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  • Count Occurence of Needle String in Haystack String, most optimally?

    - by Taranfx
    The Problem is simple Find "ABC" in "ABCDSGDABCSAGAABCCCCAAABAABC" Here is the solution I propose, I'm looking for any solutions that might be better than this one. public static void main(String[] args) { String haystack = "ABCDSGDABCSAGAABCCCCAAABAABC"; String needle = "ABC"; char [] needl = needle.toCharArray(); int needleLen = needle.length(); int found=0; char hay[] = haystack.toCharArray(); int index =0; int chMatched =0; for (int i=0; i<hay.length; i++){ if (index >= needleLen || chMatched==0) index=0; System.out.print("\nchar-->"+hay[i] + ", with->"+needl[index]); if(hay[i] == needl[index]){ chMatched++; System.out.println(", matched"); }else { chMatched=0; index=0; if(hay[i] == needl[index]){ chMatched++; System.out.print("\nchar->"+hay[i] + ", with->"+needl[index]); System.out.print(", matched"); }else continue; } if(chMatched == needleLen){ found++; System.out.println("found. Total ->"+found); } index++; } System.out.println("Result Found-->"+found); } It took me a while creating this one. Can someone suggest a better solution (if any) P.S. Drop the sysouts if they look messy to you.

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  • Echo mysql results in a loop?

    - by Roy D. Porter
    I am using turn.js to make a book. Every div within the 'deathnote' div becomes a new page. <div id="deathnote"> //starts book <div style="background-image:url(images/coverpage.jpg);"></div> //creates new page <div style="background-image:url(images/paper.jpg);"></div> //creates new page <div style="background-image:url(images/paper.jpg);"></div> //creates new page </div> //ends book What I am doing is trying to get 3 'content' (content being a name and cause of death) divs onto 1 page, and then generate a new page. So here is what i want: <div id="deathnote"> //starts book <div style="background-image:url(images/coverpage.jpg);"></div> //creates new page <div style="background-image:url(images/paper.jpg);"></div> //creates new page <div style="background-image:url(images/paper.jpg);"> //creates new page but leaves it open <div> CONTENT </div> <div> CONTENT </div> <div> CONTENT </div> </div> //ends the page </div> //ends book Seems simple enough, however the content is data from a MySQL DB, so i have to echo it in using PHP. Here is what i have so far <div id="deathnote"> <div style="background-image:url(images/coverpage.jpg);"></div> <div style="background-image:url(images/paper.jpg);"></div> <div style="background-image:url(images/paper.jpg);"></div> <div style="background-image:url(images/paper.jpg);"></div> <div style="background-image:url(images/paper.jpg);"></div> <div style="background-image:url(images/paper.jpg);"></div> <?php $pagecount = 0; $db = new mysqli('localhost', 'username', 'passw', 'DB'); if($db->connect_errno > 0){ die('Unable to connect to database [' . $db->connect_error . ']'); } $sql = <<<SQL SELECT * FROM `TABLE` SQL; if(!$result = $db->query($sql)){ die('There was an error running the query [' . $db->error . ']'); } //IGNORE ALL OF THE GARBAGE ABOVE. IT IS SIMPLE CONNECTING SCRIPT THAT I KNOW WORKS //THE METHOD I AM HAVING TROUBLE WITH IS BELOW $pagecount = 0; while($row = $result->fetch_assoc()){ //GETS THE VALUE (and makes sure it isn't nothing echo '<div style="background-image:url(images/paper.jpg);">'; //THIS OPENS A NEW PAGE while ($pagecount !== 3) { //KEEPS COUNT OF HOW MUCH CONTENT DIVS IS ON THE PAGE while($row = $result->fetch_assoc()){ //START A CONTENT DIV echo '<div class="content"><div class="name">' . $row['victim'] . '</div><div class="cod">' . $row['cod'] . '</div></div>'; //END A CONTENT DIV $pagecount++; //UP THE PAGE COUNT } } $pagecount=0; //PUT IT BACK TO 0 echo '</div>'; //END PAGE } $db->close(); ?> <div style="background-image:url(images/backpage.jpg);"></div> //BACK PAGE </div> At the moment i seem to be causing and infinite loop so the page won't load. The problem resides within the while loops. Any help is greatly appreciated. Thanks in advance guys. :)

