I have been struggling with this for some time, I need to write a recursive function, that uses divide and conquer method to interleave an array, for example:
[1,2,3,4,a,b,c,d]
is turned to
[1,a,2,b,3,c,4,d].
I've searched the internet, but only found the solution when the number of elements in array is a power of 2. The transition has to be done in place, with nlogn time. Also, is it possible to be done when the number of elements is odd? I'm writing in java but will be grateful for any help.
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I am trying to understand the speed of the Binary Search algorithm.
I understand it needs to operate on a sorted array.
However if the array comes in unsorted and performing the sorting. Wouldn't the sorting be part of the Binary Search and thus its performance would be slower?
I am confused because I think that there is very little chance to use this algorithm if the data does not come in sorted.
And if my code needs to sort it then why isn't it counting towards the search algorithm.
Sorry if I am confusing,
Thank you for helping.
You can't just point at an algorithm and say: It's got O(n^2) complexity!
That's what people usually say, sure. But that's shorthand. They're omitting things; assuming that the listener / reader will make assumptions.
You need to fully describe the exact algorithm, the conditions under which it is applied, and the precise definition of n and any other variable.
Then, you can answer that question. The problem you're having here is that the definition of 'what is the performance of binary search' is unclear. If you assume it means X whilst your buddy assumes it means Y, and you then argue about the answers, you're not actually having a constructive debate at all. You're just tilting at windmills; the real problem is that neither of you figured out the problem is communicating the basics.
Given that there is some confusion here, I'll give you 3 different more or less equally sensible more fleshed out definitions, along with the actual answer for each such definition. Hint, for one of them, 'binary search' isn't the fastest algorithm!
Given [1] a list that is already sorted, and [2] a single value, write me an algorithm that determines if this value is in the list or not.
The best answer would be: A binary sort algorithm, and its complexity would be O(log n).
Given [1] a list that is not sorted, and [2] a single value, write me an algorithm that determines if this value is in the list or not.
The best answer would be: Just iterate through the list. Its complexity would be O(n), and binary sort is not part of this answer at all.
given [1] a list that is not sorted, and [2] a list of tests, whereby each individual test is defined by a single value, but they all use the same input unsorted list, write an algorithm that will, for each test, determine if the value for that test is in the list or not, and then give me the amortized complexity (basically, the complexity of the whole thing, divided by the # of tests we ran).
Then the best answer would be: First sort the list, spending O(n log n) time to do so, but we get to amortize that over the test case count, and then use binary search for each individual test, adding an O(log n) complexity to each test. If we term n the size of the input list and t the number of tests we have, this gets us:
O( (n log n)/t + O(log n) )
Which is the actual answer to the question, complex as it may look. But, if t is large or even considered effectively infinite in size, OR we add one more rider to the question:
The list from [1] is given to you in advance and, within reasonable time and memory limits, you may preprocess this data without needing to amortize these costs across your test cases
then that boils down to just O(log n), as the large value for t makes that (n log n) / t factor approach zero.
In communicating this to your buddy, given that we don't talk in entire scientific papers, one might then say: "The algorithmic complexity of the binary sort algorithm is O(log n)", even if that omits a gigantic chunk of the full story.
You interpret the question as per the second case (input is unsorted, the input comprises both the list and the value to search for, no multi-test clause). Someone who says 'binary search is O(log n)' is labouring under either the first or third. You're both right.
NB: The third definition seems unusually complicated. However, it matches common scenarios. For example, 'we have compiled a list of folks living in town and their phone numbers, and we want to print them in a giant book with the aim of letting recipients of this book look up phone numbers. We expect over the lifetime of a single print run that the 100,000 citizens of the township will eaech do on average about 50 lookups, for a grand total of 5 million lookups for this single list. That gives you t= 5 million, n = 200,000 (let's say 200k people live here, half of which get a phonebook). Plug those numbers in and sorting the phonebook wins by a landslide vs. releasing the phonebook in arbitrary, unsorted order. Even if, yes, you start 'down' the effort of sorting it and won't make up for that loss until a few folks have speedily looked up a few phone numbers to make up for your efforts in sorting it before printing the book.
Yes. If
the data comes in unsorted
you only need to search for one element
...then you would have to first sort the data to use binary search, which would take a total of O(n log n + log n) = O(n log n) time.
But once the data is sorted, you can then binary search on that data as many times as you want. You don't have to sort it again each time.
If I have an array {2,3,4,5,6,5,7}, the output of this should be 10, as 5 is the most commonly occurring integer occurring twice. Importantly, I want to make this program optimized to its best time complexity while stating how I calculated it and also the space complexity, all in Java.
