Print a given matrix in spiral form

Given a 2D array, print it in spiral form. See the following examples.

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


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

spiral-matrix



Solution:

C/C++

/* This code is adopted from the solution given 
   @ http://effprog.blogspot.com/2011/01/spiral-printing-of-two-dimensional.html */

#include <stdio.h>
#define R 3
#define C 6

void spiralPrint(int m, int n, int a[R][C])
{
    int i, k = 0, l = 0;

    /*  k - starting row index
        m - ending row index
        l - starting column index
        n - ending column index
        i - iterator
    */

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

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

        /* Print the first column from the remaining columns */
        if (l < n)
        {
            for (i = m-1; i >= k; --i)
            {
                printf("%d ", a[i][l]);
            }
            l++;    
        }        
    }
}

/* 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}
    };

    spiralPrint(R, C, a);
    return 0;
}

Java

// Java program to print a given matrix in spiral form
import java.io.*;

class GFG 
{
    // Function print matrix in spiral form
    static void spiralPrint(int m, int n, int a[][])
    {
        int i, k = 0, l = 0;
        /*  k - starting row index
        m - ending row index
        l - starting column index
        n - ending column index
        i - iterator
        */
         
        while (k < m && l < n)
        {
            // Print the first row from the remaining rows
            for (i = l; i < n; ++i)
            {
                System.out.print(a[k][i]+" ");
            }
            k++;
 
            // Print the last column from the remaining columns 
            for (i = k; i < m; ++i)
            {
                System.out.print(a[i][n-1]+" ");
            }
            n--;
 
            // Print the last row from the remaining rows */
            if ( k < m)
            {
                for (i = n-1; i >= l; --i)
                {
                    System.out.print(a[m-1][i]+" ");
                }
                m--;
            }
 
            // Print the first column from the remaining columns */
            if (l < n)
            {
                for (i = m-1; i >= k; --i)
                {
                    System.out.print(a[i][l]+" ");
                }
                l++;    
            }        
        }
    }
    
    // driver program
    public static void main (String[] args) 
    {
        int R = 3;
        int C = 6;
        int a[][] = { {1,  2,  3,  4,  5,  6},
                      {7,  8,  9,  10, 11, 12},
                      {13, 14, 15, 16, 17, 18}
                    };
        spiralPrint(R,C,a);
    }
}

// Contributed by Pramod Kumar

Python3

# Python3 program to print 
# given matrix in spiral form
def spiralPrint(m, n, a) :
    k = 0; l = 0

    ''' k - starting row index
        m - ending row index
        l - starting column index
        n - ending column index
        i - iterator '''
    

    while (k < m and l < n) :
        
        # Print the first row from
        # the remaining rows 
        for i in range(l, n) :
            print(a[k][i], end = " ")
            
        k += 1

        # Print the last column from
        # the remaining columns 
        for i in range(k, m) :
            print(a[i][n - 1], end = " ")
            
        n -= 1

        # Print the last row from
        # the remaining rows 
        if ( k < m) :
            
            for i in range(n - 1, (l - 1), -1) :
                print(a[m - 1][i], end = " ")
            
            m -= 1
        
        # Print the first column from
        # the remaining columns 
        if (l < n) :
            for i in range(m - 1, k - 1, -1) :
                print(a[i][l], end = " ")
            
            l += 1

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

# This code is contributed by Nikita Tiwari.

C#

// C# program to print a given
// matrix in spiral form
using System;

class GFG
{
    // Function print matrix in spiral form
    static void spiralPrint(int m, int n, int [,]a)
    {
        int i, k = 0, l = 0;
        /* k - starting row index
        m - ending row index
        l - starting column index
        n - ending column index
        i - iterator
        */
        
        while (k < m && l < n)
        {
            // Print the first row from the remaining rows
            for (i = l; i < n; ++i)
            {
                Console.Write(a[k, i] +" ");
            }
            k++;

            // Print the last column from the remaining columns 
            for (i = k; i < m; ++i)
            {
                Console.Write(a[i,n - 1]+" ");
            }
            n--;

            // Print the last row from the remaining rows */
            if ( k < m)
            {
                for (i = n-1; i >= l; --i)
                {
                    Console.Write(a[m - 1, i]+" ");
                }
                m--;
            }

            // Print the first column from the remaining columns */
            if (l < n)
            {
                for (i = m-1; i >= k; --i)
                {
                    Console.Write(a[i, l] + " ");
                }
                l++; 
            }     
        }
    }
    
    // Driver program
    public static void Main () 
    {
        int R = 3;
        int C = 6;
        int [,]a = { {1, 2, 3, 4, 5, 6},
                    {7, 8, 9, 10, 11, 12},
                    {13, 14, 15, 16, 17, 18}
                    };
        spiralPrint(R,C,a);
    }
}

// This code is contributed by Sam007


Output:

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

Time Complexity: Time complexity of the above solution is O(mn).

Please write comments if you find the above code incorrect, or find other ways to solve the same problem.




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