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  • Sort vector<int>(n) in O(n) time using O(m) space?

    - by Adam
    I have a vector<unsigned int> vec of size n. Each element in vec is in the range [0, m], no duplicates, and I want to sort vec. Is it possible to do better than O(n log n) time if you're allowed to use O(m) space? In the average case m is much larger than n, in the worst case m == n. Ideally I want something O(n). I get the feeling that there's a bucket sort-ish way to do this: unsigned int aux[m]; aux[vec[i]] = i; Somehow extract the permutation and permute vec. I'm stuck on how to do 3. In my application m is on the order of 16k. However this sort is in the inner loops and accounts for a significant portion of my runtime.

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  • question about Tetration

    - by davit-datuashvili
    i have question how write program which calculates following procedures http://en.wikipedia.org/wiki/Tetration i have exponential program which returns x^n here is code public class Exp{ public static long exp(long x,long n){ long t=0; if (n==0){ t= 1; } else{ if (n %2==0){ t= exp(x,n/2)* exp(x,n/2); } else{ t= x*exp(x,n-1); } } return t; } public static void main(String[]args){ long x=5L; long n=4L; System.out.println(exp(x,n)); } } but how use it in Tetration program?please help

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  • Multiply without multiplication, division and bitwise operators, and no loops. Recursion

    - by lxx22
    public class MultiplyViaRecursion{ public static void main(String[] args){ System.out.println("8 * 9 == " + multiply(8, 9)); System.out.println("6 * 0 == " + multiply(6, 0)); System.out.println("0 * 6 == " + multiply(0, 6)); System.out.println("7 * -6 == " + multiply(7, -6)); } public static int multiply(int x, int y){ int result = 0; if(y > 0) return result = (x + multiply(x, (y-1))); if(y == 0) return result; if(y < 0) return result = -multiply(x, -y); return result; } } My question is very simple and basic, why after each "if" the "return" still cannot pass the compilation, error shows missing return.

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  • List of values as keys for a Map

    - by thr
    I have lists of variable length where each item can be one of four unique, that I need to use as keys for another object in a map. Assume that each value can be either 0, 1, 2 or 3 (it's not integer in my real code, but a lot easier to explain this way) so a few examples of key lists could be: [1, 0, 2, 3] [3, 2, 1] [1, 0, 0, 1, 1, 3] [2, 3, 1, 1, 2] [1, 2] So, to re-iterate: each item in the list can be either 0, 1, 2 or 3 and there can be any number of items in a list. My first approach was to try to hash the contents of the array, using the built in GetHashCode() in .NET to combine the hash of each element. But since this would return an int I would have to deal with collisions manually (two equal int values are identical to a Dictionary). So my second approach was to use a quad tree, breaking down each item in the list into a Node that has four pointers (one for each possible value) to the next four possible values (with the root node representing [], an empty list), inserting [1, 0, 2] => Foo, [1, 3] => Bar and [1, 0] => Baz into this tree would look like this: Grey nodes nodes being unused pointers/nodes. Though I worry about the performance of this setup, but there will be no need to deal with hash collisions and the tree won't become to deep (there will mostly be lists with 2-6 items stored, rarely over 6). Is there some other magic way to store items with lists of values as keys that I have missed?

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  • Why is i-- faster than i++ in loops? [closed]

    - by Afshin Mehrabani
    Possible Duplicate: JavaScript - Are loops really faster in reverse…? I don't know if this question is valid in other languages or not, but I'm asking this specifically for JavaScript. I see in some articles and questions that the fastest loop in JavaScript is something like: for(var i = array.length; i--; ) Also in Sublime Text 2, when you try to write a loop, it suggests: for (var i = Things.length - 1; i >= 0; i--) { Things[i] }; I want to know, why is i-- faster than i++ in loops?

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  • algorithms undirected graph twodegree[]

    - by notamathwiz
    For each node u in an undirected graph, let twodegree[u] be the sum of the degrees of u's neighbors. Show how to compute the entire array of twodegree[.] values in linear time, given a graph in adjacency list format. This is the solution for all u ? V : degree[u] = 0 for all (u; w) ? E: degree[u] = degree[u] + 1 for all u ? V : twodegree[u] = 0 for all (u; w) ? E: twodegree[u] = twodegree[u] + degree[w] can someone explain what degree[u] does in this case and how twodegree[u] = twodegree[u] + degree[w] is supposed to be the sum of the degrees of u's neighbors?

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  • Recursive solution to finding patterns

    - by user2997162
    I was solving a problem on recursion which is to count the total number of consecutive 8's in a number. For example: input: 8801 output: 2 input: 801 output: 0 input: 888 output: 3 input: 88088018 output:4 I am unable to figure out the logic of passing the information to the next recursive call about whether the previous digit was an 8. I do not want the code but I need help with the logic. For an iterative solution, I could have used a flag variable, but in recursion how do I do the work which flag variable does in an iterative solution. Also, it is not a part of any assignment. This just came to my mind because I am trying to practice coding using recursion.

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  • Enumerate all k-partitions of 1d array with N elements?

    - by user301217
    This seems like a simple request, but google is not my friend because "partition" scores a bunch of hits in database and filesystem space. I need to enumerate all partitions of an array of N values (N is constant) into k sub-arrays. The sub-arrays are just that - a starting index and ending index. The overall order of the original array will be preserved. For example, with N=4 and k=2: [ | a b c d ] (0, 4) [ a | b c d ] (1, 3) [ a b | c d ] (2, 2) [ a b c | d ] (3, 1) [ a b c d | ] (4, 0) I'm pretty sure this isn't an original problem (and no, it's not homework), but I'd like to do it for every k <= N, and it'd be great if the later passes (as k grows) took advantage of earlier results. If you've got a link, please share.

