# 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
```

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
```

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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Shashank is Passionate About Computer Science, Problem Solving & Technology,he graduated from Birla Institute of Technology Mesra. Design and Analysis of Algorithms ,Application of Data Structures are his area of Interested & he Wants to Contribute to Computer Science. You can find him more active on his personal blog "Cracking The Code" http://shashank7s.blogspot.com Cheers !!!

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