We need a sorted array to perform a binary search. In that case, the time complexity is already greater than the linear search, so isn't linear search a better option in that case?
A linear search runs in O(N) time, because it scans through the array from start to end.
On the other hand, a binary search first sorts the array in O(NlogN) time (if it is not already sorted), then performs lookups in O(logN) time.
For a small number of lookups, using a linear search would be faster than using binary search. However, whenever the number of lookups is greater than logN, binary search will theoretically have the upper hand in performance.
So, the answer to your question is: Linear search and binary search perform lookups in different ways. Linear search scans through the whole array, while binary search sorts the array first. These two search techniques have differing time complexities, but that does not mean that one will always be better than the other.
Specifically, linear search works well when the size of the list is small and/or you only need to perform a small number of lookups. Binary search should perform better in all other situations.
It'll be better if your container is sorted already or if you want to search for many values.
First of all for Binary Search the precondition is that the array is sorted, which means you do not need to resort it. Secondly if you are talking about integer arrays, you can use RadixSort O(d*n) or CountingSort O(n+l) which are similar to linear search in terms of complexity....
Binary search is faster than linear when the given array is already sorted.
For a sorted array, binary search offers an average O(log n) meanwhile linear offers O(n).
For any given array that is not sorted, linear search becomes best since O(n) is better than sorting the array ( using quicksort for example O(n log n) ) and then applying binary search after that, thus given O(n log n + log n) complexity.
Related
Why do we even say Binary Search (BS) is better than Linear Search (LS)? As when i give an unsorted array to BS and LS. We need to sort for BS. The Time complexity of all inbuild sorting algorithms are atleast O(nlogn). So Overall Time complexity to sort and find is O(nlogn+logn)=O(nlogn). As BS take O(logn). Whereas LS takes only O(n) to find a element.
So according to this statement LS is better than BS to search a element in List.
Sorting and using binary search makes sense when you have several searches to perform on the same array.
Let the cost of a single linear search be a.N, the cost of sorting b.N.lg(N) and the cost of a binary search c.lg(N).
Now you compare M.a.N to b.N.lg(N) + M.c.lg(N) for M searches. The breakeven point is at
M = b.N.lg(N) / (a.N - c.lg(N)) ≈ b.lg(N)/a
which is a small multiple of lg(N).
If given an unsorted array which among the following two scenarios would have lesser time complexity or would perform better
Binary Search - by sorting the array first and then using binary search algorithm
Sequential search - on unsorted array
So if given an unsorted array to search an element should we go for sorting it and then apply binary search or directly apply sequential search algorithm on unsorted array.
Both exist because both have their places where they are useful.
If you will search only once, sequential search is fastest. But if you do many queries, then binary search is faster. Given that sorting is O(n log(n)), binary search becomes the same number of operations if you have to do O(log(n)) searches.
BUT, operations are not created equal. In particular binary search requires yes/no questions that are hard for branch prediction. As a result, if you're searching a list of under 100 integers, a binary search is likely to be slower than a sequential search because the binary search has multiple pipeline stalls (each mispredicted binary choice) while sequential search only has one (when you find the element you're looking for).
So if you're doing many lookups, and you either have a lot of data or complex data (eg strings), binary search is better.
Given an input stream of numbers ranging from 1 to 10^5 (non-repeating) we need to be able to tell at each point how many numbers smaller than this have been previously encountered.
I tried to use the set in C++ to maintain the elements already encountered and then taking upper_bound on the set for the current number. But upper_bound gives me the iterator of the element and then again I have to iterate through the set or use std::distance which is again linear in time.
Can I maintain some other data structure or follow some other algorithm in order to achieve this task more efficiently?
EDIT : Found an older question related to fenwick trees that is helpful here. Btw I have solved this problem now using segment trees taking hints from #doynax comment.
How to use Binary Indexed tree to count the number of elements that is smaller than the value at index?
Regardless of the container you are using, it is very good idea to enter them as sorted set so at any point we can just get the element index or iterator to know how many elements are before it.
You need to implement your own binary search tree algorithm. Each node should store two counters with total number of its child nodes.
Insertion to binary tree takes O(log n). During the insertion counters of all parents of that new element should be incremented O(log n).
Number of elements that are smaller than the new element can be derived from stored counters O(log n).
So, total running time O(n log n).
Keep your table sorted at each step. Use binary search. At each point, when you are searching for the number that was just given to you by the input stream, binary search is going to find either the next greatest number, or the next smallest one. Using the comparison, you can find the current input's index, and its index will be the numbers that are less than the current one. This algorithm takes O(n^2) time.
What if you used insertion sort to store each number into a linked list? Then you can count the number of elements less than the new one when finding where to put it in the list.
It depends on whether you want to use std or not. In certain situations, some parts of std are inefficient. (For example, std::vector can be considered inefficient in some cases due to the amount of dynamic allocation that occurs.) It's a case-by-case type of thing.
One possible solution here might be to use a skip list (relative of linked lists), as it is easier and more efficient to insert an element into a skip list than into an array.
