late tide
Given a two-dimensional square matrix of order n x n, swap the elements in the upper and lower quarters and also swap the elements in the left and right quarters, excluding elements located on the diagonals. The matrix is divided into four quarters, bounded by the main and secondary diagonals: upper, lower, left, and right.
5
1 4 5 6 5
5 6 7 8 9
9 0 1 2 4
3 4 5 6 0
1 2 3 2 1
Sample Output 0
1 2 3 2 5
9 6 5 8 5
4 2 1 0 9
0 4 7 6 3
1 4 5 6 1
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balmy mist
So what part is giving you trouble?
balmy mist
yeah, at least from what you've written, it wasn't immediately clear to me what you had to do.
maybe this explains it
you have to swap the red trianges and the blue triangles
late tide
actually i got that point but the problem is that i don't know how to code something like this
balmy mist
ok, so the first thing you want to do is to iterate over the matrix
and then for each element, you want to figure out where it has to go
and then you swap it with the element at that position
there's one small catch: If you do it like I just said, you end up swapping each element twice, once on the left side and once on the right side, and so it ends up where it was originally
eh, I guess I didn't explain that last part properly
ok, so notice that element on the left of that diagonal end up on the right side of it, and elements on the right side end up on the left side:
late tide
there's one small catch: If you do it like I just said, you end up swapping each...
i have encountered this problem before
balmy mist
do you know how to solve this?
late tide
like i < n / 2?
late tide
let me understand and try if i have problem i will mention it
thanks a lot for explanation
balmy mist
np
hmm actually, I don't like this i < n / 2 approach all that much
depending on what i is here (row or column), this means you only iterate over the upper or left half of the matrix
and the idea of iterating over only half of the matrix is correct, but I wouldn't use the upper or left half
balmy mist
ok, so notice that element on the left of that diagonal end up on the right side...
instead, notice how I've drawn a diagonal in this picture
so I would iterate over the elements to the left of this diagonal
do you know how you can do that?
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@late tide Has your question been resolved? If so, type !solved :)
late tide
do you know how you can do that?
j < i ?
#include <iostream>
using namespace std;
int main() {
int n;
cin >> n;
int matrix[n][n];
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
cin >> matrix[i][j];
}
}
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
if (i != j && i + j != n - 1) {
if (i < j && i + j < n - 1) {
int new_i = n - 1 - j;
int new_j = n - 1 - i;
swap(matrix[i][j], matrix[new_i][new_j]);
} else if (i > j && i + j < n - 1) {
int new_i = j;
int new_j = i;
swap(matrix[i][j], matrix[new_i][new_j]);
}
}
}
}
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
cout << matrix[i][j];
if (j < n - 1) cout << " ";
}
cout << endl;
}
}```i managed to do something but failed
balmy mist
```cpp
#include <iostream>
using namespace std;
int main() {
int n;
cin...
the idea is to only iterate while j < i, so for (int j = 0; j < i; ++j)
late tide
the idea is to only iterate while `j < i`, so `for (int j = 0; j < i; ++j)`
it is not working 
balmy mist
ok, so assuming your code right now looks like ```c++
for (int i = 0; i < n; ++i) {
for (int j = 0; j < i; ++j) {
...
that means you're considering the entries to the left of this line:
so now you have to figure out if an entry belongs to the left triangle or to the lower triangle
can you think of a condition that would test that?
late tide
can you think of a condition that would test that?
Upper
i < j
i + j < n - 1
Lower
i > j
i + j > n - 1
Left
i > j
i + j < n - 1
Right
i < j
i + j > n - 1
balmy mist
yeah that looks correct
we're already making sure that j < i, so we can't get the upper or right triangle
which means we only need to test if i + j is less than, equal, or greater than n - 1
if it's equal, we don't do anything
if it's less than n-1, it's a left triangle, and to get to the right triangle, we need to flip the entry horizontally
which means we need to change j and keep i the same
and if it's greater than n - 1, we change i and keep j the same
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