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
Solution:
C/C++
/* This code is adopted from the solution given #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.
leave a comment
0 Comments