I am trying to use Treemap, but don't know is it the best solution or how to write the best optimized code in Java, along with other calculations. Please help.
The best solution is with a bucket array, you run on your array once O(n), and count the number of appearances of each number.
at the and you find the maximum bucket and print - bucket_value*bucket_number.
I have read many variations of the Knapsack problem, but the version I am tasked with is a little different and I don't quite understand how to solve it.
I have an array of integers that represent weights (ie. {1,4,6,12,7,2}) and need to find only one solution that adds up to the target weight.
I understand the basic algorithm, but I cannot understand how to implement it.
First, what is my base case? Is it when the array is empty? The target has been reached? The target has been over-reached? Or maybe some combination?
Second, when the target is over-reached, how do I backtrack and try the next item?
Third, what am I supposed to return? Should I be returning ints (in which case, am I supposed to print them out?)? Or do I return arrays and the final return is the solution?
Think carefully about the problem you are trying to solve. To approach this problem, I am
considering the inputs and outputs of the Knapsack algorithm.
Input: A set of integers (the knapsack) and a single integer (the proposed sum)
Output: A set of integers who add up to the proposed sum, or null.
In this way your code might look like this
public int[] knapsackSolve(int[] sack, int prospectiveSum) {
//your algorithm here
}
The recursive algorithm is straightforward. First compute the sum of the sack. If it equals
prospectiveSum then return the sack. Otherwise iterate over sack, and for each item initialise a new knapsack with that item removed. Call knapsackSolve on this. If there is a solution, return it. Otherwise proceed to the next item.
For example if we call knapsackSolve({1,2,3},5) the algorithm tries 1 + 2 + 3 = 5 which is false. So it loops through {1,2,3} and calls knapsackSolve on the sublists {2,3},{1,3} and {1,2}. knapsackSolve({2,3},5) is the one that returns a solution.
This isn't a very efficient algorithm although it illustrates fairly well how complex the Knapsack problem is!
Basically the Knapsack problem is formulated as (from Wikipedia): "Given a set of items, each with a mass and a value, determine the number of each item to include in a collection so that the total weight is less than or equal to a given limit and the total value is as large as possible. For your problem we are interested in weights only. So you can set all values to 1. Additionally we only want to know if the target weight can be reached exactly. I guess you are allowed to use a weight only once and not multiple times?
This problem can be solved nicely with dynamic programming. Are you familiar with that principle? Applying dynamic programming is much more elegant and quicker to program than backtracking. But if you are only allowed to do backtracking use the approach from user2738608 posted above.
I have an assignment in my intro to programming course that I don't understand at all. I've been falling behind because of problems at home. I'm not asking you to do my assignment for me I'm just hoping for some help for a programming boob like me.
The question is this:
Calculate the time complexity in average case for searching, adding, and removing in a
- unsorted vector
- sorted vector
- unsorted singlelinked list
- sorted singlelinked list
- hash table
Let n be the number of elements in the datastructure
and present the solution in a
table with three rows and five columns.
I'm not sure what this even means.. I've read as much as I can about time complexity but I don't understand it.. It's so confusing. I don't know where I would even start.. Remember I'm a novice programmer, as dumb as they come. I did really well last semester but had problems at home at the start of this one so I missed a lot of lectures and the first assignments so now I'm in over my head..
Maybe if someone could give me the answer and the reasoning behind it on a couple of them I could maybe understand it and do the others? I have a hard time learning through theory, examples work best.
Time complexity is a formula that describes how the cost of an operation varies related to the number of elements. It is usually expressed using "big-O" notation, for example O(1) or constant time, O(n) where cost relates linearly to n, O(n2) where cost increases as the square of the size of the input. There can be others involving exponentials or logarithms. Read up on "Big-O Notation".
You are being asked to evaluate five different data structures, and provide average cost for three different operations on each data structure (hence the table with three rows and five columns).
Time complexity is an abstract concept, that allows us to compare the complexity of various algorithms by looking at how many operations are performed in order to handle its inputs. To be precise, the exact number of operations isn't important, the bottom line is, how does the number of operations scale with increasing complexity of inputs.
Generally, the number of inputs is denoted as n and the complexity is denoted as O(p(n)), with p(n) being some kind of expression with n. If an algorithm has O(n) complexity, it means, that is scales linearly, with every new input, the time needed to run the algorithm increases by the same amount.
If an algorithm has complexity of O(n^2) it means, that the amount of operations grows as a square of number of inputs. This goes on and on, up to exponencially complex algorithms, that are effectively useless for large enough inputs.
