Spiral form of matrix

The Spiral Form of a matrix is a way of arranging the elements of a matrix in a spiral order, starting from the top-left corner and moving in a clockwise direction towards the center. This arrangement is useful in various applications such as image processing, pattern recognition, and data visualization.

Problem Statement

Given a matrix, the task is to print its elements in a spiral order, starting from the top-left corner and moving towards the center in a clockwise manner.

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 spiral order of the matrix is:

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

Idea to Solve the Problem

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

1. Traverse the top row from left to right.
2. Traverse the right column from top to bottom.
3. Traverse the bottom row from right to left.
4. Traverse the left column from bottom to top.

After completing one cycle, we repeat these steps for the inner submatrix until all elements have been visited.

Pseudocode

function spiral(matrix, startRow, startCol, endRow, endCol, element):
    Traverse top row from startCol to endCol
    Traverse right column from startRow + 1 to endRow
    Traverse bottom row from endCol - 1 to startCol if startRow < endRow
    Traverse left column from endRow - 1 to startRow + 1 if startCol < endCol
    If startRow + 1 <= endRow - 1 and startCol < 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 if startRow is less than endRow, and prints each element.

5. It traverses the left column from endRow - 1 to startRow + 1 if startCol is less than endCol, and prints each element.

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

Code Solution

1) Spiral traversal of matrix in java 2) Spiral traversal of matrix in c++ 3) Spiral traversal of matrix in c 4) Spiral traversal of matrix in c# 5) Spiral traversal of matrix in golang 6) Spiral traversal of matrix in vb.net 7) Print matrix in spiral form python 8) Spiral traversal of matrix in php 9) Spiral traversal of matrix in node js 10) Spiral traversal of matrix in typescript 11) Spiral traversal of matrix in ruby 12) Spiral traversal of matrix in scala 13) Spiral traversal of matrix in swift 14) Print matrix in spiral of matrix in kotlin

Time Complexity

The time complexity of the 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.