Javascript Program to Print a given matrix in reverse spiral form

Given a 2D array, print it in reverse spiral form. We have already discussed Print a given matrix in spiral form. This article discusses how to do the reverse printing. See the following examples. 

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

Javascript

<script>

// This is a modified code of
// https://www.geeksforgeeks.org/
// print-a-given-matrix-in-spiral-form/

let R = 3;
let C = 6;

// Function that print matrix in
// reverse spiral form.
function ReversespiralPrint(m, n, a)
{
    // Large array to initialize it
    // with elements of matrix
    let b = new Array(100);

    /* k - starting row index
    l - starting column index*/
    let i, k = 0, l = 0;

    // Counter for single dimension array
    //in which elements will be stored
    let z = 0;

    // Total elements in matrix
    let size = m*n;

    while (k < m && l < n)
    {
        // Variable to store value of matrix.
        let val;

        /* Print the first row from
           the remaining rows */
        for (i = l; i < n; ++i)
        {
            // printf("%d ", a[k][i]);
            val = a[k][i];
            b[z] = val;
            ++z;
        }
        k++;

        /* Print the last column from
           the remaining columns */
        for (i = k; i < m; ++i)
        {
            // printf("%d ", a[i][n-1]);
            val = a[i][n-1];
            b[z] = val;
            ++z;
        }
        n--;

        /* Print the last row from the
           remaining rows */
        if ( k < m)
        {
            for (i = n-1; i >= l; --i)
            {
                // printf("%d ", a[m-1][i]);
                val = a[m-1][i];
                b[z] = val;
                ++z;
            }
            m--;
        }

        /* Print the first column from the
           remaining columns */
        if (l < n)
        {
            for (i = m-1; i >= k; --i)
            {
                // printf("%d ", a[i][l]);
                val = a[i][l];
                b[z] = val;
                ++z;
            }
            l++;
        }
    }
    for (let i=size-1 ; i>=0 ; --i)
    {
        document.write(b[i] + " ");
    }
}

/* Driver program to test above functions */
    let a = [ [1, 2, 3, 4, 5, 6],
                    [7, 8, 9, 10, 11, 12],
                    [13, 14, 15, 16, 17, 18]];
    ReversespiralPrint(R, C, a);

</script>

Output:

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

Time complexity: O(R*C) where R and C are no of rows and columns in given matrix

Auxiliary Space: O(1) because using constant space

Please refer complete article on Print a given matrix in reverse spiral form for more details!