If Big-Omega is the lower bound then what does it mean to have a worst case time complexity of Big-Omega(n).
From the book "data structures and algorithms with python" by Michael T. Goodrich:
consider a dynamic array that doubles it size when the element reaches its capacity.
this is from the book:
"we fully explored the append method. In the worst case, it requires
Ω(n) time because the underlying array is resized, but it uses O(1)time in the amortized sense"
The parameterized version, pop(k), removes the element that is at index k < n
of a list, shifting all subsequent elements leftward to fill the gap that results from
the removal. The efficiency of this operation is O(n−k), as the amount of shifting
depends upon the choice of index k. Note well that this
implies that pop(0) is the most expensive call, using Ω(n) time.
how is "Ω(n)" describes the most expensive time?
The number inside the parenthesis is the number of operations you must do to actually carry out the operation, always expressed as a function of the number of items you are dealing with. You never worry about just how hard those operations are, only the total number of them.
If the array is full and has to be resized you need to copy all the elements into the new array. One operation per item in the array, thus an O(n) runtime. However, most of the time you just do one operation for an O(1) runtime.
Common values are:
O(1): One operation only, such as adding it to the list when the list isn't full.
O(log n): This typically occurs when you have a binary search or the like to find your target. Note that the base of the log isn't specified as the difference is just a constant and you always ignore constants.
O(n): One operation per item in your dataset. For example, unsorted search.
O(n log n): Commonly seen in good sort routines where you have to process every item but can divide and conquer as you go.
O(n^2): Usually encountered when you must consider every interaction of two items in your dataset and have no way to organize it. For example a routine I wrote long ago to find near-duplicate pictures. (Exact duplicates would be handled by making a dictionary of hashes and testing whether the hash existed and thus be O(n)--the two passes is a constant and discarded, you wouldn't say O(2n).)
O(n^3): By the time you're getting this high you consider it very carefully. Now you're looking at three-way interactions of items in your dataset.
Higher orders can exist but you need to consider carefully what's it's going to do. I have shipped production code that was O(n^8) but with very heavy pruning of paths and even then it took 12 hours to run. Had the nature of the data not been conductive to such pruning I wouldn't have written it at all--the code would still be running.
You will occasionally encounter even nastier stuff which needs careful consideration of whether it's going to be tolerable or not. For large datasets they're impossible:
O(2^n): Real world example: Attempting to prune paths so as to retain a minimum spanning tree--I computed all possible trees and kept the cheapest. Several experiments showed n never going above 10, I thought I was ok--until a different seed produced n = 22. I rewrote the routine for not-always-perfect answer that was O(n^2) instead.
O(n!): I don't know any examples. It blows up horribly fast.
Problem Statement:- Given an array of integers and an integer k, print all the pairs in the array whose sum is k
Method 1:-
Sort the array and maintain two pointers low and high, start iterating...
Time Complexity - O(nlogn)
Space Complexity - O(1)
Method 2:-
Keep all the elements in the dictionary and do the process
Time Complexity - O(n)
Space Complexity - O(n)
Now, out of above 2 approaches, which one is the most efficient and on what basis I am going to compare the efficiency, time (or) space in this case as both are different in both the approaches
I've left my comment above for reference.
It was hasty. You do allow O(nlogn) time for the Method 1 sort (I now think I understand?) and that's fair (apologies;-).
What happens next? If the input array must be used again, then you need a sorted copy (the sort would not be in-place) which adds an O(n) space requirement.
The "iterating" part of Method 1 also costs ~O(n) time.
But loading up the dictionary in Method 2 is also ~O(n) time (presumably a throw-away data structure?) and dictionary access - although ~O(1) - is slower (than array indexing).
Bottom line: O-notation is helpful if it can identify an "overpowering cost" (rendering others negligible by comparison), but without a hint at use-cases (typical and boundary, details like data quantities and available system resources etc), questions like this (seeking a "generalised ideal" answer) can't benefit from it.
Often some simple proof-of-concept code and performance tests on representative data can make "the right choice obvious" (more easily and often more correctly than speculative theorising).
Finally, in the absence of a clear performance winner, there is always "code readability" to help decide;-)
Do any of the sort algorithms sort a given array of integers such that in case a[i]=a[j], it is considered that a[i]>a[j] if i>j?
