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  • Code golf: combining multiple sorted lists into a single sorted list

    - by Alabaster Codify
    Implement an algorithm to merge an arbitrary number of sorted lists into one sorted list. The aim is to create the smallest working programme, in whatever language you like. For example: input: ((1, 4, 7), (2, 5, 8), (3, 6, 9)) output: (1, 2, 3, 4, 5, 6, 7, 8, 9) input: ((1, 10), (), (2, 5, 6, 7)) output: (1, 2, 5, 6, 7, 10) Note: solutions which concatenate the input lists then use a language-provided sort function are not in-keeping with the spirit of golf, and will not be accepted: sorted(sum(lists,[])) # cheating: out of bounds! Apart from anything else, your algorithm should be (but doesn't have to be) a lot faster! Clearly state the language, any foibles and the character count. Only include meaningful characters in the count, but feel free to add whitespace to the code for artistic / readability purposes. To keep things tidy, suggest improvement in comments or by editing answers where appropriate, rather than creating a new answer for each "revision". EDIT: if I was submitting this question again, I would expand on the "no language provided sort" rule to be "don't concatenate all the lists then sort the result". Existing entries which do concatenate-then-sort are actually very interesting and compact, so I won't retro-actively introduce a rule they break, but feel free to work to the more restrictive spec in new submissions. Inspired by http://stackoverflow.com/questions/464342/combining-two-sorted-lists-in-python

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  • Merge Join component sorted outputs [SSIS]

    - by jamiet
    One question that I have been asked a few times of late in regard to performance tuning SSIS data flows is this: Why isn’t the Merge Join output sorted (i.e.IsSorted=True)? This is a fair question. After all both of the Merge Join inputs are sorted, hence why wouldn’t the output be sorted as well? Well here’s a little secret, the Merge Join output IS sorted! There’s a caveat though – it is only under certain circumstances and SSIS itself doesn’t do a good job of informing you of it. Let’s take a look at an example. Here we have a dataflow that consumes data from the [AdventureWorks2008].[Sales].[SalesOrderHeader] & [AdventureWorks2008].[Sales].[SalesOrderDetail] tables then joins them using a Merge Join component: Let’s take a look inside the editor of the Merge Join: We are joining on the [SalesOrderId] field (which is what the two inputs just happen to be sorted upon). We are also putting [SalesOrderHeader].[SalesOrderId] into the output. Believe it or not the output from this Merge Join component is sorted (i.e. has IsSorted=True) but unfortunately the Merge Join component does not have an Advanced Editor hence it is hidden away from us. There are a couple of ways to prove to you that is the case; I could open up the package XML inside the .dtsx file and show you the metadata but there is an easier way than that – I can attach a Sort component to the output. Take a look: Notice that the Sort component is attempting to sort on the [SalesOrderId] column. This gives us the following warning: Validation warning. DFT Get raw data: {992B7C9A-35AD-47B9-A0B0-637F7DDF93EB}: The data is already sorted as specified so the transform can be removed. The warning proves that the output from the Merge Join is sorted! It must be noted that the Merge Join output will only have IsSorted=True if at least one of the join columns is included in the output. So there you go, the Merge Join component can indeed produce a sorted output and that’s very useful in order to avoid unnecessary expensive Sort operations downstream. Hope this is useful to someone out there! @Jamiet  P.S. Thank you to Bob Bojanic on the SSIS product team who pointed this out to me!

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  • Merge Join component sorted outputs [SSIS]

