Spiral Matrix Traversal in DSA Javascript - 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, then down, left, and up, repeating until all elements are visited.

beginner

Which four directions are used in Spiral Matrix Traversal?

Right → Down → Left → Up, then repeat these directions in the same order until all elements are covered.

intermediate

In Spiral Matrix Traversal, what do the variables 'top', 'bottom', 'left', and 'right' represent?

'top' and 'bottom' represent the current row boundaries, while 'left' and 'right' represent the current column boundaries of the matrix section that is yet to be traversed.

intermediate

Why do we update the boundaries after completing each direction in Spiral Matrix Traversal?

Because after traversing one edge (like the top row), that row or column is fully visited and should not be visited again, so we move the boundary inward to avoid repeats.

beginner

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.

What is the first direction followed in Spiral Matrix Traversal?

A Left

B Down

C Up

D Right

The traversal starts by moving right from the top-left corner.

After moving right across the top row, which boundary is updated?

A bottom boundary moves up

B top boundary moves down

C left boundary moves right

D right boundary moves left

After traversing the top row, the top boundary moves down to exclude that row.

Which condition stops the Spiral Matrix Traversal loop?

A When all rows are visited once

B When the matrix is square

C When top boundary > bottom boundary or left boundary > right boundary

D When the matrix has more columns than rows

Traversal stops when the boundaries cross, meaning no more elements remain to visit.

What data structure is typically used to store the result of Spiral Matrix Traversal?

A Array or list

B Stack

C Queue

D Hash map

The elements are collected in an array or list in the order they are visited.

If a matrix has 3 rows and 4 columns, how many elements will Spiral Matrix Traversal visit?

A 12

B 4

C 3

D 7

The traversal visits all elements: 3 rows * 4 columns = 12 elements.

Explain how you would traverse a matrix in spiral order using boundary variables.

Think about moving right, down, left, and up while shrinking the boundaries.

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

Boundaries crossing means all elements are visited.

Practice

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

easy

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

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

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

D. Traverse the matrix diagonally from top-left to bottom-right

Solution

1. Step 1: Understand traversal pattern

   Spiral traversal means moving around the edges of the matrix and then moving inward layer by layer.

2. Step 2: Compare options

   Only Traverse the matrix in a circular pattern starting from the outer layer moving inward describes this circular inward movement. Others describe linear or diagonal traversals.

3. Final Answer:

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

4. Quick Check:

   Spiral traversal = circular inward movement [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 is the correct way to initialize variables for spiral traversal boundaries in JavaScript?

easy

A. let top = 1, bottom = matrix.length, left = 1, right = matrix[0].length;

B. let top = 0, bottom = matrix.length, left = 0, right = matrix[0].length;

C. let top = 0, bottom = matrix.length - 1, left = 0, right = matrix[0].length - 1;

D. let top = -1, bottom = matrix.length - 1, left = -1, right = matrix[0].length - 1;

Solution

1. Step 1: Understand matrix indexing

   Matrix indices start at 0, so top and left start at 0, bottom and right at length - 1.

2. Step 2: Check options

   let top = 0, bottom = matrix.length - 1, left = 0, right = matrix[0].length - 1; correctly sets boundaries using zero-based indexing. Others use incorrect start or end values.

3. Final Answer:

   let top = 0, bottom = matrix.length - 1, left = 0, right = matrix[0].length - 1; -> Option C

4. Quick Check:

   Zero-based indexing for boundaries [OK]

Hint: Remember arrays start at 0 in JavaScript [OK]

Common Mistakes:

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

3. What is the output of the following code?

const matrix = [
  [1, 2, 3],
  [4, 5, 6],
  [7, 8, 9]
];

function spiralTraversal(mat) {
  const result = [];
  let top = 0, bottom = mat.length - 1;
  let left = 0, right = mat[0].length - 1;

  while (top <= bottom && left <= right) {
    for (let i = left; i <= right; i++) result.push(mat[top][i]);
    top++;
    for (let i = top; i <= bottom; i++) result.push(mat[i][right]);
    right--;
    if (top <= bottom) {
      for (let i = right; i >= left; i--) result.push(mat[bottom][i]);
      bottom--;
    }
    if (left <= right) {
      for (let i = bottom; i >= top; i--) result.push(mat[i][left]);
      left++;
    }
  }
  return result;
}

console.log(spiralTraversal(matrix));

medium

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

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

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

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

Solution

1. Step 1: Traverse top row left to right

   Elements: 1, 2, 3

2. Step 2: Traverse right column top to bottom

   Elements: 6, 9

3. Step 3: Traverse bottom row right to left

   Elements: 8, 7

4. Step 4: Traverse left column bottom to top

   Elements: 4

5. Step 5: Traverse inner matrix

   Element: 5

6. Final Answer:

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

7. 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:

• Mixing order of bottom row traversal
• Skipping the inner element 5
• Incorrect boundary updates causing repeats

4. Identify the error in this spiral traversal code snippet:

function spiral(mat) {
  let top = 0, bottom = mat.length - 1;
  let left = 0, right = mat[0].length - 1;
  const res = [];

  while (top < bottom && left < right) {
    for (let i = left; i <= right; i++) res.push(mat[top][i]);
    top++;
    for (let i = top; i <= bottom; i++) res.push(mat[i][right]);
    right--;
    for (let i = right; i >= left; i--) res.push(mat[bottom][i]);
    bottom--;
    for (let i = bottom; i >= top; i--) res.push(mat[i][left]);
    left++;
  }
  return res;
}

medium

A. The while loop condition should be <= instead of < to include single rows or columns

B. The for loops should use < instead of <= to avoid index errors

C. The variables top and bottom are swapped incorrectly

D. The result array is not initialized properly

Solution

1. Step 1: Analyze loop condition

   The condition 'top < bottom && left < right' skips the case when top equals bottom or left equals right, missing last row or column.

2. Step 2: Correct condition

   Changing to 'top <= bottom && left <= right' ensures all rows and columns are included in traversal.

3. Final Answer:

   The while loop condition should be <= instead of < to include single rows or columns -> Option A

4. Quick Check:

   Use <= to cover all layers [OK]

Hint: Use <= in loop condition to cover all matrix layers [OK]

Common Mistakes:

• Using < causes missing last row or column
• Changing for loop bounds incorrectly
• Swapping top and bottom variables

5. Given a rectangular matrix of size 3x4:

const matrix = [
  [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]

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

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

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

Solution

1. Step 1: Traverse top row left to right

   Elements: 1, 2, 3, 4

2. Step 2: Traverse right column top to bottom

   Elements: 8, 12

3. Step 3: Traverse bottom row right to left

   Elements: 11, 10, 9

4. Step 4: Traverse left column bottom to top

   Element: 5

5. Step 5: Traverse inner row left to right

   Elements: 6, 7

6. Final Answer:

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

7. Quick Check:

   Follow edges clockwise, then inner row [OK]

Hint: Remember to handle rectangular shapes carefully [OK]

Common Mistakes:

• Mixing order of inner elements
• Skipping last inner row
• Confusing left column traversal order