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):

    fn spirally_traverse(mat: &Vec<Vec<i32>>) {
        // Determine dimensions automatically
        let n = mat.len();
        let m = mat[0].len();

        //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.
        let mut a = 0;
        let mut 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.
        let mut c = 1;
        let mut d = 1;
    
        let total_num_of_elements = n * m;
        let mut counter = 0;
    
        while counter < total_num_of_elements {
            // Traverse top current row from left to right
            for j in b..(m - d + 1) {
                if counter >= total_num_of_elements { break; }
                print!("{}, ", mat[a][j]);
                counter += 1;
            }

            // Traverse last current column from top to bottom
            for i in (a + 1)..(n - c + 1) {
                if counter >= total_num_of_elements { break; }
                print!("{}, ", mat[i][m - d]);
                counter += 1;
            }

            // Traverse current bottom row from right to left
            for j in (b..(m - d)).rev() {
                if counter >= total_num_of_elements { break; }
                print!("{}, ", mat[n - c][j]);
                counter += 1;
            }

            // Traverse current first column from bottom to top
            for i in ((a + 1)..(n - c)).rev() {
                if counter >= total_num_of_elements { break; }
                print!("{}, ", mat[i][b]);
                counter += 1;
            }
        
            a+=1; b+=1; c+=1; d+=1;
        }
    
        println!();
    }

fn main() {
    let mat: Vec<Vec<i32>> = vec![
        vec![1,  2,  3,  4,  5],
        vec![14, 15, 16, 17, 6],
        vec![13, 20, 19, 18, 7],
        vec![12, 11, 10, 9,  8],
    ];

    spirally_traverse(&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.