@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.
