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):

package main

import (
	"fmt"
)

    func spirallyTraverse(mat *[4][5]int) {    //use pointer to avoid unnecessary function recopying of Array
        // Determine dimensions automatically
	var n = len(*mat)
	var m = len((*mat)[0])

        //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.
        var a = 0;
        var 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.
        var c = 1;
        var d = 1;
    
        var totalNumOfElements = n * m;
        var counter = 0;
    
        var i=0; var j=0;
        for (counter < totalNumOfElements) {
            // Traverse top current row from left to right
	    for j = b; j <= m-d; j++ {
		fmt.Print((*mat)[a][j], ", ")
		counter++
	    }

            // Traverse last current column from top to bottom
	    for i = a + 1; i <= n-c; i++ {
		fmt.Print((*mat)[i][m-d], ", ")
		counter++
	    }

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

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

func main() {
    var mat = [4][5]int{{1,  2,  3,  4,  5}, 
                        {14, 15, 16, 17, 6},
                        {13, 20, 19, 18, 7},
                        {12, 11, 10, 9,  8}};
          
    spirallyTraverse(&mat);    //use pointer to avoid unnecessary function recopying of Array
}

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.