    - by jamiet
    One question that I have been asked a few times of late in regard to performance tuning SSIS data flows is this: Why isn’t the Merge Join output sorted (i.e.IsSorted=True)? This is a fair question. After all both of the Merge Join inputs are sorted, hence why wouldn’t the output be sorted as well? Well here’s a little secret, the Merge Join output IS sorted! There’s a caveat though – it is only under certain circumstances and SSIS itself doesn’t do a good job of informing you of it. Let’s take a look at an example. Here we have a dataflow that consumes data from the [AdventureWorks2008].[Sales].[SalesOrderHeader] & [AdventureWorks2008].[Sales].[SalesOrderDetail] tables then joins them using a Merge Join component: Let’s take a look inside the editor of the Merge Join: We are joining on the [SalesOrderId] field (which is what the two inputs just happen to be sorted upon). We are also putting [SalesOrderHeader].[SalesOrderId] into the output. Believe it or not the output from this Merge Join component is sorted (i.e. has IsSorted=True) but unfortunately the Merge Join component does not have an Advanced Editor hence it is hidden away from us. There are a couple of ways to prove to you that is the case; I could open up the package XML inside the .dtsx file and show you the metadata but there is an easier way than that – I can attach a Sort component to the output. Take a look: Notice that the Sort component is attempting to sort on the [SalesOrderId] column. This gives us the following warning: Validation warning. DFT Get raw data: {992B7C9A-35AD-47B9-A0B0-637F7DDF93EB}: The data is already sorted as specified so the transform can be removed. The warning proves that the output from the Merge Join is sorted! It must be noted that the Merge Join output will only have IsSorted=True if at least one of the join columns is included in the output. So there you go, the Merge Join component can indeed produce a sorted output and that’s very useful in order to avoid unnecessary expensive Sort operations downstream. Hope this is useful to someone out there! @Jamiet  P.S. Thank you to Bob Bojanic on the SSIS product team who pointed this out to me!

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  • How to store sorted records in csv file ?

    - by Harikrishna
    I sort the records of the datatable datewise with the column TradingDate which is type of datetime. TableWithOnlyFixedColumns.DefaultView.Sort = "TradingDate asc"; Now I want to display these sorted records into csv file but it does not display records sorted by date. TableWithOnlyFixedColumns.DefaultView.Sort = "TradingDate asc";TableWithOnlyFixedColumns.Columns["TradingDate"].ColumnName + "] asc"; DataTable newTable = TableWithOnlyFixedColumns.Clone(); newTable.DefaultView.Sort = TableWithOnlyFixedColumns.DefaultView.Sort; foreach (DataRow oldRow in TableWithOnlyFixedColumns.Rows) { newTable.ImportRow(oldRow); } // we'll use these to check for rows with nulls var columns = newTable.DefaultView.Table.Columns.Cast<DataColumn>(); using (var writer = new StreamWriter(@"C:\Documents and Settings\Administrator\Desktop\New Text Document (3).csv")) { for (int i = 0; i < newTable.DefaultView.Table.Rows.Count; i++) { DataRow row = newTable.DefaultView.Table.Rows[i]; // check for any null cells if (columns.Any(column => row.IsNull(column))) continue; string[] textCells = row.ItemArray .Select(cell => cell.ToString()) // may need to pick a text qualifier here .ToArray(); // check for non-null but EMPTY cells if (textCells.Any(text => string.IsNullOrEmpty(text))) continue; writer.WriteLine(string.Join(",", textCells)); } } So how to store sorted records in csv file ?

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  • Java - sorted stack

    - by msr
    Hello, I need a sorted stack. I mean, the element removed from the stack must be the one with great priority. Stack dimension varies a lot (becomes bigger very fast). I need also to search elements in that stack. Does Java give some good implementation for this? What class or algorithm do you suggest for this? I'm using a PriorityQueue right now which I consider reasonable except for searching, so Im wondering if I can use something better. Thanks in advance!

