Search Results

Search found 26127 results on 1046 pages for 'google penguin algorithm'.

Page 328/1046 | < Previous Page | 324 325 326 327 328 329 330 331 332 333 334 335  | Next Page >

  • help implementing algorithm

    - by davit-datuashvili
    http://en.wikipedia.org/wiki/All_nearest_smaller_values this is site of the problem and here is my code but i have some trouble to implement it import java.util.*; public class stack{ public static void main(String[]args){ int x[]=new int[]{ 0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15 }; Stack<Integer> st=new Stack<Integer>(); for (int a:x){ while (!st.empty() && st.pop()>=a){ System.out.println( st.pop()); if (st.empty()){ break; } else{ st.push(a); } } } } } and here is pseudo code from site S = new empty stack data structure for x in the input sequence: while S is nonempty and the top element of S is greater than or equal to x: pop S if S is empty: x has no preceding smaller value else: the nearest smaller value to x is the top element of S push x onto S

    Read the article

  • Optimizing a Parking Lot Problem. What algorithims should I use to fit the most amount of cars in th

    - by Adam Gent
    What algorithms (brute force or not) would I use to put in as many cars (assume all cars are the same size) in a parking lot so that there is at least one exit (from the container) and a car cannot be blocked. Or can someone show me an example of this problem solved programmatically. The parking lot varies in shape would be nice but if you want to assume its some invariant shape that is fine. Another Edit: Assume that driving distance in the parking lot is not a factor (although it would be totally awesome if it was weighted factor to number of cars in lot). Another Edit: Assume 2 Dimensional (no cranes or driving over cars). Another Edit: You cannot move cars around once they are parked (its not a valet parking lot). I hope the question is specific enough now.

    Read the article

  • Is this linear search implementation actually useful?

    - by Helper Method
    In Matters Computational I found this interesting linear search implementation (it's actually my Java implementation ;-)): public static int linearSearch(int[] a, int key) { int high = a.length - 1; int tmp = a[high]; // put a sentinel at the end of the array a[high] = key; int i = 0; while (a[i] != key) { i++; } // restore original value a[high] = tmp; if (i == high && key != tmp) { return NOT_CONTAINED; } return i; } It basically uses a sentinel, which is the searched for value, so that you always find the value and don't have to check for array boundaries. The last element is stored in a temp variable, and then the sentinel is placed at the last position. When the value is found (remember, it is always found due to the sentinel), the original element is restored and the index is checked if it represents the last index and is unequal to the searched for value. If that's the case, -1 (NOT_CONTAINED) is returned, otherwise the index. While I found this implementation really clever, I wonder if it is actually useful. For small arrays, it seems to be always slower, and for large arrays it only seems to be faster when the value is not found. Any ideas?

    Read the article

  • Find a node in a Graph that minimizes the distance between two other nodes

    - by Andrés
    Here is the thing. I have a directed weighted graph G, with V vertices and E edges. Given two nodes in the graph, let's say A, and B, and given the weight of an edge A-B denoted as w(A, B), I need to find a node C so that max(w(A, C), w(B, C)) is minimal among all possibilities. By possibilities I mean all the values C can take. I don't know if it is completely clear, if it's not, I'll try to be more precise. Thanks in advance.

    Read the article

  • How to select number of lines from large text files?

