Spiral Matrix Traversal in DSA Typescript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Spiral Traversal on 3x3 Matrix

What is the output of the spiral traversal on the following 3x3 matrix?

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

DSA Typescript

function spiralOrder(matrix: number[][]): number[] {
  const result: number[] = [];
  let top = 0, bottom = matrix.length - 1;
  let left = 0, right = matrix[0].length - 1;

  while (top <= bottom && left <= right) {
    for (let i = left; i <= right; i++) {
      result.push(matrix[top][i]);
    }
    top++;

    for (let i = top; i <= bottom; i++) {
      result.push(matrix[i][right]);
    }
    right--;

    if (top <= bottom) {
      for (let i = right; i >= left; i--) {
        result.push(matrix[bottom][i]);
      }
      bottom--;
    }

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

console.log(spiralOrder([[1, 2, 3], [4, 5, 6], [7, 8, 9]]));

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

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

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

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

💡 Hint

Follow the edges of the matrix in a clockwise spiral.

The spiral traversal starts from top-left, moves right, then down, then left, then up, and repeats inward until all elements are visited.

❓ Predict Output

intermediate

Spiral Traversal Output for 2x4 Matrix

What is the spiral traversal output for this 2x4 matrix?

[[1, 2, 3, 4], [5, 6, 7, 8]]

DSA Typescript

function spiralOrder(matrix: number[][]): number[] {
  const result: number[] = [];
  let top = 0, bottom = matrix.length - 1;
  let left = 0, right = matrix[0].length - 1;

  while (top <= bottom && left <= right) {
    for (let i = left; i <= right; i++) {
      result.push(matrix[top][i]);
    }
    top++;

    for (let i = top; i <= bottom; i++) {
      result.push(matrix[i][right]);
    }
    right--;

    if (top <= bottom) {
      for (let i = right; i >= left; i--) {
        result.push(matrix[bottom][i]);
      }
      bottom--;
    }

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

console.log(spiralOrder([[1, 2, 3, 4], [5, 6, 7, 8]]));

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

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

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

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

💡 Hint

Traverse right on top row, then down right column, then left on bottom row.

The spiral traversal covers top row left to right, right column top to bottom, bottom row right to left, and then up left column if any.

🔧 Debug

advanced

Identify the Bug in Spiral Traversal Code

The following code is intended to perform spiral traversal on a matrix. What is the bug causing incorrect output?

DSA Typescript

function spiralOrder(matrix: number[][]): number[] {
  const result: number[] = [];
  let top = 0, bottom = matrix.length - 1;
  let left = 0, right = matrix[0].length - 1;

  while (top < bottom && left < right) {
    for (let i = left; i <= right; i++) {
      result.push(matrix[top][i]);
    }
    top++;

    for (let i = top; i <= bottom; i++) {
      result.push(matrix[i][right]);
    }
    right--;

    for (let i = right; i >= left; i--) {
      result.push(matrix[bottom][i]);
    }
    bottom--;

    for (let i = bottom; i >= top; i--) {
      result.push(matrix[i][left]);
    }
    left++;
  }
  return result;
}

A The while loop condition should be 'top <= bottom && left <= right' instead of 'top < bottom && left < right'.

B The for loops should use '<' instead of '<=' in their conditions.

C The variables 'top' and 'left' should be decremented instead of incremented.

D The result array should be cleared inside the while loop.

💡 Hint

Check the loop condition to ensure all elements are covered.

Using '<' excludes the last row or column when top equals bottom or left equals right, missing elements in the spiral.

❓ Predict Output

advanced

Spiral Traversal Output for 4x3 Matrix

What is the spiral traversal output for this 4x3 matrix?

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

DSA Typescript

function spiralOrder(matrix: number[][]): number[] {
  const result: number[] = [];
  let top = 0, bottom = matrix.length - 1;
  let left = 0, right = matrix[0].length - 1;

  while (top <= bottom && left <= right) {
    for (let i = left; i <= right; i++) {
      result.push(matrix[top][i]);
    }
    top++;

    for (let i = top; i <= bottom; i++) {
      result.push(matrix[i][right]);
    }
    right--;

    if (top <= bottom) {
      for (let i = right; i >= left; i--) {
        result.push(matrix[bottom][i]);
      }
      bottom--;
    }

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

console.log(spiralOrder([[1, 2, 3], [4, 5, 6], [7, 8, 9], [10, 11, 12]]));

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

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

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

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

💡 Hint

Follow the spiral edges clockwise from top-left inward.

