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
<amit.codenam...@gmail.com>wrote:
> 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 <puneet.nsi...@gmail.com> wrote:
> > 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 <amit.codenam...@gmail.com> wrote:
> >
> >
> >
> > > #include <cstdio>
> > > #include <algorithm>
> > > 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++)
> > > swap(a[i][j], a[n-i-1][m-j-1]);
> > > if(n&1)
> > > for(int j = 0; j < m/2; j++)
> > > swap(a[n/2][j], a[n/2][m-j-1]);
> >
> > > for(int i = 0; i < n; i++) {
> > > for(int j = 0; j < m; j++)
> > > printf("%d ", a[i][j]);
> > > printf("\n");
> > > }
> > > }
> >
> > > On Jul 27, 7:54 pm, Anika Jain <anika.jai...@gmail.com> wrote:
> > >> is it lyk for {1,2,3
> > >> 4,5,6,
> > >> 7,8,9}
> > >> to be {3,2,1,
> > >> 6,5,4,
> > >> 9,8,7} ??
> >
> > >> On Wed, Jul 27, 2011 at 9:37 AM, Deoki Nandan <deok...@gmail.com>
> wrote:
> > >> > rotate a 2D matrix by angle 180
> >
> > >> > --
> > >> > **With Regards
> > >> > Deoki Nandan Vishwakarma
> >
> > >> > *
> > >> > *
> >
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--
**With Regards
Deoki Nandan Vishwakarma
*
*
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