Print matrix in antispiral form - GeeksforGeeks

22 Apr, 2021
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.  
C++
Java
Python 3
C#
Javascript
// 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 Code for Print matrix in antispiral formimport 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 program to print# matrix in anti-spiral formR = 4C = 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 Codemat = [[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
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
<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 
 
Diagonal traversal of Matrix
Print matrix in spiral form
Print a given matrix in zigzag form
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