You have to use the skip list approach, so you can use a binary search to insert each new element. (One cannot use binary search on a normal linked list.) If you're tracking the length with an accumulator, returning the number of larger elements would be as simple as length-index.
One more possible bonus to using this approach is that std::set.insert() is log(n) efficient already without a hint, so efficiency is already in question.
I'm doing a project in which i need to insert data into vectors sort it and search it ...
i need fastest possible algorithms for sort and search ... i've been searching and found out that std::sort is basically quicksort which is one of the fastest sorts but i cant figure out which search algorithm is the best ? binarysearch?? can u help me with it? tnx ... So i've got 3 methods:
void addToVector(Obj o)
{
fvector.push_back(o);
}
void sortVector()
{
sort(fvector.begin(), fvector().end());
}
Obj* search(string& bla)
{
//i would write binary search here
return binarysearch(..);
}
I've been searching and found out that std::sort is basically
quicksort.
Answer: Not quite. Most implementations use a hybrid algorithm like
introsort, which combines quick-sort, heap-sort and insertion sort.
Quick-sort is one of the fastest sorting methods.
Answer: Not quite. In general it holds (i.e., in the average case quick-sort is of complexity). However, quick-sort has quadratic worst-case performance (i.e., ). Furthermore, for a small number of inputs (e.g., if you have a std::vector with a small numbers of elements) sorting with quick-sort tends to achieve worst performance than other sorting algorithms that are considered "slower" (see chart below):
I can't figure out which searching algorithm is the best. Is it binary-search?
Answer: Binary search has the same average and worst case performance (i.e., ). Also have in mind that binary-search requires that the container should be arranged in ascending or descending order. However, whether is better than other searching methods (e.g., linear search which has time complexity) depends on a number of factors. Some of them are:
The number of elements/objects (see chart below).
The type of elements/objects.
Bottom Line:
Usually looking for the "fastest" algorithm denotes premature optimization and according to one of the "great ones" (Premature optimization is the root of all evil - Donald Knuth). The "fastest", as I hope it has been clearly shown, depends on quite a number of factors.
Use std::sort to sort your std::vector.
After sorting your std::vector use std::binary_search to find out whether a certain element exists in your std::vector or use std::lower_bound or std::upper_bound to find and get an element from your std::vector.
For amortised O(1) access times, use a [std::unordered_map], maybe using a custom hash for best effects.
Sorting seems to be unneccessary extra work.
Searching and Sorting efficiency is highly dependent on the type of data, the ordering of the raw data, and the quantity of the data.
For example, for small sorted data sets, a linear search may be faster than a binary search; or the time differences between the two is negligible.
Some sort algorithms will perform horribly on inversely ordered data, such a binary tree sort. Data that does not have much variation may cause a high degree of collisions on hash algorithms.
Perhaps you need to answer the bigger question: Is search or sorting the execution bottleneck in my program? Profile and find out.
If you need the fastest or the best sorting algorithm... There is no such one. At least it haven't been found yet. There are algorithms that provide better results for different data, there are algorithms that provide good results for most of data. You either need to analyze your data and find the best one for your case or use generic algo like std::sort and expect it to provide good results but not the best.
if your elements are of integer you should use bucket sort algorithm which run at O(N) time instead of O(nlogn) average case as with qsort
[http://en.wikipedia.org/wiki/Bucket_sort]
Sorting
In case you want to know about the fastest sorting technique for integer values in a vector then I would suggest you to refer the following link:
https://github.com/fenilgmehta/Fastest-Integer-Sort
It uses radix sort and counting sort for large arrays and merge sort along with insertion sort for small arrays.
According to statistics, this sorting algorithm is way faster than C++ std::sort for integral values.
It is 6 times faster than C++ STL std::sort for "int64_t array[10000000]"
Searching
If you want to know whether a particular value is present in the vector or not, then you should use binary_search(...)
If you want to know the exact location of an element, then use lower_bound(...) and upper_bound(...)
I want to check whether an element is present in the given array(2D) and to find the count towards the left of a cell and to the right of a cell and also to the top and bottom.How can i do it without using brute force
If the array is sorted then you can find in O(nlogm) time complexity using binary search!
where n,m is the rows and columns.
In addition to the other answers (probably not very useful for any present applications, just an idea for thought) is Grover's algorithm. A exert from Wikipedia:
Grover's algorithm is a quantum algorithm for searching an unsorted
database with N entries in O(N1/2) time and using O(log N) storage
space (see big O notation). Lov Grover formulated it in 1996.
In models of classical computation, searching an unsorted database
cannot be done in less than linear time (so merely searching through
every item is optimal). Grover's algorithm illustrates that in the
quantum model searching can be done faster than this; in fact its time
complexity O(N1/2) is asymptotically the fastest possible for
searching an unsorted database in the linear quantum model.
If your matrix is unsorted, and you don't have anything like hash-tables for quick access, there is no way.
If your matrix is, for example, sorted, you can use more effective search algorithms (such as binary search, for example) to find the element faster. Don't forget that a 2D array can be represented with a vector, and a variable to hold the column count.