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  • Sorting Algorithms

    - by MarkPearl
    General Every time I go back to university I find myself wading through sorting algorithms and their implementation in C++. Up to now I haven’t really appreciated their true value. However as I discovered this last week with Dictionaries in C# – having a knowledge of some basic programming principles can greatly improve the performance of a system and make one think twice about how to tackle a problem. I’m going to cover briefly in this post the following: Selection Sort Insertion Sort Shellsort Quicksort Mergesort Heapsort (not complete) Selection Sort Array based selection sort is a simple approach to sorting an unsorted array. Simply put, it repeats two basic steps to achieve a sorted collection. It starts with a collection of data and repeatedly parses it, each time sorting out one element and reducing the size of the next iteration of parsed data by one. So the first iteration would go something like this… Go through the entire array of data and find the lowest value Place the value at the front of the array The second iteration would go something like this… Go through the array from position two (position one has already been sorted with the smallest value) and find the next lowest value in the array. Place the value at the second position in the array This process would be completed until the entire array had been sorted. A positive about selection sort is that it does not make many item movements. In fact, in a worst case scenario every items is only moved once. Selection sort is however a comparison intensive sort. If you had 10 items in a collection, just to parse the collection you would have 10+9+8+7+6+5+4+3+2=54 comparisons to sort regardless of how sorted the collection was to start with. If you think about it, if you applied selection sort to a collection already sorted, you would still perform relatively the same number of iterations as if it was not sorted at all. Many of the following algorithms try and reduce the number of comparisons if the list is already sorted – leaving one with a best case and worst case scenario for comparisons. Likewise different approaches have different levels of item movement. Depending on what is more expensive, one may give priority to one approach compared to another based on what is more expensive, a comparison or a item move. Insertion Sort Insertion sort tries to reduce the number of key comparisons it performs compared to selection sort by not “doing anything” if things are sorted. Assume you had an collection of numbers in the following order… 10 18 25 30 23 17 45 35 There are 8 elements in the list. If we were to start at the front of the list – 10 18 25 & 30 are already sorted. Element 5 (23) however is smaller than element 4 (30) and so needs to be repositioned. We do this by copying the value at element 5 to a temporary holder, and then begin shifting the elements before it up one. So… Element 5 would be copied to a temporary holder 10 18 25 30 23 17 45 35 – T 23 Element 4 would shift to Element 5 10 18 25 30 30 17 45 35 – T 23 Element 3 would shift to Element 4 10 18 25 25 30 17 45 35 – T 23 Element 2 (18) is smaller than the temporary holder so we put the temporary holder value into Element 3. 10 18 23 25 30 17 45 35 – T 23   We now have a sorted list up to element 6. And so we would repeat the same process by moving element 6 to a temporary value and then shifting everything up by one from element 2 to element 5. As you can see, one major setback for this technique is the shifting values up one – this is because up to now we have been considering the collection to be an array. If however the collection was a linked list, we would not need to shift values up, but merely remove the link from the unsorted value and “reinsert” it in a sorted position. Which would reduce the number of transactions performed on the collection. So.. Insertion sort seems to perform better than selection sort – however an implementation is slightly more complicated. This is typical with most sorting algorithms – generally, greater performance leads to greater complexity. Also, insertion sort performs better if a collection of data is already sorted. If for instance you were handed a sorted collection of size n, then only n number of comparisons would need to be performed to verify that it is sorted. It’s important to note that insertion sort (array based) performs a number item moves – every time an item is “out of place” several items before it get shifted up. Shellsort – Diminishing Increment Sort So up to now we have covered Selection Sort & Insertion Sort. Selection Sort makes many comparisons and insertion sort (with an array) has the potential of making many item movements. Shellsort is an approach that takes the normal insertion sort and tries to reduce the number of item movements. In Shellsort, elements in a collection are viewed as sub-collections of a particular size. Each sub-collection is sorted so that the elements that are far apart move closer to their final position. Suppose we had a collection of 15 elements… 10 20 15 45 36 48 7 60 18 50 2 19 43 30 55 First we may view the collection as 7 sub-collections and sort each sublist, lets say at intervals of 7 10 60 55 – 20 18 – 15 50 – 45 2 – 36 19 – 48 43 – 7 30 10 55 60 – 18 20 – 15 50 – 2 45 – 19 36 – 43 48 – 7 30 (Sorted) We then sort each sublist at a smaller inter – lets say 4 10 55 60 18 – 20 15 50 2 – 45 19 36 43 – 48 7 30 10 18 55 60 – 2 15 20 50 – 19 36 43 45 – 7 30 48 (Sorted) We then sort elements at a distance of 1 (i.e. we apply a normal insertion sort) 10 18 55 60 2 15 20 50 19 36 43 45 7 30 48 2 7 10 15 18 19 20 30 36 43 45 48 50 55 (Sorted) The important thing with shellsort is deciding on the increment sequence of each sub-collection. From what I can tell, there isn’t any definitive method and depending on the order of your elements, different increment sequences may perform better than others. There are however certain increment sequences that you may want to avoid. An even based increment sequence (e.g. 2 4 8 16 32 …) should typically be avoided because it does not allow for even elements to be compared with odd elements until the final sort phase – which in a way would negate many of the benefits of using sub-collections. The performance