Print a given matrix in spiral form

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

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

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

Solution:

C++

#include <bits/stdc++.h> 
using namespace std; 
#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)  
        {  
            cout<<a[k][i]<<" ";  
        }  
        k++;  
  
        /* Print the last column from the remaining columns */
        for (i = k; i < m; ++i)  
        {  
            cout<<a[i][n-1]<<" ";  
        }  
        n--;  
  
        /* Print the last row from the remaining rows */
        if ( k < m)  
        {  
            for (i = n-1; i >= l; --i)  
            {  
                cout<<a[m-1][i]<<" ";  
            }  
            m--;  
        }  
  
        /* Print the first column from the remaining columns */
        if (l < n)  
        {  
            for (i = m-1; i >= k; --i)  
            {  
                cout<<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;  
}  
  
// This is code is contributed by rathbhupendra 

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 

PHP

<?php  
// PHP program to print a given 
// matrix in spiral form 
$R = 3; 
$C = 6; 
  
function 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 && $l < $n) 
    { 
        /* Print the first row from 
           the remaining rows */
        for ($i = $l; $i < $n; ++$i) 
        { 
            echo $a[$k][$i] . " "; 
        } 
        $k++; 
  
        /* Print the last column  
        from the remaining columns */
        for ($i = $k; $i < $m; ++$i) 
        { 
            echo $a[$i][$n - 1] . " "; 
        } 
        $n--; 
  
        /* Print the last row from 
           the remaining rows */
        if ($k < $m) 
        { 
            for ($i = $n - 1; $i >= $l; --$i) 
            { 
                echo $a[$m - 1][$i] . " "; 
            } 
            $m--; 
        } 
  
        /* Print the first column from 
           the remaining columns */
        if ($l < $n) 
        { 
            for ($i = $m - 1; $i >= $k; --$i) 
            { 
                echo $a[$i][$l] . " "; 
            } 
            $l++;  
        }      
    } 
} 
  
// 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)); 
  
spiralPrint($R, $C, $a); 
  
// This code is contributed 
// by ChitraNayal 
?> 

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.

My Personal Notes

Suggest a Topic

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

C++

C

Java

Python3

C#

PHP

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