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Print a given matrix in anticlockwise spiral form

Printing a given matrix in anticlockwise spiral form is a common problem in computer programming. The goal is to traverse the elements of the matrix in an anticlockwise spiral order, starting from the top-left corner and moving towards the center.

Anticlockwise spiral form of matrix

Problem Statement

Given a matrix, the task is to print its elements in an anticlockwise spiral order. This means that the elements are traversed in an inward spiral manner, moving from the top-left corner to the center of the matrix.

Example

Consider the following 6x6 matrix:

1  2  3  4  5  6
22 23 24 25 26 7
21 36 37 38 27 8
20 35 42 39 28 9
19 34 41 40 29 10
18 33 32 31 30 11
17 16 15 14 13 12

The anticlockwise spiral traversal of the matrix is:

1  22  21  20  19  18  17  16  15  14  13  12  11  10  9  8  7  6  5  4  3  2  23  36  35  34  33  32  31  30  29  28  27  26  25  24  37  42  41  40  39  38

Idea to Solve the Problem

To solve this problem, we can iterate through the matrix in an anticlockwise spiral order by following these steps:

  1. Traverse the top row from startCol to endCol and print each element.
  2. Traverse the right column from startRow + 1 to endRow and print each element.
  3. Traverse the bottom row from endCol - 1 to startCol and print each element.
  4. Traverse the left column from endRow - 1 to startRow + 1 and print each element.
  5. If there is an inner submatrix (when startRow + 1 <= endRow - 1 and startCol + 1 <= endCol - 1), make a recursive call with the inner submatrix as arguments.

Pseudocode

function anticlockwiseSpiral(matrix, startRow, startCol, endRow, endCol, element):
    Traverse top row from startCol to endCol and print elements
    Traverse right column from startRow + 1 to endRow and print elements
    Traverse bottom row from endCol - 1 to startCol and print elements
    Traverse left column from endRow - 1 to startRow + 1 and print elements
    If startRow + 1 <= endRow - 1 and startCol + 1 <= endCol - 1:
        Recursive call with inner submatrix

Algorithm Explanation

  1. The algorithm takes a matrix and boundaries (startRow, startCol, endRow, endCol) as inputs.

  2. It traverses the top row from startCol to endCol and prints each element.

  3. It traverses the right column from startRow + 1 to endRow and prints each element.

  4. It traverses the bottom row from endCol - 1 to startCol and prints each element.

  5. It traverses the left column from endRow - 1 to startRow + 1 and prints each element.

  6. If there is an inner submatrix (when startRow + 1 <= endRow - 1 and startCol + 1 <= endCol - 1), the algorithm makes a recursive call to the same function with the inner submatrix as arguments.

Program List

Time Complexity

The time complexity of the anticlockwise spiral matrix traversal algorithm is O(m * n), where m is the number of rows and n is the number of columns in the matrix. This is because each element of the matrix is visited exactly once.

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