Print matrix in antispiral form

Given a 2D array, the task is to print matrix in anti spiral form:

Examples:

Output: 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1

Input : arr[][4] = {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 :arr[][6] = {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

The idea is simple, we traverse matrix in spiral form and put all traversed elements in a stack. Finally one by one elements from stack and print them. 

Implementation:

C++

// C++ program to print matrix in anti-spiral form
#include <bits/stdc++.h>
using namespace std;
#define R 4
#define C 5

void antiSpiralTraversal(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  */
    stack<int> stk;

    while (k <= m && l <= n)
    {
        /* Print the first row from the remaining rows */
        for (i = l; i <= n; ++i)
            stk.push(a[k][i]);
        k++;

        /* Print the last column from the remaining columns */
        for (i = k; i <= m; ++i)
            stk.push(a[i][n]);
        n--;

        /* Print the last row from the remaining rows */
        if ( k <= m)
        {
            for (i = n; i >= l; --i)
                stk.push(a[m][i]);
            m--;
        }

        /* Print the first column from the remaining columns */
        if (l <= n)
        {
            for (i = m; i >= k; --i)
                stk.push(a[i][l]);
            l++;
        }
    }

    while (!stk.empty())
    {
        cout << stk.top() << " ";
        stk.pop();
    }
}

/* Driver program to test above functions */
int main()
{
    int mat[R][C] =
    {
        {1,  2,  3,  4,  5},
        {6,  7,  8,  9,  10},
        {11, 12, 13, 14, 15},
        {16, 17, 18, 19, 20}
    };

    antiSpiralTraversal(R-1, C-1, mat);

    return 0;
}

Java

// Java Code for Print matrix in antispiral form
import java.util.*;

class GFG {

    public static void antiSpiralTraversal(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  */
        Stack<Integer> stk=new Stack<Integer>();

        while (k <= m && l <= n)
        {
            /* Print the first row from the remaining
             rows */
            for (i = l; i <= n; ++i)
                stk.push(a[k][i]);
            k++;

            /* Print the last column from the remaining
            columns */
            for (i = k; i <= m; ++i)
                stk.push(a[i][n]);
            n--;

            /* Print the last row from the remaining
            rows */
            if ( k <= m)
            {
                for (i = n; i >= l; --i)
                    stk.push(a[m][i]);
                m--;
            }

            /* Print the first column from the remaining
            columns */
            if (l <= n)
            {
                for (i = m; i >= k; --i)
                    stk.push(a[i][l]);
                l++;
            }
        }

        while (!stk.empty())
        {
            System.out.print(stk.peek() + " ");
            stk.pop();
        }
    }

    /* Driver program to test above function */
    public static void main(String[] args)
    {
         int mat[][] =
                {
                    {1,  2,  3,  4,  5},
                    {6,  7,  8,  9,  10},
                    {11, 12, 13, 14, 15},
                    {16, 17, 18, 19, 20}
                };

        antiSpiralTraversal(mat.length - 1, mat[0].length - 1,
                                                       mat);
    }
  }
// This code is contributed by Arnav Kr. Mandal.

Python 3

# Python 3 program to print
# matrix in anti-spiral form
R = 4
C = 5

def antiSpiralTraversal(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
    stk = []

    while (k <= m and l <= n):

        # Print the first row
        # from the remaining rows
        for i in range(l, n + 1):
            stk.append(a[k][i])
        k += 1

        # Print the last column
        # from the remaining columns
        for i in range(k, m + 1):
            stk.append(a[i][n])
        n -= 1

        # Print the last row
        # from the remaining rows
        if ( k <= m):
            for i in range(n, l - 1, -1):
                stk.append(a[m][i])
            m -= 1

        # Print the first column
        # from the remaining columns
        if (l <= n):
            for i in range(m, k - 1, -1):
                stk.append(a[i][l])
            l += 1

    while len(stk) != 0:
        print(str(stk[-1]), end = " ")
        stk.pop()

# Driver Code
mat = [[1, 2, 3, 4, 5],
       [6, 7, 8, 9, 10],
       [11, 12, 13, 14, 15],
       [16, 17, 18, 19, 20]];

antiSpiralTraversal(R - 1, C - 1, mat)

# This code is contributed
# by ChitraNayal

C#

using System;
using System.Collections.Generic;

// C# Code for Print matrix in antispiral form

public class GFG
{

    public static void antiSpiralTraversal(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  */
        Stack<int> stk = new Stack<int>();

        while (k <= m && l <= n)
        {
            /* Print the first row from the remaining
             rows */
            for (i = l; i <= n; ++i)
            {
                stk.Push(a[k][i]);
            }
            k++;

            /* Print the last column from the remaining
            columns */
            for (i = k; i <= m; ++i)
            {
                stk.Push(a[i][n]);
            }
            n--;

            /* Print the last row from the remaining
            rows */
            if (k <= m)
            {
                for (i = n; i >= l; --i)
                {
                    stk.Push(a[m][i]);
                }
                m--;
            }

            /* Print the first column from the remaining
            columns */
            if (l <= n)
            {
                for (i = m; i >= k; --i)
                {
                    stk.Push(a[i][l]);
                }
                l++;
            }
        }

        while (stk.Count > 0)
        {
            Console.Write(stk.Peek() + " ");
            stk.Pop();
        }
    }

    /* Driver program to test above function */
    public static void Main(string[] args)
    {
         int[][] mat = new int[][]
         {
             new int[] {1, 2, 3, 4, 5},
             new int[] {6, 7, 8, 9, 10},
             new int[] {11, 12, 13, 14, 15},
             new int[] {16, 17, 18, 19, 20}
         };

        antiSpiralTraversal(mat.Length - 1, mat[0].Length - 1, mat);
    }
}

// This code is contributed by Shrikant13

Javascript

<script>

// Javascript Code for Print matrix in antispiral form

    function antiSpiralTraversal(m,n,a)
    {
        let i, k = 0, l = 0;

        /*  k - starting row index
            m - ending row index
            l - starting column index
            n - ending column index
            i - iterator  */
        let stk=[];

        while (k <= m && l <= n)
        {
            /* Print the first row from the remaining
             rows */
            for (i = l; i <= n; ++i)
                stk.push(a[k][i]);
            k++;

            /* Print the last column from the remaining
            columns */
            for (i = k; i <= m; ++i)
                stk.push(a[i][n]);
            n--;

            /* Print the last row from the remaining
            rows */
            if ( k <= m)
            {
                for (i = n; i >= l; --i)
                    stk.push(a[m][i]);
                m--;
            }

            /* Print the first column from the remaining
            columns */
            if (l <= n)
            {
                for (i = m; i >= k; --i)
                    stk.push(a[i][l]);
                l++;
            }
        }

        while (stk.length!=0)
        {
            document.write(stk[stk.length-1] + " ");
            stk.pop();
        }

    }

    /* Driver program to test above function */
    let mat = [[1, 2, 3, 4, 5],
       [6, 7, 8, 9, 10],
       [11, 12, 13, 14, 15],
       [16, 17, 18, 19, 20]];
    antiSpiralTraversal(mat.length - 1, mat[0].length - 1,
                                                       mat);

    // This code is contributed by avanitrachhadiya2155
</script>

Output

12 13 14 9 8 7 6 11 16 17 18 19 20 15 10 5 4 3 2 1 

Time complexity: O(R*C)
Auxiliary Space: O(R*C)

• Diagonal traversal of Matrix
• Print matrix in spiral form
• Print a given matrix in zigzag form

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