You are probably looking for stable sorting algorithms. Stable sorting algorithms maintain the relative order of records with equal keys (i.e. values).
Concept explanation -
https://en.wikipedia.org/wiki/Sorting_algorithm#Stability
Algorithms which are stable -
https://en.wikipedia.org/wiki/Sorting_algorithm#Comparison_of_algorithms
Sorting algorithms defines the general steps to sort the input. Such edge cases are handled by actual implementation depending on the requirement of the user.
I have coded a few sorting methods in C and I would like to find the input size at which the program is optimal (i.e.) profiling each algorithm. But how do I do this? I know to time each method, but I don't know how I can find the size at which it is 'optimal'.
It depends on some factors:
Data behaviour: is your data already partially sorted? or it is very random?
Data size: for a big input (say 1 thousand or more) you can assure that O(N^2) sorting methods will lose to O(N*log(N)) methods..
Data structure of the data: is it array or list or ?. Sorting method with non sequential access to data will be slower for something like list
So the answer is by empirically running your program with some real data you will likely handle combined by varying in the input size.
When a slower method (like O(N^2)) gets beaten by some faster method (like O(N*log(N))) when input size is > X then you can say that the slower method is 'empirically optimal' for input size <= X (the value depends on the characteristics of the input data).
Sort algorithms do not have a single number at which they are optimal.
For pure execution time, almost every sort algorithm will be fastest on a set of 2 numbers, but that it not useful in most cases.
Some sort algorithms may work more efficiently on smaller data sets, but that does not mean they are 'optimal' at that size.
Some sorts may also work better on other characteristics of the data. There are sorts that can be extremely efficient if the data is almost sorted already, but may be very slow if it is not. Others will run the same on any set of a given size.
It is more useful to look at the Big O of the sort (such as O(n^2), O(n log n) etc) and any special properties the sort has, such as operating on nearly sorted data.
To find the input size at which the program is optimal (by which I assume you mean the fastest, or for which the sorting algorithm requires the fewest comparisons) you will have to test it against various inputs and graph the independent axis (input size) against the dependent axis (runtime) and find the minimum.
I have some data, up to a between a million and a billion records, each which is represented by a bitfield, about 64 bits per key. The bits are independent, you can imagine them basically as random bits.
If I have a test key and I want to find all values in my data with the same key, a hash table will spit those out very easily, in O(1).
What algorithm/data structure would efficiently find all records most similar to the query key? Here similar means that most bits are identical, but a minimal number are allowed to be wrong. This is traditionally measured by Hamming distance., which just counts the number of mismatched bits.
There's two ways this query might be made, one might be by specifying a mismatch rate like "give me a list of all existing keys which have less than 6 bits that differ from my query" or by simply best matches, like "give me a list of the 10,000 keys which have the lowest number of differing bits from my query."
You might be temped to run to k-nearest-neighbor algorithms, but here we're talking about independent bits, so it doesn't seem likely that structures like quadtrees are useful.
The problem can be solved by simple brute force testing a hash table for low numbers of differing bits. If we want to find all keys that differ by one bit from our query, for example, we can enumerate all 64 possible keys and test them all. But this explodes quickly, if we wanted to allow two bits of difference, then we'd have to probe 64*63=4032 times. It gets exponentially worse for higher numbers of bits.
So is there another data structure or strategy that makes this kind of query more efficient?
The database/structure can be preprocessed as much as you like, it's the query speed that should be optimized.
What you want is a BK-Tree. It's a tree that's ideally suited to indexing metric spaces (your problem is one), and supports both nearest-neighbour and distance queries. I wrote an article about it a while ago.
BK-Trees are generally described with reference to text and using levenshtein distance to build the tree, but it's straightforward to write one in terms of binary strings and hamming distance.
This sounds like a good fit for an S-Tree, which is like a hierarchical inverted file. Good resources on this topic include the following papers:
Hierarchical Bitmap Index: An Efficient and Scalable Indexing Technique for Set-Valued Attributes.
Improved Methods for Signature-Tree Construction (2000)
Quote from the first one:
The hierarchical bitmap index efficiently supports dif-
ferent classes of queries, including subset, superset and similarity queries.
