@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.
