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  • How to find the number of inversions in an array ?

    - by Michael
    This is an phone interview question: "Find the number of inversions in an array". I guess they mean O(N*log N) solution since O(N^2) is trivial. I guess it cannot be better than O(N*log N) since sorting is O(N*log N) I have checked a similar question from SO and can summarize the answers as follows: Calculate half the distance the elements should be moved to sort the array : copy the array and sort the copy. For each element of the original array a[i] find it's position j in the sorted copy (binary search) and sum abs(i - j)/2. Modify merge sort : modify merge to count inversions between two sorted arrays (it takes O(N)) and run merge sort with the modified merge. Does it make sense ? Are there other (maybe simpler) solution ? Isn't it too hard for a phone interview ?

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  • Number distribution

    - by Carra
    Problem: We have x checkboxes and we want to check y of them evenly. Example 1: select 50 checkboxes of 100 total. [-] [x] [-] [x] ... Example 2: select 33 checkboxes of 100 total. [-] [-] [x] [-] [-] [x] ... Example 3: select 66 checkboxes of 100 total: [-] [x] [x] [-] [x] [x] ... But we're having trouble to come up with a formula to check them in code, especially once you go 11/111 or something similar. Anyone has an idea?

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  • help implementing All Nearest Smaller Values 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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  • What are some practical uses of generating all permutations of a list, such as ['a', 'b', 'c'] ?

    - by Jian Lin
    I was asked by somebody in an interview for web front end job, to write a function that generates all permutation of a string, such as "abc" (or consider it ['a', 'b', 'c']). so the expected result from the function, when given ['a', 'b', 'c'], is abc acb bac bca cab cba Actually in my past 20 years of career, I have never needed to do something like that, especially when doing front end work for web programming. What are some practical use of this problem nowadays, in web programming, front end or back end, I wonder? As a side note, I kind of feel that expecting a result in 3 minutes might be "either he gets it or he doesn't", especially I was thinking of doing it by a procedural, non-recursive way at first. After the interview, I spent another 10 minutes and thought of how to do it using recursion, but expecting it to be solved within 3 minutes... may not be a good test of how qualified he is, especially for front end work.

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  • How do I recursively define a Hash in Ruby from supplied arguments?

    - by Sarah Beckham
    This snippet of code populates an @options hash. values is an Array which contains zero or more heterogeneous items. If you invoke populate with arguments that are Hash entries, it uses the value you specify for each entry to assume a default value. def populate(*args) args.each do |a| values = nil if (a.kind_of? Hash) # Converts {:k => "v"} to `a = :k, values = "v"` a, values = a.to_a.first end @options[:"#{a}"] ||= values ||= {} end end What I'd like to do is change populate such that it recursively populates @options. There is a special case: if the values it's about to populate a key with are an Array consisting entirely of (1) Symbols or (2) Hashes whose keys are Symbols (or some combination of the two), then they should be treated as subkeys rather than the values associated with that key, and the same logic used to evaluate the original populate arguments should be recursively re-applied. That was a little hard to put into words, so I've written some test cases. Here are some test cases and the expected value of @options afterwards: populate :a => @options is {:a => {}} populate :a => 42 => @options is {:a => 42} populate :a, :b, :c => @options is {:a => {}, :b => {}, :c => {}} populate :a, :b => "apples", :c => @options is {:a => {}, :b => "apples", :c => {}} populate :a => :b => @options is {:a => :b} # Because [:b] is an Array consisting entirely of Symbols or # Hashes whose keys are Symbols, we assume that :b is a subkey # of @options[:a], rather than the value for @options[:a]. populate :a => [:b] => @options is {:a => {:b => {}}} populate :a => [:b, :c => :d] => @options is {:a => {:b => {}, :c => :d}} populate :a => [:a, :b, :c] => @options is {:a => {:a => {}, :b => {}, :c => {}}} populate :a => [:a, :b, "c"] => @options is {:a => [:a, :b, "c"]} populate :a => [:one], :b => [:two, :three => "four"] => @options is {:a => :one, :b => {:two => {}, :three => "four"}} populate :a => [:one], :b => [:two => {:four => :five}, :three => "four"] => @options is {:a => :one, :b => { :two => { :four => :five } }, :three => "four" } } It is acceptable if the signature of populate needs to change to accommodate some kind of recursive version. There is no limit to the amount of nesting that could theoretically happen. Any thoughts on how I might pull this off?