on the number of comparisons and item movements of Shellsort is hard to determine, however it is considered to be considerably better than the normal insertion sort. Quicksort Quicksort uses a divide and conquer approach to sort a collection of items. The collection is divided into two sub-collections – and the two sub-collections are sorted and combined into one list in such a way that the combined list is sorted. The algorithm is in general pseudo code below… Divide the collection into two sub-collections Quicksort the lower sub-collection Quicksort the upper sub-collection Combine the lower & upper sub-collection together As hinted at above, quicksort uses recursion in its implementation. The real trick with quicksort is to get the lower and upper sub-collections to be of equal size. The size of a sub-collection is determined by what value the pivot is. Once a pivot is determined, one would partition to sub-collections and then repeat the process on each sub collection until you reach the base case. With quicksort, the work is done when dividing the sub-collections into lower & upper collections. The actual combining of the lower & upper sub-collections at the end is relatively simple since every element in the lower sub-collection is smaller than the smallest element in the upper sub-collection. Mergesort With quicksort, the average-case complexity was O(nlog2n) however the worst case complexity was still O(N*N). Mergesort improves on quicksort by always having a complexity of O(nlog2n) regardless of the best or worst case. So how does it do this? Mergesort makes use of the divide and conquer approach to partition a collection into two sub-collections. It then sorts each sub-collection and combines the sorted sub-collections into one sorted collection. The general algorithm for mergesort is as follows… Divide the collection into two sub-collections Mergesort the first sub-collection Mergesort the second sub-collection Merge the first sub-collection and the second sub-collection As you can see.. it still pretty much looks like quicksort – so lets see where it differs… Firstly, mergesort differs from quicksort in how it partitions the sub-collections. Instead of having a pivot – merge sort partitions each sub-collection based on size so that the first and second sub-collection of relatively the same size. This dividing keeps getting repeated until the sub-collections are the size of a single element. If a sub-collection is one element in size – it is now sorted! So the trick is how do we put all these sub-collections together so that they maintain their sorted order. Sorted sub-collections are merged into a sorted collection by comparing the elements of the sub-collection and then adjusting the sorted collection. Lets have a look at a few examples… Assume 2 sub-collections with 1 element each 10 & 20 Compare the first element of the first sub-collection with the first element of the second sub-collection. Take the smallest of the two and place it as the first element in the sorted collection. In this scenario 10 is smaller than 20 so 10 is taken from sub-collection 1 leaving that sub-collection empty, which means by default the next smallest element is in sub-collection 2 (20). So the sorted collection would be 10 20 Lets assume 2 sub-collections with 2 elements each 10 20 & 15 19 So… again we would Compare 10 with 15 – 10 is the winner so we add it to our sorted collection (10) leaving us with 20 & 15 19 Compare 20 with 15 – 15 is the winner so we add it to our sorted collection (10 15) leaving us with 20 & 19 Compare 20 with 19 – 19 is the winner so we add it to our sorted collection (10 15 19) leaving us with 20 & _ 20 is by default the winner so our sorted collection is 10 15 19 20. Make sense? Heapsort (still needs to be completed) So by now I am tired of sorting algorithms and trying to remember why they were so important. I think every year I go through this stuff I wonder to myself why are we made to learn about selection sort and insertion sort if they are so bad – why didn’t we just skip to Mergesort & Quicksort. I guess the only explanation I have for this is that sometimes you learn things so that you can implement them in future – and other times you learn things so that you know it isn’t the best way of implementing things and that you don’t need to implement it in future. Anyhow… luckily this is going to be the last one of my sorts for today. The first step in heapsort is to convert a collection of data into a heap. After the data is converted into a heap, sorting begins… So what is the definition of a heap? If we have to convert a collection of data into a heap, how do we know when it is a heap and when it is not? The definition of a heap is as follows: A heap is a list in which each element contains a key, such that the key in the element at position k in the list is at least as large as the key in the element at position 2k +1 (if it exists) and 2k + 2 (if it exists). Does that make sense? At first glance I’m thinking what the heck??? But then after re-reading my notes I see that we are doing something different – up to now we have really looked at data as an array or sequential collection of data that we need to sort – a heap represents data in a slightly different way – although the data is stored in a sequential collection, for a sequential collection of data to be in a valid heap – it is “semi sorted”. Let me try and explain a bit further with an example… Example 1 of Potential Heap Data Assume we had a collection of numbers as follows 1[1] 2[2] 3[3] 4[4] 5[5] 6[6] For this to be a valid heap element with value of 1 at position [1] needs to be greater or equal to the element at position [3] (2k +1) and position [4] (2k +2). So in the above example, the collection of numbers is not in a valid heap. Example 2 of Potential Heap Data Lets look at another collection of numbers as follows 6[1] 5[2] 4[3] 3[4] 2[5] 1[6] Is this a valid heap? Well… element with the value 6 at position 1 must be greater or equal to the element at position [3] and position [4]. Is 6 > 4 and 6 > 3? Yes it is. Lets look at element 5 as position 2. It must be greater than the values at [4] & [5]. Is 5 > 3 and 5 > 2? Yes it is. If you continued to examine this second collection of data you would find that it is in a valid heap based on the definition of a heap.