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  • Latex: vertical line in lstlistings

    - by Helltone
    I want to have a vertical line for indentation in the lstlisting environment, similar to what one can get in algorithm2e. I tried doing something like the code below, but the the |'s are not contiguous and the result is ugly. \lstset{ ... showtabs=true, tabsize=3, tab=\hfill$|$\hfill, ... }

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  • Efficient Array Storage for Binary Tree

    - by Sundararajan S
    We have to write the nodes of a binary tree to a file. What is the most space efficient way of writing a binary tree . We can store it in array format with parent in position 'i' and its childs in 2i,2i+1. But this will waste lot of space in case of sparse binary trees.

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  • Effecient data structure design

    - by Sway
    Hi there, I need to match a series of user inputed words against a large dictionary of words (to ensure the entered value exists). So if the user entered: "orange" it should match an entry "orange' in the dictionary. Now the catch is that the user can also enter a wildcard or series of wildcard characters like say "or__ge" which would also match "orange" The key requirements are: * this should be as fast as possible. * use the smallest amount of memory to achieve it. If the size of the word list was small I could use a string containing all the words and use regular expressions. however given that the word list could contain potentially hundreds of thousands of enteries I'm assuming this wouldn't work. So is some sort of 'tree' be the way to go for this...? Any thoughts or suggestions on this would be totally appreciated! Thanks in advance, Matt

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  • longest common subsequence

    - by davit-datuashvili
    i have following code public class LCS1 { public static String lcs(String a,String b) { String x; String y; int alen=a.length(); int blen=b.length(); if (alen==0 || blen==0) { return ""; } else if (a.charAt(alen-1)==b.charAt(blen-1)) { return lcs(a.substring(0,alen-1),b.substring(0,blen-1)); } else { x=lcs(a,b.substring(0,blen-1)); y=lcs(a.substring(0,alen-1),b); } return (x.length()>y.length()) ? x : y; } public static void main(String[]args){ String a="computer"; String b="houseboat"; System.out.println(lcs(a,b)); } } it should return "out" but returns nothing what is problem?

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  • make tree in scheme

    - by ???
    (define (entry tree) (car tree)) (define (left-branch tree) (cadr tree)) (define (right-branch tree) (caddr tree)) (define (make-tree entry left right) (list entry left right)) (define (mktree order items_list) (cond ((= (length items_list) 1) (make-tree (car items_list) '() '())) (else (insert2 order (car items_list) (mktree order (cdr items_list)))))) (define (insert2 order x t) (cond ((null? t) (make-tree x '() '())) ((order x (entry t)) (make-tree (entry t) (insert2 order x (left-branch t)) (right-branch t))) ((order (entry t) x ) (make-tree (entry t) (left-branch t) (insert2 order x (right-branch t)))) (else t))) The result is: (mktree (lambda (x y) (< x y)) (list 7 3 5 1 9 11)) (11 (9 (1 () (5 (3 () ()) (7 () ()))) ()) ()) But I'm trying to get: (7 (3 (1 () ()) (5 () ())) (9 () (11 () ()))) Where is the problem?

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  • Count subset of binary pattern ..

    - by mr.bio
    Hi there . I have a A=set of strings and a B=seperate string. I want to count the number of occurences in from B in A. Example : A: 10001 10011 11000 10010 10101 B: 10001 result would be 3.(10001 is a subset of 10001,10011,10101) So i need a function that takes a set and string and returns an int. int myfunc(set<string> , string){ int result; // My Brain is melting return result ; }

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  • Implementing a hilbert map of the internet

    - by Martin
    In the XKCD comic 195 a design for a map of the internet address space is suggested using a hilbert curve so that items from a similar IPs will be clustered together. Given an IP address, how would I calculate the 2D coordinates (in the range zero to one) that this IP is located on such a map?

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  • Could I do this blind relative to absolute path conversion (for perforce depot paths) better?

