Spiral Matrix Traversal in DSA Go - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is Spiral Matrix Traversal?

It is a way to visit all elements of a matrix in a spiral order, starting from the top-left corner and moving right, down, left, and up repeatedly until all elements are visited.

beginner

Which boundaries are used to control the spiral traversal of a matrix?

Four boundaries: top row, bottom row, left column, and right column. These boundaries shrink inward as the traversal progresses.

beginner

In Spiral Matrix Traversal, what happens after traversing the top row?

After traversing the top row from left to right, the top boundary moves down by one row to avoid revisiting the same row.

intermediate

Why do we check if top <= bottom and left <= right during spiral traversal?

To ensure we do not revisit rows or columns that have already been traversed and to stop the traversal when all elements are visited.

intermediate

What is the time complexity of Spiral Matrix Traversal for an m x n matrix?

O(m * n), because every element in the matrix is visited exactly once during the traversal.

Which direction is the first movement in Spiral Matrix Traversal?

A Bottom to top

B Top to bottom

C Right to left

D Left to right

The traversal starts from the top-left corner and moves left to right along the top row first.

What boundary moves after completing the traversal of the right column?

A Top boundary moves down

B Bottom boundary moves up

C Right boundary moves left

D Left boundary moves right

After traversing the right column from top to bottom, the right boundary moves left to avoid revisiting the same column.

When does the spiral traversal stop?

A When top boundary is greater than bottom boundary or left boundary is greater than right boundary

B After visiting the first row

C After visiting the first column

D When all boundaries are equal

The traversal stops when the top boundary crosses the bottom or the left boundary crosses the right, meaning all elements are visited.

What is the space complexity of spiral traversal if we store the result in a list?

A O(m * n)

B O(m + n)

C O(1)

D O(log(m * n))

We store all elements in a list, so space needed is proportional to the number of elements, O(m * n).

Which of these is NOT a correct step in spiral traversal?

A Traverse bottom row right to left

B Traverse right column bottom to top

C Traverse left column bottom to top

D Traverse top row left to right

The right column is traversed from top to bottom, not bottom to top.

Explain how to perform spiral traversal on a matrix using boundary variables.

Think about moving around the edges and shrinking the boundaries inward.

Describe the stopping condition for spiral matrix traversal and why it is necessary.

Consider what happens when boundaries cross each other.

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