    - by MiNdFrEaK
    I was wondering how to select number of lines from a certain text file. As an example: I have a text file containing the following lines: branch 27 : rect id 23400 rect: -115.475609 -115.474907 31.393650 31.411301 branch 28 : rect id 23398 rect: -115.474907 -115.472282 31.411301 31.417351 branch 29 : rect id 23396 rect: -115.472282 -115.468033 31.417351 31.427151 branch 30 : rect id 23394 rect: -115.468033 -115.458733 31.427151 31.438181 Non-Leaf Node: level=1 count=31 address=53 branch 0 : rect id 42 rect: -115.768539 -106.251556 31.425039 31.717550 branch 1 : rect id 50 rect: -109.559479 -106.009361 31.296721 31.775299 branch 2 : rect id 51 rect: -110.937401 -106.226143 31.285870 31.771971 branch 3 : rect id 54 rect: -109.584412 -106.069092 31.285240 31.775230 branch 4 : rect id 56 rect: -109.570961 -106.000954 31.296721 31.780769 branch 5 : rect id 58 rect: -115.806213 -106.366188 31.400450 31.687519 branch 6 : rect id 59 rect: -113.173859 -106.244057 31.297440 31.627750 branch 7 : rect id 60 rect: -115.811478 -106.278252 31.400450 31.679470 branch 8 : rect id 61 rect: -109.953888 -106.020111 31.325319 31.775270 branch 9 : rect id 64 rect: -113.070969 -106.015968 31.331841 31.704750 branch 10 : rect id 68 rect: -113.065689 -107.034576 31.326300 31.770809 branch 11 : rect id 71 rect: -112.333344 -106.059860 31.284081 31.662920 branch 12 : rect id 73 rect: -115.071083 -106.309677 31.267879 31.466850 branch 13 : rect id 74 rect: -116.094414 -106.286308 31.236290 31.424770 branch 14 : rect id 75 rect: -115.423264 -106.286308 31.229691 31.415510 branch 15 : rect id 76 rect: -116.111656 -106.313110 31.259390 31.478300 branch 16 : rect id 77 rect: -116.247467 -106.309677 31.240231 31.451799 branch 17 : rect id 78 rect: -116.170792 -106.094543 31.156429 31.391781 branch 18 : rect id 79 rect: -116.225723 -106.292709 31.239960 31.442850 branch 19 : rect id 80 rect: -116.268013 -105.769913 31.157240 31.378111 branch 20 : rect id 82 rect: -116.215424 -105.827202 31.198441 31.383421 branch 21 : rect id 83 rect: -116.095734 -105.826439 31.197460 31.373819 branch 22 : rect id 84 rect: -115.423264 -105.815018 31.182640 31.368891 branch 23 : rect id 85 rect: -116.221527 -105.776512 31.160931 31.389830 branch 24 : rect id 86 rect: -116.203369 -106.473831 31.168350 31.367611 branch 25 : rect id 87 rect: -115.727631 -106.501587 31.189100 31.395941 branch 26 : rect id 88 rect: -116.237289 -105.790756 31.164780 31.358959 branch 27 : rect id 89 rect: -115.791344 -105.990044 31.072620 31.349529 branch 28 : rect id 90 rect: -115.736847 -106.495079 31.187969 31.376900 branch 29 : rect id 91 rect: -115.721710 -106.000130 31.160351 31.354601 branch 30 : rect id 92 rect: -115.792236 -106.000793 31.166620 31.378811 Leaf Node: level=0 count=21 address=42 branch 0 : rect id 18312 rect: -106.412270 -106.401367 31.704750 31.717550 branch 1 : rect id 18288 rect: -106.278252 -106.253387 31.520321 31.548361 I just want those lines which are in between Non-Leaf Node level=1 to Leaf Node Level=0 and also there are a lot of segments like this and I need them all.

    Read the article

  • C++ priority queue structure userd ?

    - by John Retallack
    While searching for some functions in C++ STL documentation I read that push and pop for priority queues needs constant time. "Constant (in the priority_queue). Although notice that push_heap operates in logarithmic time." My question is what kind of data structure is used to mantain a priority queue with O(1) for push and pop ?

    Read the article

  • Looking for ideas on automatically arranging a set of objects (furniture) in a virtual room in AS3

    - by raf
    First of all, I don't want to visually arrange 3D models dragging them with the mouse, all I want is: Given a room of certain dimensions (L,W,H) and given a set of elements like beds, chairs, etc (with L,W,H dimensions, of course) I want to automatically arrange those elements to take advantage of the space as much as I can. So I want to be able to put as much furniture as I can in a given room. At the end I need to represent the arranged items visually, inside the room. My first thought was to use an array of items and sorting it with array.sortOn(["l","w","h"] Array.NUMERIC) and then define a gap between the objects and make the maths to put the objects one next to another, etc. but that isn't a good approach because some items may be placed on top of another ones (boxes of the same size, boxes on top of tables, etc). I really don't have experience on 3D programming, that's why I'm asking for help. Thanks in advance.

