Problem link : contest practice Difficulty : Medium Prerequisites : Lowest Common Ancestor, Trees Problem : Given a tree, you have to answer the question of the format "r u v" which means which is the LCA of u and v if the root of the tree is at r. ExplanationAt first, let's consider some partial solutions. How to get 20 pointsFor this subtask you an find the LCA in any way you want as long as the complexity is not slower than O(N). For example, by DFS from the root, you can number the vertices so that given two arbitrary vertices, you can check whether they are ancestor and descendant(for this, you can store T_in and T_out for each node. T_in is the time when the DFS for that node was begun. T_out is the time when the DFS was over). How to get 60 pointsHere there are not more than 10 different roots, but the queries are quite high, so you should know the fast way to find LCA. More specifically O(Nlog(N)) is enough. Notice that there will be no more than 10 different roots so your complexity will be O(10 × Nlog(N)). How to get 100 pointsThere are two interesting observations that you can make:
With this two observations you need to implement two function: finding LCA and distance of the two vertices in the tree. Proof for these two observation is not hard but too long to be mentioned here. It is left as an exercise for you.
This question is marked "community wiki".
asked 27 Jul '14, 14:46

This might be a much simpler way for unrooted queries: link answered 27 Oct '14, 15:49
@sudeepdino008 did you find the proof why it works?
(30 Nov '17, 23:53)

For Query(root,u,v): Let a=LCA(u,v),b=LCA(root,u) and c=LCA(root,v) and the answer for the query is the one value that is different from other two if all of them are not equal i.e
answered 24 Jul '17, 19:46

Infact, the only possible answers are LCA(r, u), LCA(r, v), LCA(u, v). I proved it by drawing the diagrams corresponding to all possible scenarios for the arrangements of the 3 nodes and the node no. 1. answered 06 Sep '18, 12:58

Can anyone tell me why am i getting TLE for the two test cases? My sol link: here Is there any special case that i need to handle. Any suggestion would be appreciated. Thanks answered 26 Sep '15, 23:28

How do we keep the parentchild relationship for different roots in the 20 points solution? answered 24 Jun '16, 11:44

IS THIS A Binary TREE?OR it can be a tree with any number of children? answered 24 Jul '17, 22:06

What is wrong with my solution: What I am doing is : If r is not in subtree of orig_lca then origlca is the answer else { if(lca(u,r)==origlca and lca(v,r)==origlca){ then answer = origlca } if(lca(u,r)==origlca){ then answer = lca(v,r); } if(none of above){ then answer = lca(u,r); } }
link
This answer is marked "community wiki".
answered 16 Aug '18, 15:53

How this works? Can anyone tell the proving of this: ............................................................................. Let LCA(u, v, w) be the LCA of v and w with respect to root u. To compute LCA(u, v, w), we can compute, for any fixed r,
and take the "odd man out", i.e., if two are equal and the third is different, then take the third, else they're all equal, so take that node. answered 25 Oct '18, 23:27
