Print a given matrix in reverse spiral form

14 Jun, 2022
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
 
 
C++
Java
Python3
C#
PHP
Javascript
// This is a modified code of// https://www.geeksforgeeks.org/print-a-given-matrix-in-spiral-form/#include <iostream>#define R 3#define C 6using 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 Code for Print a given matrix in// reverse spiral formclass 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 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 Codea = [[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# Code for Print a given matrix in// reverse spiral formusing 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 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?>
<script> // This is a modified code of// https://www.geeksforgeeks.org/// 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.
This article is contributed by Sahil Rajput. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
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