thanks amit for this code
i tried this for the first time and am still not over finding out my
mistakes (got another one while trying it out on an array)
yours was an easy one :)
--
You received this message because you are subscribed to the Google Groups
"Algorithm Geeks" group.
To post to thi
i am sorry for givin such a naive solution and making so many mistakes.
i still left a mistake in dir==4
*
col=n-1-(col+1);*
--
You received this message because you are subscribed to the Google Groups
"Algorithm Geeks" group.
To post to this group, send email to algogeeks@googlegroups.com.
To u
There will be following changes to this
while(noOfCellsVisitedhttp://groups.google.com/group/algogeeks?hl=en.
I think you meant spiral traversal,
#include
#include
using namespace std;
#define REP(i, n) for(int i = 0; i < n; i++)
const int MX = 1000;
int a[MX][MX], seen[MX][MX], n;
int dx[] = {0, 1, 0, -1};
int dy[] = {1, 0, -1, 0};
int main() {
scanf("%d", &n);
REP(i, n) REP(j, n) scanf("%d
If this is the case, we can follow this method:
take variables
1. dir = 1
2. st_col=0
3. end_col = n-1
4. st_row = 1
5. end_row = n-1
6. row = 0
7. col = n-1
8. noOfCellsVisited = 0
dir are as follows:
1 - left ro right
2 - top to bottom
3 - right to left
4 - bottom to top
row will keep account
I think that we have to make a spiral traversal
So o/p for your matrix would be:
1, 2, 3, 4, 8, 12, 16, 15, 14, 13, 9, 5, 6, 7, 11, 10
On Thu, Jul 28, 2011 at 1:56 PM, Rajeev Kumar wrote:
> @puneet
>
> Can you give us an example what u mean by 180 degree rotation...
> For given matrix lik
@puneet
Can you give us an example what u mean by 180 degree rotation...
For given matrix like
1, 2, 3, 4
5, 6, 7, 8
9,10,11, 12
13,14,15,16
What is the expected output?
On Wed, Jul 27, 2011 at 11:29 PM, Puneet Gautam wrote:
> Can anyone give an O(n) solution pls...??
> I t
there is no specification on complexity . if input matrix is
1 2 3
4 5 6
7 8 9
then after 180 rotation output should be
1 2 3 6 9 8 7 4 5
On Wed, Jul 27, 2011 at 11:34 PM, amit karmakar
wrote:
> If you meant "rotate a 2D matrix by angle 180" of order n x n
> Then you cannot have a O(n) algo, Eac
If you meant "rotate a 2D matrix by angle 180" of order n x n
Then you cannot have a O(n) algo, Each of the n^2 elements must be
accessed so you cannot have anything less than n^2
On Jul 27, 10:59 pm, Puneet Gautam wrote:
> Can anyone give an O(n) solution pls...??
> I think the above code is an
Can anyone give an O(n) solution pls...??
I think the above code is an O(n^2) solution..
if i am not wrong...!!!
On 7/27/11, amit wrote:
> #include
> #include
> using namespace std;
>
> const int MX = 1000;
> int n, m;
> int a[MX][MX];
>
> int main() {
> scanf("%d%d", &n, &m);
> for(i
#include
#include
using namespace std;
const int MX = 1000;
int n, m;
int a[MX][MX];
int main() {
scanf("%d%d", &n, &m);
for(int i = 0; i < n; i++)
for(int j = 0; j < m; j++)
scanf("%d", &a[i][j]);
for(int i = 0; i < n/2; i++)
for(int j = 0; j < m; j++)
11 matches
Mail list logo
