making a circle around a two dimensional array - c++

Question

I am having trouble with making a turn inside a two dimensional array to output the elements in spiral. I tried this code, but it is outputting not enough elements, I tried to make some if statements outside of the loop to cover all cases for which the general algorithm doesn't output. Can you help suggesting some way to manage the correct output.
CODE
#include <iostream>
#include <algorithm>
//#include <cmath>
using namespace std;
int main() {
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
int r, c;
cin >> r >> c;
int r_beg = 0, r_end = r - 1, c_beg = 0, c_end = c - 1;
int **m = new int*[r];
for (int i = 0; i < r; i++)
{
m[i] = new int[c];
for (int j = 0; j < c; j++)
{
cin >> m[i][j];
}
}
for (int runs = min(r, c) / 2; runs--;) {
for (int i = c_beg; i < c_end; i++)
cout << m[r_beg][i] << " ";
for (int i = r_beg; i < r_end; i++)
cout << m[i][c_end] << " ";
for (int i = c_end; i > c_beg; i--)
cout << m[r_end][i] << " ";
for (int i = r_end; i > r_beg; i--)
cout << m[i][c_beg] << " ";
r_beg++;
c_beg++;
r_end--;
c_end--;
}
if (r <= c && c_beg <= c_end) {
for (int i = c_beg; i <= c_end; i++)
cout << m[r_end][i] << " ";
}
else if (r >= c && r_beg <= r_end) {
for (int i = r_beg; i <= r_end; i++)
cout << m[i][c_end] << " ";
}
for (int i = 0; i < r; i++)
delete[] m[i];
delete[] m;
return 0;
}
Example:
Input:
3 3
1 2 3
4 5 6
7 8 9
Output: 1 2 3 6 9 8 7 4 5
If you have for example 3x10 matrix. It doesn't output.
Input:
3
7
1 2 3 4 5 6 7
8 9 10 11 12 13 14
15 16 17 18 19 20 21
Output: should get to 13, but it stops at 8.

Answer 1

The code, as presented, loops towards center until the smaller of the two dimensions is consumed up. However, if that smaller dimension has odd size, then parts of the central row or column respectively haven't been printed out. You can cover that one with some special case handling after your outer loop:
for (int runs = std::min(r, c) / 2; runs--;)
{
// ...
}
if(c < r)
{
if(c & 1)
{
for (int i = r_beg; i <= r_end; i++)
// ^ (!)
// don't forget to print last element: there's no second loop
// that would print the corner element a second time now!
std::cout << m[i][c_end] << " ";
}
}
else
{
// handles the square matrix case as well
if(r & 1)
{
for (int i = c_beg; i <= c_end; i++)
std::cout << m[r_beg][i] << " ";
}
}

Answer 2

This can be solved by carefully fine-tuning the bail-out conditions / ranges of the for loops:
#include <iostream>
using namespace std; // :-(
int main() {
//freopen("input.txt", "r", stdin);
//freopen("output.txt", "w", stdout);
int r, c;
cin >> r >> c;
int r_beg = 0, r_end = r - 1, c_beg = 0, c_end = c - 1;
int **m = new int*[r];
for (int i = 0; i < r; i++)
{
m[i] = new int[c];
for (int j = 0; j < c; j++)
{
cin >> m[i][j];
}
}
for (int runs = min(r, c);;)
{
for (int i = c_beg; i <= c_end; i++)
cout << " " << m[r_beg][i];
++r_beg;
for (int i = r_beg; i <= r_end; i++)
cout << " " << m[i][c_end];
--c_end;
if (!--runs) break;
for (int i = c_end; i >= c_beg; i--)
cout << " " << m[r_end][i];
--r_end;
for (int i = r_end; i >= r_beg; i--)
cout << " " << m[i][c_beg];
++c_beg;
if (!--runs) break;
}
for (int i = 0; i < r; i++)
delete[] m[i];
delete[] m;
return 0;
}
Input:
3 7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21
Output:
1 2 3 4 5 6 7 14 21 20 19 18 17 16 15 8 9 10 11 12 13
Live Demo on ideone
Input:
4 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Output:
1 2 3 4 8 12 16 15 14 13 9 5 6 7 11 10
Notes:
I changed the bail-out of the for loop:
Instead of min(r, c) / 2, I use min(r, c) and decrement/check runs twice in the body.
I adjusted the update of r_beg, r_end, c_beg, and c_end.