If an additional storage is used to store the frequency / marker searching the frequency/marker in the array would require an additional nested loop. Would it still be O(n)?
On Fri, Aug 5, 2011 at 8:36 PM, kartik sachan <[email protected]>wrote: > I think in this case count sort type thing would work better way > > just take a array of max 26(as 26 CHARACTER ARE THERE) > > and using index as count['M']=store how many times M has comes > > similary store other character > > now print array whose value >0 > > but IN this case u might lost the ORDER......................(but that can > be done using binary search using T(log(n)) > > > the total time complexcity is O(n) > > -- > 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.