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  • sloving Algorithm notation

    - by neednewname
    Use big-O notation to classify the traditional grade school algorithms for addition and multiplication. That is, if asked to add two numbers each having N digits, how many individual additions must be performed? If asked to multiply two N-digit numbers, how many individual multiplications are required Suppose f is a function that returns the result of reversing the string of symbols given as its input, and g is a function that returns the concatenation of the two strings given as its input. If x is the string hrwa, what is returned by g(f(x),x)? Explain your answer - don't just provide the result!

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  • Counting bits set in a .Net BitArray Class

    - by Sam
    I am implementing a library where I am extensively using the .Net BitArray class and need an equivalent to the Java BitSet.Cardinality() method, i.e. a method which returns the number of bits set. I was thinking of implementing it as an extension method for the BitArray class. The trivial implementation is to iterate and count the bits set (like below), but I wanted a faster implementation as I would be performing thousands of set operations and counting the answer. Is there a faster way than the example below? count = 0; for (int i = 0; i < mybitarray.Length; i++) { if (mybitarray [i]) count++; }

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  • DFS Backtracking with java

    - by Cláudio Ribeiro
    I'm having problems with DFS backtracking in an adjacency matrix. Here's my code: (i added the test to the main in case someone wants to test it) public class Graph { private int numVertex; private int numEdges; private boolean[][] adj; public Graph(int numVertex, int numEdges) { this.numVertex = numVertex; this.numEdges = numEdges; this.adj = new boolean[numVertex][numVertex]; } public void addEdge(int start, int end){ adj[start-1][end-1] = true; adj[end-1][start-1] = true; } List<Integer> visited = new ArrayList<Integer>(); public Integer DFS(Graph G, int startVertex){ int i=0; if(pilha.isEmpty()) pilha.push(startVertex); for(i=1; i<G.numVertex; i++){ pilha.push(i); if(G.adj[i-1][startVertex-1] != false){ G.adj[i-1][startVertex-1] = false; G.adj[startVertex-1][i-1] = false; DFS(G,i); break; }else{ visited.add(pilha.pop()); } System.out.println("Stack: " + pilha); } return -1; } Stack<Integer> pilha = new Stack(); public static void main(String[] args) { Graph g = new Graph(6, 9); g.addEdge(1, 2); g.addEdge(1, 5); g.addEdge(2, 4); g.addEdge(2, 5); g.addEdge(2, 6); g.addEdge(3, 4); g.addEdge(3, 5); g.addEdge(4, 5); g.addEdge(6, 4); g.DFS(g, 1); } } I'm trying to solve the euler path problem. the program solves basic graphs but when it needs to backtrack, it just does not do it. I think the problem might be in the stack manipulations or in the recursive dfs call. I've tried a lot of things, but still can't seem to figure out why it does not backtrack. Can somebody help me ?

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  • How to perform a Depth First Search iteratively using async/parallel processing?

    - by Prabhu
    Here is a method that does a DFS search and returns a list of all items given a top level item id. How could I modify this to take advantage of parallel processing? Currently, the call to get the sub items is made one by one for each item in the stack. It would be nice if I could get the sub items for multiple items in the stack at the same time, and populate my return list faster. How could I do this (either using async/await or TPL, or anything else) in a thread safe manner? private async Task<IList<Item>> GetItemsAsync(string topItemId) { var items = new List<Item>(); var topItem = await GetItemAsync(topItemId); Stack<Item> stack = new Stack<Item>(); stack.Push(topItem); while (stack.Count > 0) { var item = stack.Pop(); items.Add(item); var subItems = await GetSubItemsAsync(item.SubId); foreach (var subItem in subItems) { stack.Push(subItem); } } return items; } EDIT: I was thinking of something along these lines, but it's not coming together: var tasks = stack.Select(async item => { items.Add(item); var subItems = await GetSubItemsAsync(item.SubId); foreach (var subItem in subItems) { stack.Push(subItem); } }).ToList(); if (tasks.Any()) await Task.WhenAll(tasks); UPDATE: If I wanted to chunk the tasks, would something like this work? foreach (var batch in items.BatchesOf(100)) { var tasks = batch.Select(async item => { await DoSomething(item); }).ToList(); if (tasks.Any()) { await Task.WhenAll(tasks); } } The language I'm using is C#.

