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 #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#
// 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 // 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.