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

Implementation:

C++

// This is a modified code of
// https://www.geeksforgeeks.org/print-a-given-matrix-in-spiral-form/
#include <iostream>
#define R 3
#define C 6
using namespace std;

// Function that print matrix in reverse spiral form.
void ReversespiralPrint(int m, int n, int a[R][C])
{
    // Large array to initialize it
    // with elements of matrix
    long int b[100];

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

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

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

    while (k < m && l < n)
    {
        // Variable to store value of matrix.
        int 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 (int i=size-1 ; i>=0 ; --i)
    {
        cout<<b[i]<<" ";
    }
}

/* Driver program to test above functions */
int main()
{
    int a[R][C] = { {1, 2, 3, 4, 5, 6},
                    {7, 8, 9, 10, 11, 12},
                    {13, 14, 15, 16, 17, 18}};
    ReversespiralPrint(R, C, a);
    return 0;
}

Java

// JAVA Code for Print a given matrix in
// reverse spiral form
import java.io.*;
class GFG {

    public static int R = 3, C = 6;

    // Function that print matrix in reverse spiral form.
    public static void ReversespiralPrint(int m, int n,
                                             int a[][])
    {
        // Large array to initialize it
        // with elements of matrix
        long b[] = new long[100];

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

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

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

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

            /* Print the first row from the remaining
            rows */
            for (i = l; i < n; ++i)
            {

                val = a[k][i];
                b[z] = val;
                ++z;
            }
            k++;

            /* Print the last column from the remaining
            columns */
            for (i = k; i < m; ++i)
            {

                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)
                {

                    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)
                {

                    val = a[i][l];
                    b[z] = val;
                    ++z;
                }
                l++;
            }
        }

        for (int x = size-1 ; x>=0 ; --x)
        {
            System.out.print(b[x]+" ");
        }
    }

    /* Driver program to test above function */
    public static void main(String[] args)
    {
        int a[][] = { {1, 2, 3, 4, 5, 6},
                    {7, 8, 9, 10, 11, 12},
                    {13, 14, 15, 16, 17, 18}};

        ReversespiralPrint(R, C, a);

    }
  }
// This code is contributed by Arnav Kr. Mandal.

Python3

# Python3 Code to Print a given
# matrix in reverse spiral form

# This is a modified code of
# https:#www.geeksforgeeks.org/print-a-given-matrix-in-spiral-form/
R, C = 3, 6

def ReversespiralPrint(m, n, a):

    # Large array to initialize it
    # with elements of matrix
    b = [0 for i in range(100)]

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

    # Counter for single dimension array
    # in which elements will be stored
    z = 0

    # Total elements in matrix
    size = m * n

    while (k < m and l < n):

        # Variable to store value of matrix.
        val = 0

        # Print the first row
        # from the remaining rows
        for i in range(l, n):

            # printf("%d ", a[k][i])
            val = a[k][i]
            b[z] = val
            z += 1
        k += 1

        # Print the last column
        # from the remaining columns
        for i in range(k, m):

            # printf("%d ", a[i][n-1])
            val = a[i][n - 1]
            b[z] = val
            z += 1

        n -= 1

        # Print the last row
        # from the remaining rows
        if (k < m):
            for i in range(n - 1, l - 1, -1):

                # printf("%d ", a[m-1][i])
                val = a[m - 1][i]
                b[z] = val
                z += 1

        m -= 1

        # Print the first column
        # from the remaining columns
        if (l < n):
            for i in range(m - 1, k - 1, -1):

                # printf("%d ", a[i][l])
                val = a[i][l]
                b[z] = val
                z += 1
            l += 1

    for i in range(size - 1, -1, -1):
        print(b[i], end = " ")

# Driver Code
a = [[1, 2, 3, 4, 5, 6],
     [7, 8, 9, 10, 11, 12],
     [13, 14, 15, 16, 17, 18]]

ReversespiralPrint(R, C, a)

# This code is contributed by mohit kumar

C#

// C# Code for Print a given matrix in
// reverse spiral form
using System;
class GFG {

    public static int R = 3, C = 6;

    // Function that print matrix in reverse spiral form.
    public static void ReversespiralPrint(int m, int n,
                                            int [,]a)
    {
        // Large array to initialize it
        // with elements of matrix
        long []b = new long[100];

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

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

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

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

            /* Print the first row from the remaining
            rows */
            for (i = l; i < n; ++i)
            {

                val = a[k,i];
                b[z] = val;
                ++z;
            }
            k++;

            /* Print the last column from the remaining
            columns */
            for (i = k; i < m; ++i)
            {

                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)
                {

                    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)
                {

                    val = a[i,l];
                    b[z] = val;
                    ++z;
                }
                l++;
            }
        }

        for (int x = size-1 ; x>=0 ; --x)
        {
        Console.Write(b[x]+" ");
        }
    }

    /* Driver program to test above function */
    public static void Main()
    {
        int [ ,]a = { {1, 2, 3, 4, 5, 6},
                    {7, 8, 9, 10, 11, 12},
                    {13, 14, 15, 16, 17, 18}};

        ReversespiralPrint(R, C, a);

    }
}
// This code is contributed by vt_m.

PHP

<?php
// PHP Code for Print a given
// matrix in reverse spiral form
$R=3;
$C=6;

// Function that print matrix
// in reverse spiral form.
function ReversespiralPrint($m, $n, array $a)
{

    // Large array to initialize it
    // with elements of matrix
    $b;

    // k - starting row index
    // l - starting column index
    $k = 0;
    $l = 0;

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

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

    while ($k < $m && $l < $n)
    {

        // Variable to store
        // value of matrix.
        $val;

        // Print the first row from
        // the remaining rows
        for ($i = $l; $i < $n; ++$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)
            {
                $val = $a[$i][$l];
                $b[$z] = $val;
                ++$z;
            }
            $l++;
        }
    }
    for ($i = $size - 1; $i >= 0; --$i)
    {
        echo $b[$i]." ";
    }
}

    // Driver Code
    $a= array(array(1, 2, 3, 4, 5, 6),
              array(7, 8, 9, 10, 11, 12),
              array(13, 14, 15, 16, 17, 18));
    ReversespiralPrint($R, $C, $a);

// This Code is contributed by mits
?>

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(m*n), To traverse the matrix O(m*n) time is required.
Auxiliary Space: O(1), No extra space is required.

This article is contributed by Sahil Rajput. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.