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  • Number of the different elements in an array.

    - by AB
    Is it possible to compute the number of the different elements in an array in linear time and constant space? Let us say it's an array of long integers, and you can not allocate an array of length sizeof(long). P.S. Not homework, just curious. I've got a book that sort of implies that it is possible.

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  • Calculating color shades

    - by matejv
    I have the next problem. I have a base color with couple of different shades of that color. Example: Base color: #4085c5 Shade: #005cb1 Now, I have a different color (let's say #d60620), but no shades of it. From the color I would like to calculate shades, that have similar difference as colors mentioned in first paragraph. First I tried calculating difference of RGB elements and applying them to second color, but the result was not like I expected to be. Than I tried with converting color to HSV, reading saturation value and applying the difference to second color, but again the resulting color was still weird. The formula was something like: (HSV(BaseColor)[S] - HSV(Shade)[S]) + HSV(SecondColor)[H] Does anyone know how this problem could be solved? I know I am doing something wrong, but I don't know what. :)

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  • Find largest rectangle containing only zeros in an N&times;N binary matrix

    - by Rajendra
    Given an NxN binary matrix (containing only 0's or 1's), how can we go about finding largest rectangle containing all 0's? Example: I 0 0 0 0 1 0 0 0 1 0 0 1 II->0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 <--IV 0 0 1 0 0 0 IV is a 6×6 binary matrix; the return value in this case will be Cell 1: (2, 1) and Cell 2: (4, 4). The resulting sub-matrix can be square or rectangular. The return value can also be the size of the largest sub-matrix of all 0's, in this example 3 × 4.

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  • Reversing a hash function

    - by martani_net
    Hi, I have the following hash function, and I'm trying to get my way to reverse it, so that I can find the key from a hashed value. uint Hash(string s) { uint result = 0; for (int i = 0; i < s.Length; i++) { result = ((result << 5) + result) + s[i]; } return result; } The code is in C# but I assume it is clear. I am aware that for one hashed value, there can be more than one key, but my intent is not to find them all, just one that satisfies the hash function suffices. Any ideas? Thank you.

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  • Minimum range of 3 sets

    - by user343882
    We have three sets S1, S2, S3. I need to find x,y,z such that x E S1 y E S2 z E S3 let min denote the minimum value out of x,y,z let max denote the maximum value out of x,y,z The range denoted by max-min should be the MINIMUM possible value

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  • Need to sync two lists with atrribute time, but times aren't equal

    - by virgula24
    I gonna try to describe my problem the best i can. I have two lists, one with audio frames and other with color frames (not relevant). Both of them have timestamps, they were captured at the same moment but at different instants. So, i have like this: index COLOR AUDIO 0 841 846 1 873 897 2 905 948 3 940 1000 the frames start at high numbers because they were captured and then trimmed to specific parts, but im shot, frame 0 is synced with only 5ms apart(timestamp in ms). On every case i have, the audio frames count is less than the color count. I need to make them have the same count. The stating frames may be coloraudio, color

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  • Metamorphic generator

    - by user222094
    I am trying to find references about different designs of metamorphic generators can someone point me to the right direction. I have gone through some papers in ACM but couldn't find what I am looking for.

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  • Python: Determine whether list of lists contains a defined sequence

    - by duhaime
    I have a list of sublists, and I want to see if any of the integer values from the first sublist plus one are contained in the second sublist. For all such values, I want to see if that value plus one is contained in the third sublist, and so on, proceeding in this fashion across all sublists. If there is a way of proceeding in this fashion from the first sublist to the last sublist, I wish to return True; otherwise I wish to return False. In other words, for each value in sublist one, for each "step" in a "walk" across all sublists read left to right, if that value + n (where n = number of steps taken) is contained in the current sublist, the function should return True; otherwise it should return False. (Sorry for the clumsy phrasing--I'm not sure how to clean up my language without using many more words.) Here's what I wrote. a = [ [1,3],[2,4],[3,5],[6],[7] ] def find_list_traversing_walk(l): for i in l[0]: index_position = 0 first_pass = 1 walking_current_path = 1 while walking_current_path == 1: if first_pass == 1: first_pass = 0 walking_value = i if walking_value+1 in l[index_position + 1]: index_position += 1 walking_value += 1 if index_position+1 == len(l): print "There is a walk across the sublists for initial value ", walking_value - index_position return True else: walking_current_path = 0 return False print find_list_traversing_walk(a) My question is: Have I overlooked something simple here, or will this function return True for all true positives and False for all true negatives? Are there easier ways to accomplish the intended task? I would be grateful for any feedback others can offer!

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  • How to determine if two strings are sufficiently close?

    - by A.06
    We say that we can "hop" from the word w1 to the word w2 if they are "sufficiently close". We define w2 to be sufficiently close to w1 if one of the following holds: w2 is obtained from w1 by deleting one letter. w2 is obtained from w1 by replacing one of the letters in w1 by some letter that appears to its right in w1 and which is also to its right in alphabetical order. I have no idea how to check if 2. is fulfilled. To check if 1. is possible this is my function: bool check1(string w1, string w2){ if(w2.length - w1.length != 1){ return false; } for(int i = 0,int j = 0;i < w2.length;i++;j++){ if(w2[i] == w1[j]){//do nothing } else if(i == j){ j++; } else{ return false; } } return true; } Given two words w1 and w2, how do we check if we can 'hop' from w1 to w2?

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