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  • Given a few strings, how many strings can be lexicographically least by modifying the alphabet?

    - by Jackson W
    Number of strings can be huge as in 30000. Given N strings, output which ones can be lexicographically least after modifying the english alphabet. e.g. acdbe...... for example if the strings were: omm moo mom ommnom "mom" is already lexicographically least with the original english alphabet. we can make the word "omm" least by switching "m" and "o" in the alphabet ("abcdefghijklonmpqrstuvwxyz"). the other ones you cant make lexicographically last, no matter what you do. any fast way to do this? I have no ways to approach this except try every single possible alphabet

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  • generic binary Search in c#

    - by Pro_Zeck
    Below is my Generic Binary Search it works ok with the intgers type array it finds all the elements in it . But the Problem Arises when i use a string array to find any string data. It runs ok for the first index and last index elements but i cant find the middle elements. Stringarray = new string[] { "b", "a", "ab", "abc", "c" }; public static void BinarySearch<T>(T[] array, T searchFor, Comparer<T> comparer) { int high, low, mid; high = array.Length - 1; low = 0; if (array[0].Equals(searchFor)) Console.WriteLine("Value {0} Found At Index {1}",array[0],0); else if (array[high].Equals(searchFor)) Console.WriteLine("Value {0} Found At Index {1}", array[high], high); else { while (low <= high) { mid = (high + low) / 2; if (comparer.Compare(array[mid], searchFor) == 0) { Console.WriteLine("Value {0} Found At Index {1}", array[mid], mid); break; } else { if (comparer.Compare(searchFor, array[mid]) > 0) high = mid + 1; else low = mid + 1; } } if (low > high) { Console.WriteLine("Value Not Found In the Collection"); } } }

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  • Another Nim's Game Variant

    - by Terry Smith
    Given N binary sequence Example : given one sequence 101001 means player 0 can only choose a position with 0 element and cut the sequence from that position {1,101,1010} player 1 can only choose a position with 1 element ans cut the sequence from that position {null,10,101000} now player 0 and player 1 take turn cutting the sequence, on each turn they can cut any one non-null sequence, if a player k can't make a move because there's no more k element on any sequence, he lose. Assume both player play optimally, who will win ? I tried to solve this problem with grundy but i'm unable to reduce the sequence to a grundy number because it both player don't have the same option to move. Can anyone give me a hint to solve this problem ?

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  • Generating different randoms valid for a day on different independent devices?

    - by Pentium10
    Let me describe the system. There are several mobile devices, each independent from each other, and they are generating content for the same record id. I want to avoid generating the same content for the same record on different devices, for this I though I would use a random and make it so too cluster the content pool based on these randoms. Suppose you have choices from 1 to 100. Day 1 Device#1 will choose for the record#33 between 1-10 Device#2 will choose for the record#33 between 40-50 Device#3 will choose for the record#33 between 50-60 Device#1 will choose for the record#55 between 40-50 Device#2 will choose for the record#55 between 1-10 Device#3 will choose for the record#55 between 10-20 Device#1 will choose for the record#11 between 1-10 Device#2 will choose for the record#22 between 1-10 Device#3 will choose for the record#99 between 1-10 Day 2 Device#1 will choose for the record#33 between 90-100 Device#2 will choose for the record#33 between 1-10 Device#3 will choose for the record#33 between 50-60 They don't have access to a central server. Data available for each of them: IMEI (unique per mobile) Date of today (same on all devices) Record id (same on all devices) What do you think, how is it possible? ps. tags can be edited

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  • An O(1) Sort ~~~

    - by FlySwat
    Before you stone me for being a heretic, There is a sort that proclaims to be O(1), called "Bead Sort" (http://en.wikipedia.org/wiki/Bead_sort) , however that is pure theory, when actually applied I found that it was actually O(N * M), which is pretty pathetic. That said, Lets list out some of the fastest sorts, and their best case and worst case speed in Big O notation. ~~ FlySwat ~~

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  • What Software Engineering Areas should be stressed upon while Interviewing Candidate for Fulltime So

    - by Rachel
    Hi, This question is somewhat related to other posts which I found on Stackoverflow but not exactly and so am prompted to ask about it. I know we must ask for Data-Structures and Algorithms but what specific data-structures or Algorithms or other CS Concepts should be asked while interviewing Sr. Software Engineering Fulltime Position as compared with Software Engineering Position. Thanks.

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  • Tracking/Counting Word Frequency

    - by Joel Martinez
    I'd like to get some community consensus on a good design to be able to store and query word frequency counts. I'm building an application in which I have to parse text inputs and store how many times a word has appeared (over time). So given the following inputs: "To Kill a Mocking Bird" "Mocking a piano player" Would store the following values: Word Count ------------- To 1 Kill 1 A 2 Mocking 2 Bird 1 Piano 1 Player 1 And later be able to quickly query for the count value of a given arbitrary word. My current plan is to simply store the words and counts in a database, and rely on caching word count values ... But I suspect that I won't get enough cache hits to make this a viable solution long term. Can anyone suggest algorithms, or data structures, or any other idea that might make this a well-performing solution?

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  • Can my tortoise vs. hare race be improved?

    - by FredOverflow
    Here is my code for detecting cycles in a linked list: do { hare = hare.next(); if (hare == back) return; hare = hare.next(); if (hare == back) return; tortoise = tortoise.next(); } while (tortoise != hare); throw new AssertionError("cyclic linkage"); Is there a way to get rid of the code duplication inside the loop? Am I right in assuming that I don't need a check after making the tortoise take a step forward? As I see it, the tortoise can never reach the end of the list before the hare (contrary to the fable). Any other ways to simplify/beautify this code?

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  • 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

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  • Picking apples off a tree

    - by John Retallack
    I have the following problem: I am given a tree with N apples, for each apple I am given it's weight and height. I can pick apples up to a given height H, each time I pick an apple the height of every apple is increased with U. I have to find out the maximum weight of apples I can pick. 1 = N = 100000 0 < {H, U, apples' weight and height, maximum weight} < 231 Example: N=4 H=100 U=10 height weight 82 30 91 10 93 5 94 15 The answer is 45: first pick the apple with the weight of 15 then the one with the weight of 30. Could someone help me approach this problem?

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  • 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?

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  • 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

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  • 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)

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  • 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

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  • 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...

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  • Using c#,c/c++ or java to improve BBN with GA

    - by madicemickael
    I have a little problem in my little project , I wish that someone here could help me! I am planning to use a bayesian network as a decision factor in my game AI and I want to improve the decision making every step of the way , anyone knows how to do that ? Any tutorials / existing implementations will be very good,I hope some of you could help me. I heard that a programmer in this community did a good implementation of this put together for poker game AI.I am planning to use it like him ,but in another poker(Texas) or maybe Rentz. Looking for C/c++ or c# or java code. Thanks , Mike

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  • 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!)?

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  • 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.

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