What your professor asks from you is to have a look at the given operations and judge, how are going to scale with increasing size of lists, you are handling. Basically this is done by looking at the algorithm and imagining, what kinds of cycles are going to be necessary. For example, if the task is to pick the first element, the complexity is O(1), meaning that it doesn't depend on the size of input. However, if you want to find a given element in the list, you already need to scan the whole list and this costs you depending on the list size. Hope this gives you a bit of an idea how algorithm complexity works, good luck with your assignment.
Ok, well there are a few things you have to start with first. Algorithmic complexity has a lot of heavy math behind it and so it is hard for novices to understand, especially if you try to look up Wikipedia definitions or more-formal definitions.
A simple definition is that time-complexity is basically a way to measure how much an operation costs to perform. Alternatively, you can also use it to see how long a particular algorithm can take to run.
Complexity is described using what is known as big-O notation. You'll usually end up seeing things like O(1) and O(n). n is usually the number of elements (possibly in a structure) on which the algorithm is operating.
So let's look at a few big-O notations:
O(1): This means that the operation runs in constant time. What this means is that regardless of the number of elements, the operation always runs in constant time. An example is looking at the first element in a non-empty array (arr[0]). This will always run in constant time because you only have to directly look at the very first element in an array.
O(n): This means that the time required for the operation increases linearly with the number of elements. An example is if you have an array of numbers and you want to find the largest number. To do this, you will have to, in the worst case, look at every single number in the array until you find the largest one. Why is that? This is because you can have a case where the largest number is the last number in the array. So you cannot be sure until you have examined every number in the array. This is why the cost of this operation is O(n).
O(n^2): This means that the time required for the operation increases quadratically with the number of elements. This usually means that for each element in the set of elements, you are running through the entire set. So that is n x n or n^2. A well-known example is the bubble-sort algorithm. In this algorithm you run through and swap adjacent elements to ensure that the array is sorted according to the order you need. The array is sorted when no-more swaps need to be made. So you have multiple passes through the array, which in the worst case is equal to the number of elements in the array.
Now there are certain things in code that you can look at to get a hint to see if the algorithm is O(n) or O(n^2).
Single loops are usually O(n), since it means you are iterating over a set of elements once:
for(int i = 0; i < n; i++) {
...
}
Doubly-nested loops are usually O(n^2), since you are iterating over an entire set of elements for each element in the set:
for(int i = 0; i < n; i++) {
for(j = 0; j < n; j++) {
...
}
}
Now how does this apply to your homework? I'm not going to give you the answer directly but I will give you enough and more hints to figure it out :). What I wrote above, describing big-O, should also help you. Your homework asks you to apply runtime analyses to different data structures. Well, certain data structures have certain runtime properties based on how they are set up.
For example, in a linked list, the only way you can get to an element in the middle of the list, is by starting with the first element and then following the next pointer until you find the element that you want. Think about that. How many steps would it take for you to find the element that you need? What do you think those steps are related to? Do the number of elements in the list have any bearing on that? How can you represent the cost of this function using big-O notation?
For each datastructure that your teacher has asked you about, try to figure out how they are set up and try to work out manually what each operation (searching, adding, removing) entails. I'm talking about writing the steps out and drawing pictures of the strucutres on paper! This will help you out immensely! Looking at that, you should have enough information to figure out the number of steps required and how it relates to the number of elements in the set.
Using this approach you should be able to solve your homework. Good luck!
I'm working on some java code for some research I'm working on, and need to have a way to iterate over all permutations of an ArrayList. I've looked over some previous questions asked here, but most were not quite what I want to do, and the ones that were close had answers dealing with strings and example code written in Perl, or in the case of the one implementation that seemed like it would work ... do not actually work.
Ideally I'm looking for tips/code snippets to help me write a function permute(list, i) that as i goes from 0 to list.size()! gives me every permutation of my ArrayList.
There is a way of counting from 0 to (n! - 1) that will list off all permutations of a list of n elements. The idea is to rewrite the numbers as you go using the factorial number system and interpreting the number as an encoded way of determining which permutation to use. If you're curious about this, I have a C++ implementation of this algorithm. I also once gave a talk about this, in case you'd like some visuals on the topic.
Hope this helps!
If iterating over all permutations is enough for you, see this answer: Stepping through all permutations one swap at a time .
For a given n the iterator produces all permutations of numbers 0 to (n-1).
You can simply wrap it into another iterator that converts the permutation of numbers into a permutation of your array elements. (Note that you cannot just replace int[] within the iterator with an arbitrary array/list. The algorithm needs to work with numbers.)