Can anyone explain the difference between binary search and binary search tree with example?Are they the same?Reading the internet it seems the second is only for trees and binary search does not follow this rule.And how to check the presence of a number in O(log(n)) time?
Binary search is an algorithm used on straightforward sorted arrays, which runs in O(log n).
Updating a sorted array is O(n).
A binary tree is a data structure that has both O(log n) search and update (Ignoring problems of balancing).
An interesting comparison is at the end of this chapter.
Related
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).
We know that an in order traversal of a binary search tree will list the elements of the tree in sorted order in O(n) time.
but how fast can a comparison based binary search tree construction algorithm take?
The longest run time would be O (log n) in the worst case I think.
I have an input stream of integers coming in an ascending order, my task is to create a balance binary search tree out of that stream, on the fly. I have gone through the link:BBST from a stream of integers and understood that we can make use of Red-Black trees. The thing is, I am looking for more optimal solutions that use 'sorted information' from the input data.
If you use a red-black tree but always start your insertion at the last node inserted, rather than the root, and use a bottom up insertion algorithm, insertion is O(1) amortized. This means constructing the tree will cost O(n), not Ω(n log n).
If the elements are coming in sorted order, then probably the simplest and most efficient thing to do is just to push each one to the end of a dynamic array (an array that doubles its size whenever it becomes full, for example).
Pushing into the array is amortized O(1).
Searches in a sorted array are O(log(n)) using binary search. Moreover, it's logarithmic time with very low constants.
Despite its simplicity, a sorted array is a type of a blanced binary search tree.
The problem is that the amount of data is unknown. Therefore, this tree must be self balancing, regardless of whether the input has a pattern or not (ascending order).
If O(log n) insertion times are not acceptable in this case (say for red-black trees) then I'm at a loss as to why you need to store this in balanced binary tree form in the first place. A binary search on a dynamic array is just as fast as a tree, with amortized O(1) insertion time.
If the size of the stream is known before hand, then of course building the tree would be much easier.
What is the time-complexity for insertion,searching an element in almost complete and complete binary tree and skewed binary tree??
Assuming you mean a sorted binary tree then both insertion and search are O(log n) worst case.
We always see operations on a (binary search) tree has O(logn) worst case running time because of the tree height is logn. I wonder if we are told that an algorithm has running time as a function of logn, e.g m + nlogn, can we conclude it must involve an (augmented) tree?
EDIT:
Thanks to your comments, I now realize divide-conquer and binary tree are so similar visually/conceptually. I had never made a connection between the two. But I think of a case where O(logn) is not a divide-conquer algo which involves a tree which has no property of a BST/AVL/red-black tree.
That's the disjoint set data structure with Find/Union operations, whose running time is O(N + MlogN), with N being the # of elements and M the number of Find operations.
Please let me know if I'm missing sth, but I cannot see how divide-conquer comes into play here. I just see in this (disjoint set) case that it has a tree with no BST property and a running time being a function of logN. So my question is about why/why not I can make a generalization from this case.
What you have is exactly backwards. O(lg N) generally means some sort of divide and conquer algorithm, and one common way of implementing divide and conquer is a binary tree. While binary trees are a substantial subset of all divide-and-conquer algorithms, the are a subset anyway.
In some cases, you can transform other divide and conquer algorithms fairly directly into binary trees (e.g. comments on another answer have already made an attempt at claiming a binary search is similar). Just for another obvious example, however, a multiway tree (e.g. a B-tree, B+ tree or B* tree), while clearly a tree is just as clearly not a binary tree.
Again, if you want to badly enough, you can stretch the point that a multiway tree can be represented as sort of a warped version of a binary tree. If you want to, you can probably stretch all the exceptions to the point of saying that all of them are (at least something like) binary trees. At least to me, however, all that does is make "binary tree" synonymous with "divide and conquer". In other words, all you accomplish is warping the vocabulary and essentially obliterating a term that's both distinct and useful.
No, you can also binary search a sorted array (for instance). But don't take my word for it http://en.wikipedia.org/wiki/Binary_search_algorithm
As a counter example:
given array 'a' with length 'n'
y = 0
for x = 0 to log(length(a))
y = y + 1
return y
The run time is O(log(n)), but no tree here!
Answer is no. Binary search of a sorted array is O(log(n)).
Algorithms taking logarithmic time are commonly found in operations on binary trees.
Examples of O(logn):
Finding an item in a sorted array with a binary search or a balanced search tree.
Look up a value in a sorted input array by bisection.
As O(log(n)) is only an upper bound also all O(1) algorithms like function (a, b) return a+b; satisfy the condition.
But I have to agree all Theta(log(n)) algorithms kinda look like tree algorithms or at least can be abstracted to a tree.
Short Answer:
Just because an algorithm has log(n) as part of its analysis does not mean that a tree is involved. For example, the following is a very simple algorithm that is O(log(n)
for(int i = 1; i < n; i = i * 2)
print "hello";
As you can see, no tree was involved. John, also provides a good example on how binary search can be done on a sorted array. These both take O(log(n)) time, and there are of other code examples that could be created or referenced. So don't make assumptions based on the asymptotic time complexity, look at the code to know for sure.
More On Trees:
Just because an algorithm involves "trees" doesn't imply O(logn) either. You need to know the tree type and how the operation affects the tree.
Some Examples:
Example 1)
Inserting or searching the following unbalanced tree would be O(n).
Example 2)
Inserting or search the following balanced trees would both by O(log(n)).
Balanced Binary Tree:
Balanced Tree of Degree 3:
Additional Comments
If the trees you are using don't have a way to "balance" than there is a good chance that your operations will be O(n) time not O(logn). If you use trees that are self balancing, then inserts normally take more time, as the balancing of the trees normally occur during the insert phase.