@Lucifer : if for the same question , we consider a Max heap instead of Min heap then complexity of the same algo will be O(N) ... right???
On Fri, Dec 16, 2011 at 8:50 PM, Lucifer <sourabhd2...@gmail.com> wrote: > @my previous post.. > > While explaining the run-time I have made an editing mistake... > instead of 'N' nodes it should be 'K' nodes.. > i.e. > Hence, for any given bin- tree having 'K' nodes, the number of null > nodes is 'K+1'. > These null nodes are nothing but the nodes where the check nodeValue < > X failed while traversing the original tree. > Hence, the total number of checks will be 2K+1 = O(2K) > > On Dec 16, 1:04 am, sunny agrawal <sunny816.i...@gmail.com> wrote: > > oops... > > wanted to write the same but yeah its meaning turns out to be totally > > different :( > > anyways very well explained by Lucifier > > > > @shashank > > i think now u will be able to get why there will be only 2k comparisons > in > > the worst case > > > > On Thu, Dec 15, 2011 at 10:51 PM, atul anand <atul.87fri...@gmail.com > >wrote: > > > > > > > > > > > > > > > > > > > > > @Lucifer : yes even i found flaw in the above algo when i gave it a > second > > > thought but didnt get time to post it. > > > bcoz min heap has property that the parent node is less than its both > > > child(subtree to be more precise ) but it does not confirm that left > child > > > is always smaller than right child of the node. > > > > > On Thu, Dec 15, 2011 at 10:31 PM, Lucifer <sourabhd2...@gmail.com> > wrote: > > > > >> @All > > > > >> I don't think the algo given above is entirely correct.. Or may be i > > >> didn't it properly... > > > > >> So basically say a preorder traversal of the heap is done based on > > >> whether the current root value < X. As the algo says that at any point > > >> if k>0 and we hit a node value which is >=X , then we are done. If i > > >> understood it properly then thats not correct. > > > > >> The reason being that say on the left subtree we end up at a node > > >> whose value is >=x and say k > 0. Now until and unless we don't parse > > >> the right subtree (or basically the right half which was neglected as > > >> part of pre-order traversal or say was to be considered later) we are > > >> not sure if the current node is actually withing the first smallest K > > >> nos. It may happen that previously neglected (or rather later to be > > >> processed) half has the kth smallest element which is actually < X. > > >> The reason being that a heap is not a binary search tree where there > > >> is a strict relation between the left and the right half so that we > > >> can say that if say a condition P is true in the left half then it > > >> will be false in the right half and vice versa. > > > > >> To solve the problem we need to do a pre-order traversal keeping the > > >> following conditions in mind: (pass K and root node) > > > > >> 1) If current node is >= X then skip the processing of the tree rooted > > >> at the current node. > > > > >> 2) If current node is < X , then decrement K by 1 and process its > > >> childs ( i.e take step 1 for rach child). > > > > >> The result will depend on: > > > > >> a) If at any stage recursion ends and the value of K>0 then the kth > > >> smallest element is >= X. > > >> b) If during tree traversal the value of K reaches 0, that means > > >> there are atleast k elements which are < X. Hence, at this point > > >> terminate the recursion ( as in no need to continue). This result > > >> signifies that the kth smallest element < X. > > > > >> Therefore to generalize... > > > > >> Perform a preorder traversal for root node < X, and keep decrementing > > >> the count K by 1. > > >> If K reaches 0 during traversal then end the recursion. > > > > >> After the call to the recursive traversal is over, check for the value > > >> of K. If greater than 0, then the kth smallest element >= X otherwise > > >> its not. > > > > >> The time complexity will always be 2K ( in the worst case basically > > >> when K value reaches 0 ). If u analyze it closely we are making 2 > > >> checks when at particular node for its children. Hence, whether both > > >> the child nodes have value < X or one of them or node, at the end we > > >> always end up making 2 checks for the children (left and right child). > > >> So given any tree one can think of a null node as a leaf node > > >> ( depicting that the node has a value >=X) . Hence, for any given bin- > > >> tree having nodes 'N', the number null nodes is 'N+1'. Hence, the > > >> total number of checks will be 2N+1 = O(2N) , > > > > >> On Dec 15, 1:00 pm, WgpShashank <shashank7andr...@gmail.com> wrote: > > >> > @sunny why we look at all k number which are greater then x , > correct ? > > >> > Lets think in this way > > > > >> > we wants to check if kth smallest element in heap thats >=x isn't > it ? > > >> so > > >> > if root of mean heap is greater then x then none other elements will > > >> less > > >> > then x so we terminate . > > >> > else our algorithm will search children of all the nodes which are > less > > >> > then x till either we have found k of them or we are exhausted e.g. > > >> when > > >> > k=0 . so we will cal our function to both left & right children ? > > > > >> > so now think we are looking for children's of only nodes which are > less > > >> > then x and at most k of these in tottal . each have atmost two > visited > > >> > childrens so we have visted at-most 3K nodes isn;t it ? for total > time > > >> O(K) > > >> > ? > > > > >> > let me know where i am wrong ? i am not getting for uy k nodes > greater > > >> then > > >> > x ? why we will do that & then how much comparisons u needs for > that ? > > > > >> > Thanks > > >> > Shashank Mani > > >> > CSE, BIT Mesrahttp://shashank7s.blogspot.com/ > > > > >> -- > > >> 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. > > > > > -- > > > 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. > > > > -- > > Sunny Aggrawal > > B.Tech. V year,CSI > > Indian Institute Of Technology,Roorkee > > -- > 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. > > -- 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.
