In work case (ie max value of i,j and k ll be)
for(i=0;i<n;i++) //i=n
{
for(j=0;j<n*n;j++) // j=i=n;
{
for(k=0;k<n*n;k++) // k=j=i=n;
}
}
So Ans O(n^5)
-- Prashant Kulkarni
On Sat, Sep 10, 2011 at 12:05 PM, deepikaanand <[email protected]>wrote:
> ans : O(n^5)
>
> inner most loop rums for j times
> then middle loop sums 'j ' from j=0 to j= i^2 so ans is proportional
> to i^4
> and outer loop again sums it for n times so ans : O(n^5)
>
> --
> You received this message because you are subscribed to the Google Groups
> "Algorithm Geeks" group.
> To post to this group, send email to [email protected].
> To unsubscribe from this group, send email to
> [email protected].
> For more options, visit this group at
> http://groups.google.com/group/algogeeks?hl=en.
>
>
--
You received this message because you are subscribed to the Google Groups
"Algorithm Geeks" group.
To post to this group, send email to [email protected].
To unsubscribe from this group, send email to
[email protected].
For more options, visit this group at
http://groups.google.com/group/algogeeks?hl=en.