    Read the article

  • Find recipes that can be cooked from provided ingridients

    - by skaurus
    Sorry for bad English :( Suppose i can preliminary organize recipes and ingredients data in any way. How can i effectively conduct search of recipes by user-provided ingredients, preferably sorted by max match - so, first going recipes that use maximum of provided ingridients and do not contain any other ingrs, after them recipes that uses less of provided set and still not any other ingrs, after them recipes with minimum additional requirements and so on? All i can think about is represent recipe ingridients like bitmasks, and compare required bitmask with all recipes, but it is obviously a bad way to go. And related things like Levenstein distance i don't see how to use here. I believe it should be quite common task...

    Read the article

  • What is the optimum way to select the most dissimilar individuals from a population?

    - by Aaron D
    I have tried to use k-means clustering to select the most diverse markers in my population, for example, if we want to select 100 lines I cluster the whole population to 100 clusters then select the closest marker to the centroid from each cluster. The problem with my solution is it takes too much time (probably my function needs optimization), especially when the number of markers exceeds 100000. So, I will appreciate it so much if anyone can show me a new way to select markers that maximize diversity in my population and/or help me optimize my function to make it work faster. Thank you # example: library(BLR) data(wheat) dim(X) mdf<-mostdiff(t(X), 100,1,nstart=1000) Here is the mostdiff function that i used: mostdiff <- function(markers, nClust, nMrkPerClust, nstart=1000) { transposedMarkers <- as.array(markers) mrkClust <- kmeans(transposedMarkers, nClust, nstart=nstart) save(mrkClust, file="markerCluster.Rdata") # within clusters, pick the markers that are closest to the cluster centroid # turn the vector of which markers belong to which clusters into a list nClust long # each element of the list is a vector of the markers in that cluster clustersToList <- function(nClust, clusters) { vecOfCluster <- function(whichClust, clusters) { return(which(whichClust == clusters)) } return(apply(as.array(1:nClust), 1, vecOfCluster, clusters)) } pickCloseToCenter <- function(vecOfCluster, whichClust, transposedMarkers, centers, pickHowMany) { clustSize <- length(vecOfCluster) # if there are fewer than three markers, the center is equally distant from all so don't bother if (clustSize < 3) return(vecOfCluster[1:min(pickHowMany, clustSize)]) # figure out the distance (squared) between each marker in the cluster and the cluster center distToCenter <- function(marker, center){ diff <- center - marker return(sum(diff*diff)) } dists <- apply(transposedMarkers[vecOfCluster,], 1, distToCenter, center=centers[whichClust,]) return(vecOfCluster[order(dists)[1:min(pickHowMany, clustSize)]]) } }

    Read the article

  • Formula for popularity? (based on "like it", "comments", "views")

    - by paullb
    I have some pages on a website and I have to create an ordering based on "popularity"/"activity" The parameters that I have to use are: views to the page comments made on the page (there is a form at the bottom where uses can make comments) clicks made to the "like it" icon Are there any standards for what a formula for popularity would be? (if not opinions are good too) (initially I thought of views + 10*comments + 10*likeit)

    Read the article

  • LinkedList.contains execution speed

    - by Le_Coeur
    Why Methode LinkedList.contains() runs quickly than such implementation: for (String s : list) if (s.equals(element)) return true; return false; I don't see great difference between this to implementations(i consider that search objects aren't nulls), same iterator and equals operation

    Read the article

  • A quick way to map unordered list of longs to buffer location ?

    - by alhazen
    I have a large number of points (indexed by long) that are processed by multiple threads and I'm using a buffer to hold the output results in order. As the number of points processed is huge, what would be an efficient way to map the indexes of the points to the corresponding ordered position in the buffer ? Example: long bufferIndex bufferIndex index (if BufferSize = 2) (if BufferSize = 4) ---------------------------------------------- 2938 0 0 2939 1 1 2941 1 3 2940 0 2 Thanks.

    Read the article

  • Big-O for Eight Year Olds?

