C++ Program to Print matrix in antispiral form
Last Updated :
20 Feb, 2023
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;
}
Output:
12 13 14 9 8 7 6 11 16 17 18 19 20 15 10 5 4 3 2 1
Time Complexity: O(m * n) .
Space Complexity: O(m*n) as stack has been created.
Please refer complete article on Print matrix in antispiral form for more details!
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