# 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 ` `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 stk=``new` `Stack(); ` `      `  `        ``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

 ``

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)

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