Spiral Matrix Traversal in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the starting row index for spiral traversal.

DSA Go

top := [1]

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

Alen(matrix)

B1

C0

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Starting top at 1 causes skipping the first row.

Using negative indices causes runtime errors.

The top row index starts at 0 because matrix indexing begins at 0.

2fill in blank

medium

Complete the code to update the left column index after traversing the top row.

DSA Go

left [1]= 1

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

A/

B-

C*

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Decrementing left moves boundary left, causing infinite loops.

Multiplying or dividing left is incorrect for index updates.

After traversing the top row, we move the left boundary right by increasing it by 1.

3fill in blank

hard

Fix the error in the loop condition to continue spiral traversal while boundaries are valid.

DSA Go

for top [1] bottom && left [2] right {

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

A>=

B<=

C<

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using && with wrong comparison signs causes skipping or infinite loops.

Using || instead of && causes incorrect loop behavior.

The loop continues while top is less than or equal to bottom and left is less than or equal to right.

4fill in blank

hard

Fill both blanks to correctly traverse the right column from top to bottom.

DSA Go

for i := [1]; i [2]= bottom; i++ {
    result = append(result, matrix[i][right])
}

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

Atop

Bbottom

C<=

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Starting from bottom or using wrong comparison signs causes no traversal or errors.

Using >= causes the loop to not run correctly.

The loop starts at top and continues while i is less than or equal to bottom to traverse down the right column.

5fill in blank

hard

Fill all three blanks to traverse the bottom row from right to left correctly.

DSA Go

for i := [1]; i [2] [3]; i-- {
    result = append(result, matrix[bottom][i])
}

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

Aright

Bleft

C>=

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using <= with decrement causes no iterations.

Starting from left or wrong comparison signs causes errors.

The loop starts at right, continues while i is greater than or equal to left, and decrements i to move leftwards.

Practice

1. What is the main idea behind Spiral Matrix Traversal?

easy

A. Traverse the matrix in a circular pattern starting from the outer layer moving inward

B. Traverse the matrix row by row from top to bottom

C. Traverse the matrix column by column from left to right

D. Traverse only the diagonal elements of the matrix

Solution

1. Step 1: Understand traversal pattern

   Spiral traversal means moving around the edges of the matrix and gradually moving inward in a circular way.

2. Step 2: Compare options with spiral pattern

   Only Traverse the matrix in a circular pattern starting from the outer layer moving inward describes moving in a circular pattern from outer to inner layers, which matches spiral traversal.

3. Final Answer:

   Traverse the matrix in a circular pattern starting from the outer layer moving inward -> Option A

4. Quick Check:

   Spiral traversal = circular outer to inner [OK]

Hint: Think of peeling an onion layer by layer [OK]

Common Mistakes:

• Confusing spiral with row-wise traversal
• Thinking spiral means diagonal traversal
• Ignoring the inward movement after outer layer

2. Which of the following Go code snippets correctly initializes the boundaries for spiral traversal of a matrix named matrix?

easy

A. top, bottom := 0, len(matrix)-1; left, right := 0, len(matrix[0])-1

B. top, bottom := 1, len(matrix); left, right := 1, len(matrix[0])

C. top, bottom := 0, len(matrix); left, right := 0, len(matrix[0])

D. top, bottom := -1, len(matrix)-1; left, right := -1, len(matrix[0])-1

Solution

1. Step 1: Identify correct boundary initialization

   For spiral traversal, top and left start at 0, bottom and right are last indices (length - 1).

2. Step 2: Check each option

   top, bottom := 0, len(matrix)-1; left, right := 0, len(matrix[0])-1 correctly sets top=0, bottom=len(matrix)-1, left=0, right=len(matrix[0])-1. Others have off-by-one or negative values.

