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C++ Program to Print a given matrix in reverse spiral form

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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

 

 

C++




// This is a modified code of
#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;
}


Output: 

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

Time complexity: O(m*n) where m is number of rows and n is number of columns of a given matrix

Auxiliary Space: O(100)

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



Last Updated : 17 Aug, 2023
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