In what situation, searching a term using a binary search tree requires a time complexity that is linear to the size of the term vocabulary (say M)? How to ensure a worst time complexity of log M?
A complete binary tree is one for which every level, except possibly the last, is completely filled. The worst case search peformance is the height of the tree, which in this case would be O(lgM), assuming M vocabulay terms in the tree.
One way to ensure this performance would be to use a self-balancing tree, e.g. a red-black tree.
Since binary search is a divide-and-conquer algorithm, we can ensure O(log M) if the tree is balanced, with equal number of terms under the sub-trees of any node. O(log M) basically means that time goes up linearly while M goes up exponentially. If it takes 1 second to search a balanced binary tree that is 10 nodes, it’d take 2 seconds to search an equally balanced tree with 100 nodes, 3 seconds to search 1000 nodes, and so on.
But if the binary search tree is extremely unbalanced to the point where it looks a lot like a linked list, we would have to go through every node, requiring a time complexity that is linear to M.
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What I mean by general tree is an unbalanced tree with multiple child nodes (not restricted to 2 child node for each branch like Binary Tree). What is the Big-O complexity of remove node, insert node, find node
The average time complexity of searching in balanced BST in O(log(n)). The worst case complexity of searching in unbalanced binary tree is O(n).
If you're talking about a regular k-ary tree that does nothing special with its data, then to find any one node in the tree would take O(n) time assuming there are n nodes.
Inserting a node would be O(1) since you can store it wherever you want, and removing a node would be O(n) since you'd have to look at every node (worst case) to find the one to delete, and since there's no order to the data you don't have to do anything with the rest of the nodes.
for the time efficiency of inserting into binary search tree,
I know that the best/average case of insertion is O(log n), where as the worst case is O(N).
What I'm wondering is if there is any way to ensure that we will always have best/average case when inserting besides implementing an AVL (Balanced BST)?
Thanks!
There is no guaranteed log n complexity without balancing a binary search tree. While searching/inserting/deleting, you have to navigate through the tree in order to position yourself at the right place and perform the operation. The key question is - what is the number of steps needed to get at the right position? If BST is balanced, you can expect on average 2^(i-1) nodes at the level i. This further means, if the tree has k levels (kis called the height of tree), the expected number of nodes in the tree is 1 + 2 + 4 + .. + 2^(k-1) = 2^k - 1 = n, which gives k = log n, and that is the average number of steps needed to navigate from the root to the leaf.
Having said that, there are various implementations of balanced BST. You mentioned AVL, the other very popular is red-black tree, which is used e.g. in C++ for implementing std::map or in Java for implementing TreeMap.
The worst case, O(n), can happen when you don't balance BST and your tree degenerates into a linked list. It is clear that in order to position at the end of the list (which is a worst case), you have to iterate through the whole list, and this requires n steps.
The best case running time for binary search is O(log(n)), if the binary tree is balanced. The worst case would be, if the binary tree is so unbalanced, that it basically represents a linked list. In that case the running time of a binary search would be O(n).
However, what if the tree is only slightly unbalanced, as is teh case for this tree:
Best case would still be O(log n) if I am not mistaken. But what would be the worst case?
Typically, when we say something like "the cost of looking up an element in a balanced binary search tree is O(log n)," what we mean is "in the worst case, we have to do O(log n) work in the course of performing a search on a balanced binary search tree." And since we're talking about big-O notation here, the previous statement is meant to be taken about balanced trees in general rather than a specific concrete tree.
If you have a specific BST in mind, you can work out the maximum number of comparisons required to find any element. Just find the deepest node in the tree, then imagine searching for a value that's bigger than that value but smaller than the next value in the tree. That will cause you to walk all the way down the tree as deeply as possible, making the maximum number of comparisons possible (specifically, h + 1 of them, where h is the height of the tree).
To be able to talk about the big-O cost of performing lookups in a tree, you'd need to talk about a family of trees of different numbers of nodes. You could imagine "kinda balanced" trees whose depth is Θ(√n), for example, where lookups would take time O(√n), for example. However, it's uncommon to encounter trees like that in practice, since generally you'd either (1) have a totally imbalanced tree or (2) use some sort of balanced tree that would prevent the height from getting that high.
In a sorted array of n values, the run-time of binary search for a value, is
O(log n), in the worst case. In the best case, the element you are searching for, is in the exact middle, and it can finish up in constant-time. In the average case too, the run-time is O(log n).
http://www.geeksforgeeks.org/group-multiple-occurrence-of-array-elements-ordered-by-first-occurrence/
Please check this question.
How to do BST method of this problem.
They have mentioned that total time complexity will be O(NLogN).
How is time complexity of tree is LogN for traversal?
Please help
search, delete and insert running time all depend on the height of tree, or O(h) for BST. A degenerate tree almost looks like a linked list can produce a running time of O(N).
On the other hand, consider a self-balancing tree such as AVL tree, the running time for lookup is lower bounded by O(logN) because like Binary Search, we divide the search space by half each time as in left and right subtree have almost identical height.
What's the worst case time complexity in a log-structured merge tree for a simple search query (like querying a single WHERE clause)?
Is it O(log N)? O(N*Log N)? Something else?
How about for a multiple query, like searching for multiple WHERE clauses in a key-value database?
The wikipedia page on LSM trees is currently lacking this info.
And I'm trying to make sense of the original paper.
I have been wondering the same.
If you have a series of trees, getting smaller by a constant factor each time, and you need to search them all for a single key, the cost seems to be O(log(N)^2).
Say the first (binary) tree takes log_2(N) branches to reach a node. The second might be half the size, and take (log_2(N) - 1) branches to find a node. The smallest tree will be some O(1) constant in size and there are roughly log_2(N) trees total. Summing the series gives O(log_2(N)^2).
However, I'm wondering if there is some more clever scheme where arbitrary single-key lookups, insertions or deletions have amortized cost O(log(N)), but haven't been able to find an answer (yet).
For a simple search indexed by a LSM tree, it is O(log n). This is because the biggest tree in the LSM tree is a B tree, which is O(log n), and the other trees are subsets of B trees or in the case of in memory trees, more efficient trees, which are no worse than O(log n). The number of trees is a constant, so it doesn't affect the order of the search time.