Print a given matrix in reverse spiral form

Last Updated : 23 Jul, 2025

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/dsa/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# 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.

Analysis of Algorithms

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