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  • what is itunes result sorted by ?

    - by NemesisII
    This is my App on Itunes: http://itunes.apple.com/us/app/buddy-calculator/id445261163?mt=8 My app Key word is : "calculator scientific calc equation math mathematics unit converter conversion statistic algebra" But when I search "Calculator" on Itunes, my app is not appeared (in first two pages). So I want to ask a question that how to improve the rank of my app , and what is the searched result sorted by (new first, vote or downloaded or ...) ? If I want to improve the rank, how can I do, does it cost fee or something ? Thanks you very much ^^ !

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  • Jquery UI Sortable - Get the item being sorted.

    - by Smickie
    Hi, When using Jquery UI Sortable (which is great by the way) how do you get the item that is currently being sorted. When you use $(this); it return the actual sortable list, not the current sorted item. I want to do fancy-pants things with the widget when the user is dragging it around. E.g. Animate it when dragging between two lists. So how do I get the current item being sorted? There a little code below just to explain a little more... $(function() { $("#sortable_1").sortable({ start : function(event, ui){ //get current element being sorted }, stop : function(event, ui){ //get current element being sorted } }).disableSelection(); });

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  • Opening images sorted by modification date/size/type/etc

    - by menino bolinho
    Suppose I have a folder with pictures in them. If I sort them by name once I open one with Image Viewer and navigate to the others the order is respected. But if I sort my files by modification date, for example, I can't do that. Basically, the default Image Viewer only lets you navigate images by name. According to this post on the ubuntuforums this has been an issue since 2007! Is there a good/easy way to fix it? Seems like such a trivial thing to me.

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  • OData Query Option top Forces Data To Be Sorted By Primary Key

    This post show a simple WCF Data Service (Formerly known as ADO.NET Data Services) project that retrieves data using the Reflection Provider for accessing data. It goes on to show that using $top... This site is a resource for asp.net web programming. It has examples by Peter Kellner of techniques for high performance programming...Did you know that DotNetSlackers also publishes .net articles written by top known .net Authors? We already have over 80 articles in several categories including Silverlight. Take a look: here.

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  • Displaying a Sorted, Paged, and Filtered Grid of Data in ASP.NET MVC

    Over the past couple of months I've authored five articles on displaying a grid of data in an ASP.NET MVC application. The first article in the series focused on simply displaying data. This was followed by articles showing how to sort, page, and filter a grid of data. We then examined how to both sort and page a single grid of data. This article looks at how to add the final piece to the puzzle: we'll see how to combine sorting, paging and filtering when displaying data in a single grid. Like with its predecessors, this article offers step-by-step instructions and includes a complete, working demo available for download at the end of the article. Read on to learn more! Read More >

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  • Implement Partial sorted query in sql server 2005