The traversal covers top row left to right, right column top to bottom, bottom row right to left, and left column bottom to top, repeating inward.

🧠 Conceptual

expert

Time Complexity of Spiral Matrix Traversal

What is the time complexity of the spiral matrix traversal algorithm for an m x n matrix?

A O(m^2 + n^2) because of nested loops over rows and columns.

B O(m + n) because traversal happens along rows and columns separately.

C O(m * n) because each element is visited exactly once.

D O(1) because traversal uses constant extra space.

💡 Hint

Consider how many elements are processed in total.

The algorithm visits each element once, so the time complexity is proportional to the total number of elements m*n.

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 first, 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 A

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 only
• Ignoring inward movement after outer layer

2. Which of the following TypeScript code snippets correctly initializes variables for spiral traversal boundaries?

easy

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

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

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

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

Solution

1. Step 1: Identify correct boundary initialization

   Top and left start at 0 (first row and column). Bottom and right are last indices, so length - 1.

2. Step 2: Check options

   let top = 0, bottom = matrix.length - 1, left = 0, right = matrix[0].length - 1; correctly sets top=0, bottom=rows-1, left=0, right=cols-1. Others have off-by-one or negative values.

3. Final Answer:

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

4. Quick Check:

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

Hint: Remember array indices start at 0 [OK]

Common Mistakes:

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

3. Given a rectangular matrix with 3 rows and 5 columns, how many elements will be printed after a complete spiral traversal?

easy

A. 15

B. 8

C. 12

D. 10

Solution

1. Step 1: Calculate total elements in matrix

   Number of elements = rows x columns = 3 x 5 = 15.

2. Step 2: Understand spiral traversal covers all elements

   Spiral traversal visits every element exactly once, so all 15 elements are printed.

3. Final Answer:

   15 -> Option A

4. Quick Check:

   Total elements = rows x columns = 15 [OK]

Hint: Spiral prints all elements once [OK]

Common Mistakes:

• Confusing spiral length with perimeter
• Counting only outer layer elements
• Ignoring rectangular shape

4. What is the output of the following TypeScript code for spiral traversal?

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

function spiralOrder(matrix: number[][]): number[] {
  const result: number[] = [];
  let top = 0, bottom = matrix.length - 1;
  let left = 0, right = matrix[0].length - 1;

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

console.log(spiralOrder(matrix));

medium

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

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

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

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

Solution

1. Step 1: Trace the spiral traversal on the matrix

   Start top row left to right: 1,2,3
   Right column top to bottom: 6,9
   Bottom row right to left: 8,7
   Left column bottom to top: 4
   Now inner layer: 5

2. Step 2: Combine all elements in order

   Result array: [1,2,3,6,9,8,7,4,5]

3. Final Answer:

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

4. Quick Check:

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

Hint: Visualize peeling matrix edges in order [OK]

Common Mistakes:

• Reading rows only left to right
• Skipping inner elements
• Mixing traversal directions

5. Identify the bug in this spiral traversal code snippet:

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

medium

A. The for loops should use < instead of <= to avoid index out of bounds

B. The variables top, bottom, left, right are not initialized

C. The while loop condition should use <= instead of < to include last row and column

D. The result array is not declared before the loop

Solution

1. Step 1: Analyze the while loop condition

   Using < excludes the last row and column when top == bottom or left == right, missing elements.

2. Step 2: Correct the condition

   Change while (top < bottom && left < right) to while (top <= bottom && left <= right) to include all layers.

3. Final Answer:

   The while loop condition should use <= instead of < to include last row and column -> Option C

4. Quick Check:

   Use <= in loop condition to cover all edges [OK]

Hint: Use <= to include boundary rows and columns [OK]

Common Mistakes:

• Using < excludes last layer elements
• Not updating boundaries inside loop
• Forgetting to check conditions before loops