Hi,
Start from right top corner
If matrix element key move to previous column
else if matrix element key move to next row
int search(int** a,int key,int m,int n)
{
if (key a[0][0] || key a[m-1][n-1]) return 0;
int min=a[0][n-1],i=0,j=n-1;
while(i=m-1 j=0)
{
why not start from middle(m/2, n/2)
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
On Wed, Apr 4, 2012 at 12:29 AM, Karthikeyan V.B kartmu...@gmail.comwrote:
Hi,
Start from right top corner
If matrix element key move to previous column
else if
we can also apply binary search on rows and columns separately by which we
will decide that in which 1/4th part we need to search.So in this case
complexity will be O(log m + log n).
just check and let me know...
On Wed, Apr 4, 2012 at 5:27 AM, Ashish Goel ashg...@gmail.com wrote:
why not
This search can be done easily in O(n+m) start from top right corner. chek
out this link you will understand!!
http://www.geeksforgeeks.org/archives/11337
On Sunday, 1 April 2012 21:18:26 UTC+5:30, ashgoel wrote:
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