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  • Efficient mapping of game entity positions in Java

    - by byte
    In Java (Swing), say I've got a 2D game where I have various types of entities on the screen, such as a player, bad guys, powerups, etc. When the player moves across the screen, in order to do efficient checking of what is in the immediate vicinity of the player, I would think I'd want indexed access to the things that are near the character based on their position. For example, if player 'P' steps onto element 'E' in the following example... | | | | | | | | | |P| | | | |E| | | | | | | | | ... would be to do something like: if(player.getPosition().x == entity.getPosition().x && entity.getPosition.y == thing.getPosition().y) { //do something } And thats fine, but that implies that the entities hold their positions, and therefor if I had MANY entities on the screen I would have to loop through all possible entities available and check each ones position against the player position. This seems really inefficient especially if you start getting tons of entities. So, I would suspect I'd want some sort of map like Map<Point, Entity> map = new HashMap<Point, Entity>(); And store my point information there, so that I could access these entities in constant time. The only problem with that approach is that, if I want to move an entity to a different point on the screen, I'd have to search through the values of the HashMap for the entity I want to move (inefficient since I dont know its Point position ahead of time), and then once I've found it remove it from the HashMap, and re-insert it with the new position information. Any suggestions or advice on what sort of data structure / storage format I ought to be using here in order to have efficient access to Entities based on their position, as well as Position's based on the Entity?

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  • Elegant Method of Inserting Code Between Loops

    - by DeathMagus
    In web development, I often find I need to format and print various arrays of data, and separate these blocks of data in some manner. In other words, I need to be able to insert code between each loop, without said code being inserted before the first entry or after the last one. The most elegant way I've found to accomplish this is as follows: function echoWithBreaks($array){ for($i=0; $i<count($array); $i++){ //Echo an item if($i<count($array)-1){ //Echo "between code" } } } Unfortunately, there's no way that I can see to implement this solution with foreach instead of for. Does anyone know of a more elegant solution that will work with foreach?

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  • Creating objects makes the VM faster?

    - by Sudhir Jonathan
    Look at this piece of code: MessageParser parser = new MessageParser(); for (int i = 0; i < 10000; i++) { parser.parse(plainMessage, user); } For some reason, it runs SLOWER (by about 100ms) than for (int i = 0; i < 10000; i++) { MessageParser parser = new MessageParser(); parser.parse(plainMessage, user); } Any ideas why? The tests were repeated a lot of times, so it wasn't just random. How could creating an object 10000 times be faster than creating it once?

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  • Porting Python algorithm to C++ - different solution

    - by cb0
    Hello, I have written a little brute string generation script in python to generate all possible combinations of an alphabet within a given length. It works quite nice, but for the reason I wan't it to be faster I try to port it to C++. The problem is that my C++ Code is creating far too much combination for one word. Heres my example in python: ./test.py gives me aaa aab aac aad aa aba .... while ./test (the c++ programm gives me) aaa aaa aaa aaa aa Here I also get all possible combinations, but I get them twice ore more often. Here is the Code for both programms: #!/usr/bin/env python import sys #Brute String Generator #Start it with ./brutestringer.py 4 6 "abcdefghijklmnopqrstuvwxyz1234567890" "" #will produce all strings with length 4 to 6 and chars from a to z and numbers 0 to 9 def rec(w, p, baseString): for c in "abcd": if (p<w - 1): rec(w, p + 1, baseString + "%c" % c) print baseString for b in range(3,4): rec(b, 0, "") And here the C++ Code #include <iostream> using namespace std; string chars="abcd"; void rec(int w,int b,string p){ unsigned int i; for(i=0;i<chars.size();i++){ if(b < (w-1)){ rec(w, (b+1), p+chars[i]); } cout << p << "\n"; } } int main () { int a=3, b=0; rec (a+1,b, ""); return 0; } Does anybody see my fault ? I don't have much experience with C++. Thanks indeed

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  • A data structure based on the R-Tree: creating new child nodes when a node is full, but what if I ha

    - by Tom
    I realize my title is not very clear, but I am having trouble thinking of a better one. If anyone wants to correct it, please do. I'm developing a data structure for my 2 dimensional game with an infinite universe. The data structure is based on a simple (!) node/leaf system, like the R-Tree. This is the basic concept: you set howmany childs you want a node (a container) to have maximum. If you want to add a leaf, but the node the leaf should be in is full, then it will create a new set of nodes within this node and move all current leafs to their new (more exact) node. This way, very populated areas will have a lot more subdivisions than a very big but rarely visited area. This works for normal objects. The only problem arises when I have more than maxChildsPerNode objects with the exact same X,Y location: because the node is full, it will create more exact subnodes, but the old leafs will all be put in the exact same node again because they have the exact same position -- resulting in an infinite loop of creating more nodes and more nodes. So, what should I do when I want to add more leafs than maxChildsPerNode with the exact same position to my tree? PS. if I failed to explain my problem, please tell me, so I can try to improve the explanation.