    - by wonderfulthunk
    I need to "blindly" (i.e. without access to the filesystem, in this case the source control server) convert some relative paths to absolute paths. So I'm playing with dotdots and indices. For those that are curious I have a log file produced by someone else's tool that sometimes outputs relative paths, and for performance reasons I don't want to access the source control server where the paths are located to check if they're valid and more easily convert them to their absolute path equivalents. I've gone through a number of (probably foolish) iterations trying to get it to work - mostly a few variations of iterating over the array of folders and trying delete_at(index) and delete_at(index-1) but my index kept incrementing while I was deleting elements of the array out from under myself, which didn't work for cases with multiple dotdots. Any tips on improving it in general or specifically the lack of non-consecutive dotdot support would be welcome. Currently this is working with my limited examples, but I think it could be improved. It can't handle non-consecutive '..' directories, and I am probably doing a lot of wasteful (and error-prone) things that I probably don't need to do because I'm a bit of a hack. I've found a lot of examples of converting other types of relative paths using other languages, but none of them seemed to fit my situation. These are my example paths that I need to convert, from: //depot/foo/../bar/single.c //depot/foo/docs/../../other/double.c //depot/foo/usr/bin/../../../else/more/triple.c to: //depot/bar/single.c //depot/other/double.c //depot/else/more/triple.c And my script: begin paths = File.open(ARGV[0]).readlines puts(paths) new_paths = Array.new paths.each { |path| folders = path.split('/') if ( folders.include?('..') ) num_dotdots = 0 first_dotdot = folders.index('..') last_dotdot = folders.rindex('..') folders.each { |item| if ( item == '..' ) num_dotdots += 1 end } if ( first_dotdot and ( num_dotdots > 0 ) ) # this might be redundant? folders.slice!(first_dotdot - num_dotdots..last_dotdot) # dependent on consecutive dotdots only end end folders.map! { |elem| if ( elem !~ /\n/ ) elem = elem + '/' else elem = elem end } new_paths << folders.to_s } puts(new_paths) end

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  • building objects from xml file at runtime and intializing, in one pass?

    - by KaluSingh Gabbar
    I have to parse the XML file and build objects representation based on that, now once I get all these data I create entries in various database for these data objects. I have to do second pass over that for value as in the first pass all I could do is build the assets in various databases. and in second pass I get the values for all the data and put it in the database. I have a feeling that this can be done in a single pass but I just want to see what are your opinions. As I am just a student who started with professional work, experienced ppl please help. Can someone who have ideas or done similar work, please provide some light on the topic so that I can think over the possibility of the work and get the prototype going based on your suggestion. Thanks a lot for your precious time, I honestly appreciate it.

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  • BFS algorithm problem

    - by Gorkamorka
    The problem is as follows: A wanderer begins on the grid coordinates (x,y) and wants to reach the coordinates (0,0). From every gridpoint, the wanderer can go 8 steps north OR 3 steps south OR 5 steps east OR 6 steps west (8N/3S/5E/6W). How can I find the shortest route from (X,Y) to (0,0) using breadth-first search? Clarifications: Unlimited grid Negative coordinates are allowed A queue (linked list or array) must be used No obstacles present

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  • True random number generator

    - by goldenmean
    Sorry for this not being a "real" question, but Sometime back i remember seeing a post here about randomizing a randomizer randomly to generate truly random numbers, not just pseudo random. I dont see it if i search for it. Does anybody know about that article?

