

A b

B c

C d

D a

💡 Hint

Array indices start at 0.

Index 3 is the fourth element, which is "d".

❓ Predict Output

advanced

Effect of Multiple Updates on Array

What is the final state of the array after all updates?

DSA Go

package main
import "fmt"
func main() {
  arr := [3]int{1, 2, 3}
  arr[0] = arr[1] + arr[2]
  arr[2] = arr[0] - arr[1]
  arr[1] = arr[2] * 2
  fmt.Println(arr)
}

A [6 4 3]

B [5 6 2]

C [5 2 3]

D [5 6 3]

💡 Hint

Update step by step and keep track of changes carefully.

Step 1: arr[0] = 2 + 3 = 5 Step 2: arr[2] = 5 - 2 = 3 Step 3: arr[1] = 3 * 2 = 6 But note arr[1] was 2 at step 2, so step 2 uses old arr[1]. Final array is [5 6 3]. Check carefully.

❓ Predict Output

advanced

What error occurs when accessing invalid index?

What error will this Go program produce?

DSA Go

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

A panic: nil pointer dereference

B runtime error: index out of range [3] with length 2

C no error, prints 0

D compile error: invalid array index

💡 Hint

Accessing an index outside array bounds causes runtime panic.

Index 3 is out of range for array of length 2, causing runtime panic.

🧠 Conceptual

expert

Array Update and Memory Behavior in Go

Consider this Go code snippet. What is the output and why?

DSA Go

package main
import "fmt"
func main() {
  arr := [3]int{1, 2, 3}
  slice := arr[:]
  slice[1] = 99
  fmt.Println(arr)
}

A [1 99 3]

B [1 2 3]

C [99 2 3]

D compile error

💡 Hint

Slices share the underlying array memory.

The slice references the original array memory. Changing slice[1] changes arr[1].

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