Array Traversal Patterns in DSA Go - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of nested loop array traversal

What is the output of the following Go code that traverses a 2D array?

DSA Go

package main
import "fmt"
func main() {
	arr := [][]int{{1, 2}, {3, 4}}
	for i := 0; i < len(arr); i++ {
		for j := 0; j < len(arr[i]); j++ {
			fmt.Print(arr[i][j], " ")
		}
	}
}

A 1 2 3 4

B 1 3 2 4

C 3 4 1 2

D 2 1 4 3

💡 Hint

Think about how nested loops iterate row by row and column by column.

The outer loop goes through each row, and the inner loop goes through each element in that row. So the elements print in order: 1, 2, then 3, 4.

❓ Predict Output

intermediate

Output of reverse traversal of an array

What is the output of this Go code that traverses an array backwards?

DSA Go

package main
import "fmt"
func main() {
	arr := []int{5, 10, 15, 20}
	for i := len(arr) - 1; i >= 0; i-- {
		fmt.Print(arr[i], " ")
	}
}

A 20 15 10 5

B 5 10 15 20

C 20 10 15 5

D 15 20 10 5

💡 Hint

The loop starts from the last index and moves to the first.

The loop starts at the last element (index 3) and moves backward to index 0, printing elements in reverse order.

🔧 Debug

advanced

Identify the runtime error in array traversal

What error will this Go code produce when run?

DSA Go

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

A prints: 1 2 3

B runtime error: index out of range

C prints: 1 2 3 0

D compilation error: invalid loop condition

💡 Hint

Check the loop condition and array indexing carefully.

The loop runs from 0 to len(arr) inclusive, but valid indices are 0 to len(arr)-1. Accessing arr[len(arr)] causes an index out of range runtime error.

🧠 Conceptual

advanced

Understanding stride in array traversal

In Go, what will be the output of this code that traverses an array with a step of 2?

DSA Go

package main
import "fmt"
func main() {
	arr := []int{10, 20, 30, 40, 50}
	for i := 0; i < len(arr); i += 2 {
		fmt.Print(arr[i], " ")
	}
}

A 50 30 10

B 20 40

C 10 20 30 40 50

D 10 30 50

💡 Hint

The loop increments index by 2 each time, skipping every other element.

The loop prints elements at indices 0, 2, and 4, which are 10, 30, and 50 respectively.

🚀 Application

expert

Find the sum of elements at even indices

What is the value of variable sum after running this Go code that sums elements at even indices?

DSA Go

package main
import "fmt"
func main() {
	arr := []int{3, 6, 9, 12, 15, 18}
	sum := 0
	for i := 0; i < len(arr); i++ {
		if i%2 == 0 {
			sum += arr[i]
		}
	}
	fmt.Print(sum)
}

A 33

B 63

C 27

D 54

💡 Hint

Add elements at indices 0, 2, and 4 only.

Elements at even indices are arr[0]=3, arr[2]=9, arr[4]=15. Their sum is 3+9+15=27.

Practice

1. What does the following Go code snippet do?
for i := 0; i < len(arr); i++ { fmt.Println(arr[i]) }

easy

A. Prints each element of the array from start to end

B. Prints the array elements in reverse order

C. Prints only the first element of the array

D. Causes a runtime error due to index out of range

Solution

1. Step 1: Understand the loop structure

   The loop starts at index 0 and runs while i is less than the length of the array, increasing i by 1 each time.

2. Step 2: Analyze the print statement inside the loop

   Each iteration prints the element at index i, so all elements from start to end are printed in order.

3. Final Answer:

   Prints each element of the array from start to end -> Option A

4. Quick Check:

   Loop from 0 to len(arr)-1 prints all elements [OK]

Hint: Loop from 0 to len(arr) prints all elements in order [OK]

Common Mistakes:

• Thinking it prints in reverse order
• Assuming it prints only one element
• Confusing loop condition causing error

2. Which of the following is the correct syntax to traverse an array arr using a range-based for loop in Go?

easy

A. for value in arr { fmt.Println(value) }

B. for i := 0; i < len(arr); i++ { fmt.Println(arr[i]) }

C. for index, value := range arr { fmt.Println(index, value) }

D. foreach i in arr { fmt.Println(arr[i]) }

Solution

1. Step 1: Recall Go's range syntax

   Go uses for index, value := range arr to iterate over arrays or slices.

