Spiral Matrix

 

  1. Clarify the problem:

    • The problem requires us to traverse a given matrix in a spiral order and return the elements in a list.
    • Spiral order means starting from the top-left corner, we move in a clockwise direction until we have visited all elements in the matrix.
    • We need to return the elements in the order in which they are visited.

  2. Analyze the problem:

    • Input: A matrix of integers.
    • Output: A list of integers representing the elements visited in a spiral order.
    • Constraints:
      • The number of rows and columns in the matrix is in the range [1, 10].
      • The matrix does not contain any empty rows or columns.

  3. Design an algorithm:

    • We can solve this problem by simulating the spiral traversal using iterative steps.
    • We initialize four variables: top, bottom, left, and right to keep track of the boundaries of the matrix.
    • We start with top = 0, bottom = m-1, left = 0, and right = n-1, where m is the number of rows and n is the number of columns.
    • We also initialize an empty result list to store the visited elements.
    • We repeatedly traverse the matrix in a clockwise spiral order until the boundaries collapse.
    • At each step, we add the elements in the current row or column to the result list and update the boundaries accordingly.
    • We continue this process until all elements in the matrix are visited.

  4. Explain your approach:

    • We will implement an iterative solution that simulates the spiral traversal of the matrix.
    • We use the variables top, bottom, left, and right to represent the boundaries of the matrix.
    • We initialize the result list as an empty list.
    • We use a while loop to continue the traversal until the boundaries collapse.
    • Inside the loop, we traverse the top row from left to right and add the elements to the result list.
    • We increment top to move to the next row.
    • Then, we traverse the right column from top to bottom and add the elements to the result list.
    • We decrement right to move to the previous column.
    • Next, we check if the top and bottom boundaries have crossed or not.
    • If they have not crossed, we traverse the bottom row from right to left and add the elements to the result list.
    • We decrement bottom to move to the previous row.
    • Finally, we check if the left and right boundaries have crossed or not.
    • If they have not crossed, we traverse the left column from bottom to top and add the elements to the result list.
    • We increment left to move to the next column.
    • We repeat this process until the boundaries collapse.
    • Finally, we return the result list containing all visited elements in spiral order.

  5. Write clean and readable code:

    python
  6. def spiralOrder(matrix):
    if not matrix:
        return []

    m, n = len(matrix), len(matrix[0])
    top, bottom, left, right = 0, m - 1, 0, n - 1
    result = []

    while top <= bottom and left <= right:
        # Traverse top row
        for j in range(left, right + 1):
            result.append(matrix[top][j])
        top += 1

        # Traverse right column
        for i in range(top, bottom + 1):
            result.append(matrix[i][right])
        right -= 1

        if top <= bottom:
            # Traverse bottom row
            for j in range(right, left - 1, -1):
                result.append(matrix[bottom][j])
            bottom -= 1

        if left <= right:
            # Traverse left column
            for i in range(bottom, top - 1, -1):
                result.append(matrix[i][left])
            left += 1

    return result

  7. Test your code:

    • We can test the code with multiple test cases, including edge cases and corner cases, to ensure its correctness.
    • For example:
      python
    • # Example 1
matrix = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
print(spiralOrder(matrix))
# Output: [1, 2, 3, 6, 9, 8, 7, 4, 5]

# Example 2
matrix = [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12]]
print(spiralOrder(matrix))
# Output: [1, 2, 3, 4, 8, 12, 11, 10, 9, 5, 6, 7]

  9. Optimize if necessary:

    • The solution has a time complexity of O(m * n) since we need to visit each element once.
    • The space complexity is O(1) since we only use a constant amount of extra space for variables.

  10. Handle error cases:

    • If the input matrix is empty (matrix = []), we can return an empty list since there are no elements to visit.

  11. Discuss complexity analysis:

    • The time complexity of the solution is O(m * n), where m is the number of rows and n is the number of columns in the matrix.
    • This is because we need to visit each element once during the spiral traversal.
    • The space complexity is O(1) since we only use a constant amount of extra space for variables.
    • The solution is efficient and optimal considering the requirements of the problem.
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