Print a given matrix in reverse spiral form
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.w3wiki.net/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.