    - by Brij
    I have to show records in such a way that some selected records should come first. After this, another records come in sorted manner from the same table. For example, If I select state having stateID = 5 then the corresponding record should come first. after this another records should come in sorted manner. For this, I tried union but it shows all in sorted. select state from statemaster where stateid=5 union all select state from statemaster where not stateid =5 order by state Thanks

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  • A good Sorted List for Java

    - by Phuong Nguyen de ManCity fan
    I'm looking for a good sorted list for java. Googling around give me some hints about using TreeSet/TreeMap. But these components is lack of one thing: random access to an element in the set. For example, I want to access nth element in the sorted set, but with TreeSet, I must iterate over other n-1 elements before I can get there. It would be a waste since I would have upto several thousands elements in my Set. Basically, I'm looking for some thing similar to a sorted list in .NET, with ability to add element fast, remove element fast, and have random access to any element in the list. Has this kind of sorted list implemented somewhere? Thanks.

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  • Interview Q: sorting an almost sorted array (elements misplaced by no more than k)

    - by polygenelubricants
    I was asked this interview question recently: You're given an array that is almost sorted, in that each of the N elements may be misplaced by no more than k positions from the correct sorted order. Find a space-and-time efficient algorithm to sort the array. I have an O(N log k) solution as follows. Let's denote arr[0..n) to mean the elements of the array from index 0 (inclusive) to N (exclusive). Sort arr[0..2k) Now we know that arr[0..k) are in their final sorted positions... ...but arr[k..2k) may still be misplaced by k! Sort arr[k..3k) Now we know that arr[k..2k) are in their final sorted positions... ...but arr[2k..3k) may still be misplaced by k Sort arr[2k..4k) .... Until you sort arr[ik..N), then you're done! This final step may be cheaper than the other steps when you have less than 2k elements left In each step, you sort at most 2k elements in O(k log k), putting at least k elements in their final sorted positions at the end of each step. There are O(N/k) steps, so the overall complexity is O(N log k). My questions are: Is O(N log k) optimal? Can this be improved upon? Can you do this without (partially) re-sorting the same elements?

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  • Iterating over key/value pairs in a dict sorted by keys

    - by Helper Method
    I have the following code, which just print the key/value pairs in a dict (the pairs are sorted by keys): for word, count in sorted(count_words(filename).items()): print word, count However, calling iteritems() instead of items() produces the same output for word, count in sorted(count_words(filename).iteritems()): print word, count Now, which one should I choose in this situation? I consulted the Python tutorial but it doesn't really answer my question.

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  • Multicore system. How are the cores "sorted"

    - by asdrubael
    If I have a system with let's say two Quadcore CPUs and HyperThreading enabled, how do I know how the phsyical and virtual cores are sorted? For example if I force a process to run on "CPU 0" how to know which Core this really is? I remember a complex drawing about this, but didn't find anything useful.

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  • Most efficient algorithm for merging sorted IEnumerable<T>

    - by franck
    Hello, I have several huge sorted enumerable sequences that I want to merge. Theses lists are manipulated as IEnumerable but are already sorted. Since input lists are sorted, it should be possible to merge them in one trip, without re-sorting anything. I would like to keep the defered execution behavior. I tried to write a naive algorithm which do that (see below). However, it looks pretty ugly and I'm sure it can be optimized. It may exist a more academical algorithm... IEnumerable<T> MergeOrderedLists<T, TOrder>(IEnumerable<IEnumerable<T>> orderedlists, Func<T, TOrder> orderBy) { var enumerators = orderedlists.ToDictionary(l => l.GetEnumerator(), l => default(T)); IEnumerator<T> tag = null; var firstRun = true; while (true) { var toRemove = new List<IEnumerator<T>>(); var toAdd = new List<KeyValuePair<IEnumerator<T>, T>>(); foreach (var pair in enumerators.Where(pair => firstRun || tag == pair.Key)) { if (pair.Key.MoveNext()) toAdd.Add(pair); else toRemove.Add(pair.Key); } foreach (var enumerator in toRemove) enumerators.Remove(enumerator); foreach (var pair in toAdd) enumerators[pair.Key] = pair.Key.Current; if (enumerators.Count == 0) yield break; var min = enumerators.OrderBy(t => orderBy(t.Value)).FirstOrDefault(); tag = min.Key; yield return min.Value; firstRun = false; } } The method can be used like that: // Person lists are already sorted by age MergeOrderedLists(orderedList, p => p.Age); assuming the following Person class exists somewhere: public class Person { public int Age { get; set; } } Duplicates should be conserved, we don't care about their order in the new sequence. Do you see any obvious optimization I could use?

