@atul.. Complexity would be 2(n-k+1) = O(2*(n-k)) .... Basically the condition based on which the traversal is performed will change. I have modified some part of the original post to show that:
Instead of having the initial count as K have it as N-K+1 .. when taking a max-heap.. 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 the initial count i.e (n-k +1) 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 initial count >0 then the kth smallest element is < X. b) If during tree traversal the value of initial count reaches 0, that means there are atleast (n-k+1) 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. On Dec 17, 1:28 pm, atul anand <atul.87fri...@gmail.com> wrote: > @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.