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  • Searching algorithmics: Parsing and processing a request

    - by James P.
    Say you were to create a search engine that can accept a query statement under the form of a String. The statement can be used to retrieve different types of objects with a given set of characteristics and possibly linked to other objects. In plain english or pseudo-code using an OOP approach, how would you go about parsing and processing statements as follows to get the series of desired objects ? get fruit with colour green get variety of apples, pears from Andy get strawberry with colour "deep red" and origin not Spain get total of sales of melons between 2010-10-10 and 2010-12-30 get last deliverydate of bananas from "Pete" and state not sold Hope the question is clear. If not I'll be more than happy to reformulate. P.S: This isn't homework ;)

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  • Java - Removing duplicates in an ArrayList

    - by Will
    I'm working on a program that uses an ArrayList to store Strings. The program prompts the user with a menu and allows the user to choose an operation to perform. Such operations are adding Strings to the List, printing the entries etc. What I want to be able to do is create a method called removeDuplicates().This method will search the ArrayList and remove any duplicated values. I want to leave one instance of the duplicated value(s) within the list. I also want this method to return the total number of duplicates removed. I've been trying to use nested loops to accomplish this but I've been running into trouble because when entries get deleted, the indexing of the ArrayList gets altered and things don't work as they should. I know conceptually what I need to do but I'm having trouble implementing this idea in code. Here is some pseudo code: start with first entry; check each subsequent entry in the list and see if it matches the first entry; remove each subsequent entry in the list that matches the first entry; after all entries have been examined, move on to the second entry; check each entry in the list and see if it matches the second entry; remove each entry in the list that matches the second entry; repeat for entry in the list Here's the code I have so far: public int removeDuplicates() { int duplicates = 0; for ( int i = 0; i < strings.size(); i++ ) { for ( int j = 0; j < strings.size(); j++ ) { if ( i == j ) { // i & j refer to same entry so do nothing } else if ( strings.get( j ).equals( strings.get( i ) ) ) { strings.remove( j ); duplicates++; } } } return duplicates; }

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  • Extracting a given number of the highest values in a List

    - by James P.
    I'm seeking to display a fixed number of items on a web page according to their respective weight (represented by an Integer). The List where these items are found can be of virtually any size. The first solution that comes to mind is to do a Collections.sort() and to get the items one by one by going through the List. Is there a more elegant solution though that could be used to prepare, say, the top eight items?

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  • How to compute palindrome from a stream of characters in sub-linear space/time?

    - by wrick
    I don't even know if a solution exists or not. Here is the problem in detail. You are a program that is accepting an infinitely long stream of characters (for simplicity you can assume characters are either 1 or 0). At any point, I can stop the stream (let's say after N characters were passed through) and ask you if the string received so far is a palindrome or not. How can you do this using less sub-linear space and/or time.

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  • K-Means Algorithm and java code

    - by Thandar
    Hi all, I need to calculate for grouping objects according to their size. I got k-means algorithms in java which calculate mostly for classifying according to their two or more features and the results are not satisfy for me.I only want to calculate for grouping objects based on one feature.Pseudocode or code would be helpful, too. Thanks u all for helping.

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  • Explanation needed for sum of prime below n numbers

    - by Bala Krishnan
    Today I solved a problem given in Project Euler its problem no 10 and it took 7 hrs for my python program to show the result. But in that forum itself a person named lassevk posted solution for this and it took only 4 sec. And its not possible for me to post this question in that forum because its not discussion forum. So, think about this if you want to mark this question as non-constructive. marked = [0] * 2000000 value = 3 s = 2 while value < 2000000: if marked[value] == 0: s += value i = value while i < 2000000: marked[i] = 1 i += value value += 2 print s If any one understand this code please explain it simple as possible. Link to the Problem 10 question.

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  • Facebook Hacker Cup: Power Overwhelming

    - by marcog
    A lot of people at Facebook like to play Starcraft II™. Some of them have made a custom game using the Starcraft II™ map editor. In this game, you play as the noble Protoss defending your adopted homeworld of Shakuras from a massive Zerg army. You must do as much damage to the Zerg as possible before getting overwhelmed. You can only build two types of units, shield generators and warriors. Shield generators do no damage, but your army survives for one second per shield generator that you build. Warriors do one damage every second. Your army is instantly overrun after your shield generators expire. How many shield generators and how many warriors should you build to inflict the maximum amount of damage on the Zerg before your army is overrun? Because the Protoss value bravery, if there is more than one solution you should return the one that uses the most warriors. Constraints 1 = G (cost for one shield generator) = 100 1 = W (cost for one warrior) = 100 G + W = M (available funds) = 1000000000000 (1012)

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