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  • Running time for Dijkstra's algorithm on a priority queue implemented by sorted list/array

    - by jay
    So I'm curious to know what the running time for the algorithm is on on priority queue implemented by a sorted list/array. I know for an unsorted list/array it is O((n^2+m)) where n is the number of vertices and m the number of edges. Thus that equates to O(n^2) time. But would it be faster if i used an sorted list/array...What would the running time be? I know extractmin would be constant time.

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  • Java - PriorityQueue vs sorted LinkedList

    - by msr
    Hello, Which implementation is less "heavy": PriorityQueue or a sorted LinkedList (using a Comparator)? I want to have all the items sorted. The insertion will be very frequent and ocasionally I will have to run all the list to make some operations. Thank you!

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  • How to let the UITable sorted by number?

    - by Tattat
    I loaded a plist from the UITable, but there is something wrong. my plist file is some data like that: 3 3uyyhuhu34 ..... 5 5uyyhuhu34 ..... 11 11uyyhuhu34 ..... I found that the list is sorted from 11, then 3, after that is 5. But I want to sorted from 3, 5 to 11. What can I do to change the behaviors? thx

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  • Time complexity for Search and Insert operation in sorted and unsorted arrays that includes duplicat

    - by iecut
    1-)For sorted array I have used Binary Search. We know that the worst case complexity for SEARCH operation in sorted array is O(lg N), if we use Binary Search, where N are the number of items in an array. What is the worst case complexity for the search operation in the array that includes duplicate values, using binary search?? Will it be the be the same O(lg N)?? Please correct me if I am wrong!! Also what is the worst case for INSERT operation in sorted array using binary search?? My guess is O(N).... is that right?? 2-) For unsorted array I have used Linear search. Now we have an unsorted array that also accepts duplicate element/values. What are the best worst case complexity for both SEARCH and INSERT operation. I think that we can use linear search that will give us O(N) worst case time for both search and delete operations. Can we do better than this for unsorted array and does the complexity changes if we accepts duplicates in the array.

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  • Properly maintain sorted state of Array/Set

    - by Jeff
    I'm trying to get data out of my MOC and then create some new objects based on those objects, and put it all back together, while keeping my sort state. The securities come out of the MOC in proper order. And everything seems to be fine until I do the assignment to the game at the bottom from setWithArray. The documentation says that setWithArray removed the duplicate objects, if there are any. I'm wonder if that's messing up my data, but I don't see a good alternative. The data is ultimately being pulled out into a UITableView. When I add items to the game manually, then they stay sorted, so I don't think the breaking of the sort is beyond the scope of what I've included here. NSError *error; NSArray *allTheSecurities = [managedObjectContext executeFetchRequest:request error:&error]; if (allTheSecurities == nil) { // Handle the error. } [request release]; /**/ NSLog( @"Enumerate..." ); NSEnumerator *enumerator = [allTheSecurities objectEnumerator]; id anObject; NSMutableArray *portfolioStocks = [[NSMutableArray alloc] init]; while (anObject = [enumerator nextObject]) { NSLog( @"Iteration... %@", [anObject name] ); NSLog( @"Build a stock..." ); PortfolioStocks *this_stock = (PortfolioStocks *)[NSEntityDescription insertNewObjectForEntityForName:@"PortfolioStocks" inManagedObjectContext:context]; NSLog( @"Set a value..." ); [this_stock setSecurity:(Security *)anObject]; [this_stock setQuantity:[NSNumber numberWithInt:0]]; NSLog( @"Add to portfolioStocks..." ); [portfolioStocks addObject:this_stock]; } //Sorted properly up to here! NSLog( @"Add to portfolio..." ); [game setPortfolio:[NSSet setWithArray:portfolioStocks]]; // <-- This is where it's not sorted anymore.

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