    - by Jason Baker
    I'm asking more about what this means to my code. I understand the concepts mathematically, I just have a hard time wrapping my head around what they mean conceptually. For example, if one were to perform an O(1) operation on a data structure, I understand that the amount of operations it has to perform won't grow because there are more items. And an O(n) operation would mean that you would perform a set of operations on each element. Could somebody fill in the blanks here? Like what exactly would an O(n^2) operation do? And what the heck does it mean if an operation is O(n log(n))? And does somebody have to smoke crack to write an O(x!)?

    Read the article

  • Given a number N, find the number of ways to write it as a sum of two or more consecutive integers

    - by hilal
    Here is the problem (Given a number N, find the number of ways to write it as a sum of two or more consecutive integers) and example 15 = 7+8, 1+2+3+4+5, 4+5+6 I solved with math like that : a + (a + 1) + (a + 2) + (a + 3) + ... + (a + k) = N (k + 1)*a + (1 + 2 + 3 + ... + k) = N (k + 1)a + k(k+1)/2 = N (k + 1)*(2*a + k)/2 = N Then check that if N divisible by (k+1) and (2*a+k) then I can find answer in O(N) time Here is my question how can you solve this by dynamic-programming ? and what is the complexity (O) ? P.S : excuse me, if it is a duplicate question. I searched but I can find

    Read the article

  • Faster way to compare two sets of points in N-dimensional space?

    - by Amit
    List1 contains a high number (~7^10) of N-dimensional points (N <=10), List2 contains the same or fewer number of N-dimensional points (N <=10). My task is this: I want to check which point in List2 is closest (euclidean distance) to a point in List1 for every point in List1 and subsequently perform some operation on it. I have been doing it the simple- the nested loop way when I didn't have more than 50 points in List1, but with 7^10 points, this obviously takes up a lot of time. What is the fastest way to do this? Any concepts from Computational Geometry might help?

    Read the article

  • Sort string based upon the count of characters Options

    - by prp
    Sample Data : input : "abcdacdc" Output : "cadb" here we have to sort strings in order of count of characters. If the count is same for characters. maintain the original order of the characters from input string. my approach: i have used array of 26 for maintaining occurrence of all characters and sort it then print it.But while doing so i am not able to maintain order in case if two characters have same count. please suggest any improvement or any other algo.

    Read the article

  • Setting last N bits in an array

    - by Martin
    I'm sure this is fairly simple, however I have a major mental block on it, so I need a little help here! I have an array of 5 integers, the array is already filled with some data. I want to set the last N bits of the array to be random noise. [int][int][int][int][int] set last 40 bits [unchanged][unchanged][unchanged][24 bits of old data followed 8 bits of randomness][all random] This is largely language agnostic, but I'm working in C# so bonus points for answers in C#

    Read the article

  • How to detect if a certain range resides (partly) within an other range?

    - by Tom
    Lets say I've got two squares and I know their positions, a red and blue square: redTopX; redTopY; redBotX; redBotY; blueTopX; blueTopY; blueBotX; blueBotY; Now, I want to check if square blue resides (partly) within (or around) square red. This can happen in a lot of situations, as you can see in this image I created to illustrate my situation better: Note that there's always only one blue and one red square, I just added multiple so I didn't have to redraw 18 times. My original logic was simple, I'd check all corners of square blue and see if any of them are inside square red: if ( ((redTopX >= blueTopX) && (redTopY >= blueTopY) && (redTopX <= blueBotX) && (redTopY <= blueBotY)) || //top left ((redBotX >= blueTopX) && (redTopY >= blueTopY) && (redBotX <= blueBotX) && (redTopY <= blueBotY)) || //top right ((redTopX >= blueTopX) && (redBotY >= blueTopY) && (redTopX <= blueBotX) && (redBotY <= blueBotY)) || //bottom left ((redBotX >= blueTopX) && (redBotY >= blueTopY) && (redBotX <= blueBotX) && (redBotY <= blueBotY)) //bottom right ) { //blue resides in red } Unfortunately, there are a couple of flaws in this logic. For example, what if red surrounds blue (like in situation 1)? I thought this would be pretty easy but am having trouble coming up with a good way of covering all these situations.. can anyone help me out here? Regards, Tom

    Read the article

< Previous Page | 324 325 326 327 328 329 330 331 332 333 334 335  | Next Page >