I am considering.. Maxheapify... A[parent(i)]>=A[i] kth largest element... therefore O(klogn)... k times u have to extract the largest element and logn to maintain the maxheapify everytime.....
minheapify....A[parent(i)]<=A[i] kth largest element.... that means ... (n-k) smallest element..... therefoe... O((n-k)logn)... With regards, Praveen Raj DCE-IT 3rd yr -- 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 unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