2. Step 2: Check each option

   for index, value := range arr { fmt.Println(index, value) } matches the correct syntax. for i := 0; i < len(arr); i++ { fmt.Println(arr[i]) } is valid but not range-based. Options C and D are invalid Go syntax.

3. Final Answer:

   for index, value := range arr { fmt.Println(index, value) } -> Option C

4. Quick Check:

   Range loop syntax = for index, value := range arr { fmt.Println(index, value) } [OK]

Hint: Use 'for index, value := range arr' for range loops in Go [OK]

Common Mistakes:

• Using Python or other language syntax
• Confusing traditional for loop with range loop
• Using invalid keywords like foreach

3. What is the output of the following Go code?

arr := []int{10, 20, 30, 40}
for i := len(arr) - 1; i >= 0; i-- {
    fmt.Print(arr[i], " ")
}

medium

A. 10 20 30 40

B. 40 30 20 10

C. Runtime error due to invalid index

D. No output

Solution

1. Step 1: Understand the loop range

   The loop starts at index len(arr)-1 (3) and decrements i until it reaches 0.

2. Step 2: Trace the printed values

   Indexes printed: 3, 2, 1, 0 corresponding to values 40, 30, 20, 10 printed with spaces.

3. Final Answer:

   40 30 20 10 -> Option B

4. Quick Check:

   Reverse loop prints elements backward [OK]

Hint: Loop from len(arr)-1 down to 0 prints array in reverse [OK]

Common Mistakes:

• Assuming forward order output
• Off-by-one errors causing runtime error
• Confusing loop condition direction

4. Identify the error in this Go code snippet that tries to print all elements of an array arr:

for i := 0; i <= len(arr); i++ {
    fmt.Println(arr[i])
}

medium

A. The print statement syntax is incorrect

B. The loop variable should start from 1 instead of 0

C. The array arr is not declared

D. The loop condition should be i < len(arr) to avoid index out of range

Solution

1. Step 1: Analyze the loop condition

   The loop runs while i <= len(arr), which means i can equal len(arr), an invalid index.

2. Step 2: Understand array indexing

   Array indexes go from 0 to len(arr)-1. Accessing arr[len(arr)] causes runtime error.

3. Final Answer:

   The loop condition should be i < len(arr) to avoid index out of range -> Option D

4. Quick Check:

   Index must be less than length, not less or equal [OK]

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

Common Mistakes:

• Using <= instead of < in loop condition
• Starting loop at 1 instead of 0
• Ignoring array declaration errors

5. Given an array arr := []int{1, 2, 3, 4, 5}, which code snippet correctly prints only the elements at even indexes (0, 2, 4)?

hard

A. for i := 0; i < len(arr); i += 2 { fmt.Print(arr[i], " ") }

B. for i := 1; i < len(arr); i += 2 { fmt.Print(arr[i], " ") }

C. for _, v := range arr { if v%2 == 0 { fmt.Print(v, " ") } }

D. for i := len(arr)-1; i >= 0; i-- { if i%2 == 1 { fmt.Print(arr[i], " ") } }

Solution

1. Step 1: Understand the requirement

   We want to print elements at even indexes: 0, 2, 4.

2. Step 2: Check each option

   for i := 0; i < len(arr); i += 2 { fmt.Print(arr[i], " ") } increments i by 2 starting at 0, correctly accessing even indexes. for i := 1; i < len(arr); i += 2 { fmt.Print(arr[i], " ") } accesses odd indexes. for _, v := range arr { if v%2 == 0 { fmt.Print(v, " ") } } prints even values, not indexes. for i := len(arr)-1; i >= 0; i-- { if i%2 == 1 { fmt.Print(arr[i], " ") } } prints odd indexes in reverse.

3. Final Answer:

   for i := 0; i < len(arr); i += 2 { fmt.Print(arr[i], " ") } -> Option A

4. Quick Check:

   Increment by 2 from 0 prints even indexes [OK]

Hint: Increment index by 2 starting at 0 for even indexes [OK]

Common Mistakes:

• Confusing even indexes with even values
• Starting loop at 1 instead of 0
• Using incorrect loop increments