3. Final Answer:

   top, bottom := 0, len(matrix)-1; left, right := 0, len(matrix[0])-1 -> Option A

4. Quick Check:

   Boundaries start at 0 and end at length-1 [OK]

Hint: Remember Go slices are 0-indexed, so last index is length-1 [OK]

Common Mistakes:

• Using length instead of length-1 for boundaries
• Starting boundaries at 1 instead of 0
• Using negative indices incorrectly

3. What is the output of the following Go code snippet for spiral traversal?

matrix := [][]int{
  {1, 2, 3},
  {4, 5, 6},
  {7, 8, 9},
}
result := []int{}
top, bottom := 0, 2
left, right := 0, 2
for top <= bottom && left <= right {
  for i := left; i <= right; i++ {
    result = append(result, matrix[top][i])
  }
  top++
  for i := top; i <= bottom; i++ {
    result = append(result, matrix[i][right])
  }
  right--
  for i := right; i >= left; i-- {
    result = append(result, matrix[bottom][i])
  }
  bottom--
  for i := bottom; i >= top; i-- {
    result = append(result, matrix[i][left])
  }
  left++
}
fmt.Println(result)

medium

A. [9 8 7 6 5 4 3 2 1]

B. [1 2 3 4 5 6 7 8 9]

C. [1 4 7 8 9 6 3 2 5]

D. [1 2 3 6 9 8 7 4 5]

Solution

1. Step 1: Trace the spiral traversal steps

   Traverse top row left to right: 1,2,3; then right column top to bottom: 6,9; then bottom row right to left: 8,7; then left column bottom to top: 4; then inner layer top row: 5.

2. Step 2: Collect elements in order

   Order is [1,2,3,6,9,8,7,4,5].

3. Final Answer:

   [1 2 3 6 9 8 7 4 5] -> Option D

4. Quick Check:

   Spiral order matches [1 2 3 6 9 8 7 4 5] [OK]

Hint: Follow edges clockwise, then move inward [OK]

Common Mistakes:

• Not updating boundaries correctly after each loop
• Repeating elements or missing inner elements
• Confusing row and column indices

4. Identify the bug in this Go code snippet for spiral traversal:

top, bottom := 0, len(matrix)
left, right := 0, len(matrix[0])
for top <= bottom && left <= right {
  for i := left; i <= right; i++ {
    fmt.Print(matrix[top][i], " ")
  }
  top++
  // ... rest of traversal
}

medium

A. Missing initialization of top and left variables

B. Using len(matrix) and len(matrix[0]) instead of len(matrix)-1 and len(matrix[0])-1 for boundaries

C. Incorrect loop condition using <= instead of <

D. Using fmt.Print instead of fmt.Println

Solution

1. Step 1: Check boundary initialization

   len(matrix) gives count, but last index is len(matrix)-1. Using len(matrix) causes out-of-range error.

2. Step 2: Confirm loop indices

   Loop uses i <= right, so right must be last valid index, not length.

3. Final Answer:

   Using len(matrix) and len(matrix[0]) instead of len(matrix)-1 and len(matrix[0])-1 for boundaries -> Option B

4. Quick Check:

   Boundaries must be last index = length-1 [OK]

Hint: Remember last index is length minus one [OK]

Common Mistakes:

• Setting boundaries to length instead of length-1
• Looping beyond matrix size causing runtime error
• Ignoring zero-based indexing

5. Given a rectangular matrix:

matrix := [][]int{
  {1, 2, 3, 4},
  {5, 6, 7, 8},
  {9, 10, 11, 12},
}

What is the spiral traversal output?

hard

A. [1 2 3 4 8 12 11 10 9 5 6 7 0]

B. [1 5 9 10 11 12 8 4 3 2 6 7]

C. [1 2 3 4 8 12 11 10 9 5 6 7]

D. [1 2 3 4 5 6 7 8 9 10 11 12]

Solution

1. Step 1: Traverse top row left to right

   Elements: 1, 2, 3, 4

2. Step 2: Traverse right column top to bottom (excluding top row)

   Elements: 8, 12

3. Step 3: Traverse bottom row right to left (excluding right column)

   Elements: 11, 10, 9

4. Step 4: Traverse left column bottom to top (excluding top and bottom rows)

   Element: 5

5. Step 5: Traverse inner top row left to right

   Elements: 6, 7

6. Final Answer:

   [1 2 3 4 8 12 11 10 9 5 6 7] -> Option C

7. Quick Check:

   Spiral order matches [1 2 3 4 8 12 11 10 9 5 6 7] [OK]

Hint: Follow edges clockwise, then move inward layer by layer [OK]

Common Mistakes:

• Adding extra elements not in matrix
• Mixing row and column traversal order
• Forgetting to shrink boundaries after each layer