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 = 6spiralPrint(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.
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