The reader is advised to read all the lessons (tutorials) in this full course, in the order presented.

Problem

Given a matrix mat[][] of size n x m, the task is to print all elements of the matrix in spiral form, as shown in the following figure:

Employ boundary traversal technique.

Example: 
Input: mat[][] = {{1,  2,  3,  4,  5}, 
                  {14, 15, 16, 17, 6},
                  {13, 20, 19, 18, 7},
                  {12, 11, 10, 9,  8}};

The input numbers have been given in spiral form, and should just be read.

Output : 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20

Solution

The algorithm is as follows:

- Traverse the top row from j=0 to j=m-2.
- Traverse the rightmost column from i=0 to i=n-2.
- Traverse the bottom row from j=m-1 to j=1.
- Traverse the leftmost column from i=n-1 to i=1.

- Traverse the second row from j=2 to j=m-3.
- Traverse the last-but-one rightmost column from i=1 to i=n-3.
- Traverse the last-but-one bottom row from j=m-2 to j=2.
- Traverse the second leftmost column from i=n-2 to i=2.

- Traverse the third row from j=3 to j=m-4.
- Continue the traversing in the spiral form. - - -
- - - - - -

The program

The program is (read through the code and comments):

public class temp {

    public static void spirallyTraverse(int[][] mat) {
        // Determine dimensions automatically
        int n = mat.length;
        int m = mat[0].length;

        //Let a be the difference between the row being traversed, and i=0.
        //Let b be the difference between the column being traversed, and j=0.
        int a = 0;
        int b = 0;
        //Let c be the difference between the bottom of the matrix and the traversing column end element index.
        //Let d be the difference between the right end of the matrix and the traversing row end element index.
        int c = 1;
        int d = 1;
    
        int totalNumOfElements = n * m;
        int counter = 0;
    
        int i=0, j=0;
        while (counter < totalNumOfElements) {
            // Traverse top current row from left to right
            for (j=j+b; j < m-d; j++) {
                System.out.print(mat[i+b][j] + ", ");
                counter++;
            }

            // Traverse last current column from top to bottom
            for (i=i+a; i < n-c; i++) {
                System.out.print(mat[i][j] + ", ");
                counter++;
            }

            // Traverse current bottom row from right to left
            if (n > 1) { // Check to avoid duplicating row in single-row matrix
                for (j = m - d; j > b; j--) {
                    System.out.print(mat[n-c][j] + ", ");
                    counter++;
                }
            }

            // Traverse current first column from bottom to top
            if (m > 1) { // Check to avoid duplicating column in single-column matrix
                for (i = n - c; i > a; i--) {
                    System.out.print(mat[i][j] + ", ");
                    counter++;
                }
            }
        
            a++; b++; c++; d++;
        }
    
        System.out.println();
    }

    public static void main(String[] args) {
        int[][] mat = {{1,  2,  3,  4,  5}, 
                       {14, 15, 16, 17, 6},
                       {13, 20, 19, 18, 7},
                       {12, 11, 10, 9,  8}};
          
        spirallyTraverse(mat);
    }
}

The output is:

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20,

The time complexity is actually O(2 n x m), with the first O(n x m) for populating the input matrix with all the elements, and the second O(n x m) for iterating over all the elements in the matrix in the called function. Since the coefficient (multiplicand of 2) is usually omitted, the time complexity is quoted as O(n x m).

The space complexity is O(n x m), for the input matrix, which is the same as the matrix in the called function (same reference).

Thanks for reading.