Our experiments show that the hierarchical bitmap index outperforms
other set indexing techniques significantly.
These papers include references to other research that you might find useful, such as M-Trees.
Create a binary tree (specifically a trie) representing each key in your start set in the following way: The root node is the empty word, moving down the tree to the left appends a 0 and moving down the right appends a 1. The tree will only have as many leaves as your start set has elements, so the size should stay manageable.
Now you can do a recursive traversal of this tree, allowing at most n "deviations" from the query key in each recursive line of execution, until you have found all of the nodes in the start set which are within that number of deviations.
I'd go with an inverted index, like a search engine. You've basically got a fixed vocabulary of 64 words. Then similarity is measured by hamming distance, instead of cosine similarity like a search engine would want to use. Constructing the index will be slow, but you ought to be able to query it with normal search enginey speeds.
The book Introduction to Information Retrieval covers the efficient construction, storage, compression and querying of inverted indexes.
"Near-optimal hashing algorithms for approximate nearest neighbor in high dimensions", from 2008, seems to be the best result as of then. I won't try to summarize since I read it over a year ago and it's hairy. That's from a page on locality-sensitive hashing, along with an implementation of an earlier version of the scheme. For more general pointers, read up on nearest neighbor search.
This kind of question has been asked before: Fastest way to find most similar string to an input?
The database/structure can be
preprocessed as much as you like
Well...IF that is true. Then all you need is a similarity matrix of your hamming distances. Make the matrix sparse by pruning out large distances. It doesn't get any faster and not that much of a memory hog.
Well, you could insert all of the neighbor keys along with the original key. That would mean that you store (64 choose k) times as much data, for k differing bits, and it will require that you decide k beforehand. Though you could always extend k by brute force querying neighbors, and this will automatically query the neighbors of your neighbors that you inserted. This also gives you a time-space tradeoff: for example, if you accept a 64 x data blowup and 64 times slower you can get two bits of distance.
I haven't completely thought this through, but I have an idea of where I'd start.
You could divide the search space up into a number of buckets where each bucket has a bucket key and the keys in the bucket are the keys that are more similar to this bucket key than any other bucket key. To create the bucket keys, you could randomly generate 64 bit keys and discard any that are too close to any previously created bucket key, or you could work out some algorithm that generates keys that are all dissimilar enough. To find the closest key to a test key, first find the bucket key that is closest, and then test each key in the bucket. (Actually, it's possible, but not likely, for the closest key to be in another bucket - do you need to find the closest key, or would a very close key be good enough?)
If you're ok with doing it probabilistically, I think there's a good way to solve question 2. I assume you have 2^30 data and cutoff and you want to find all points within cutoff distance from test.
One_Try()
1. Generate randomly a 20-bit subset S of 64 bits
2. Ask for a list of elements that agree with test on S (about 2^10 elements)
3. Sort that list by Hamming distance from test
4. Discard the part of list after cutoff
You repeat One_Try as much as you need while merging the lists. The more tries you have, the more points you find. For example, if x is within 5 bits, you'll find it in one try with about (2/3)^5 = 13% probability. Therefore if you repeat 100 tries you find all but roughly 10^{-6} of such x. Total time: 100*(1000*log 1000).
The main advantage of this is that you're able to output answers to question 2 as you proceed, since after the first few tries you'll certainly find everything within distance not more than 3 bits, etc.
If you have many computers, you give each of them several tries, since they are perfectly parallelizable: each computer saves some hash tables in advance.
Data structures for large sets described here: Detecting Near-Duplicates for Web Crawling
or
in memory trie: Judy-arrays at sourceforge.net
Assuming you have to visit each row to test its value (or if you index on the bitfield then each index entry), then you can write the actual test quite efficiently using
A xor B
To find the difference bits, then bit-count the result, using a technique like this.
This effectively gives you the hamming distance.
Since this can compile down to tens of instructions per test, this can run pretty fast.
If you are okay with a randomized algorithm (monte carlo in this case), you can use the minhash.
If the data weren't so sparse, a graph with keys as the vertices and edges linking 'adjacent' (Hamming distance = 1) nodes would probably be very efficient time-wise. The space would be very large though, so in your case, I don't think it would be a worthwhile tradeoff.