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.
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; } |
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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. |
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Python 3
# Python 3 program to print # matrix in anti-spiral form R = 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 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 |
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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 |
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Output:
12 13 14 9 8 7 6 11 16 17 18 19 20 15 10 5 4 3 2 1
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