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  • Are fragments of hashes collision-resistent?

    - by Mark
    Let me see if someone would mind clearing up this elementary point about md5 and hashing. If you only use the first 4 bytes of an md5 hash, would that mean theoretically only 1 in 255^4 chance of collision. iow is that the intention with it (and other hash algorithms) - that you only have to use a small portion of the returned hash (say the hash is of a file of some size).

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  • Did I implement this correctly?

    - by user146780
    I'm trying to implement line thickness as denoted here: start = line start = vector(x1, y1) end = line end = vector(x2, y2) dir = line direction = end - start = vector(x2-x1, y2-y1) ndir = normalized direction = dir*1.0/length(dir) perp = perpendicular to direction = vector(dir.x, -dir.y) nperp = normalized perpendicular = perp*1.0/length(perp) perpoffset = nperp*w*0.5 diroffset = ndir*w*0.5 p0, p1, p2, p3 = polygon points: p0 = start + perpoffset - diroffset p1 = start - perpoffset - diroffset p2 = end + perpoffset + diroffset p3 = end - perpoffset + diroffset I'v implemented this like so: void OGLENGINEFUNCTIONS::GenerateLinePoly(const std::vector<std::vector<GLdouble>> &input, std::vector<GLfloat> &output, int width) { output.clear(); float temp; float dirlen; float perplen; POINTFLOAT start; POINTFLOAT end; POINTFLOAT dir; POINTFLOAT ndir; POINTFLOAT perp; POINTFLOAT nperp; POINTFLOAT perpoffset; POINTFLOAT diroffset; POINTFLOAT p0, p1, p2, p3; for(int i = 0; i < input.size() - 1; ++i) { start.x = input[i][0]; start.y = input[i][1]; end.x = input[i + 1][0]; end.y = input[i + 1][1]; dir.x = end.x - start.x; dir.y = end.y - start.y; dirlen = sqrt((dir.x * dir.x) + (dir.y * dir.y)); ndir.x = dir.x * (1.0 / dirlen); ndir.y = dir.y * (1.0 / dirlen); perp.x = dir.x; perp.y = -dir.y; perplen = sqrt((perp.x * perp.x) + (perp.y * perp.y)); nperp.x = perp.x * (1.0 / perplen); nperp.y = perp.y * (1.0 / perplen); perpoffset.x = nperp.x * width * 0.5; perpoffset.y = nperp.y * width * 0.5; diroffset.x = ndir.x * width * 0.5; diroffset.y = ndir.x * width * 0.5; // p0 = start + perpoffset - diroffset //p1 = start - perpoffset - diroffset //p2 = end + perpoffset + diroffset // p3 = end - perpoffset + diroffset p0.x = start.x + perpoffset.x - diroffset.x; p0.y = start.y + perpoffset.y - diroffset.y; p1.x = start.x - perpoffset.x - diroffset.x; p1.y = start.y - perpoffset.y - diroffset.y; p2.x = end.x + perpoffset.x + diroffset.x; p2.y = end.y + perpoffset.y + diroffset.y; p3.x = end.x - perpoffset.x + diroffset.x; p3.y = end.y - perpoffset.y + diroffset.y; output.push_back(p0.x); output.push_back(p0.y); output.push_back(p1.x); output.push_back(p1.y); output.push_back(p2.x); output.push_back(p2.y); output.push_back(p3.x); output.push_back(p3.y); } } But right now the lines look perpendicular and wrong, it should be giving me quads to render which is what i'm rendering, but the points it is outputing are strange. Have I done this wrong? Thanks

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  • Recursive breadth-first travel function in Java or C++?

    - by joejax
    Here is a java code for breadth-first travel: void breadthFirstNonRecursive(){ Queue<Node> queue = new java.util.LinkedList<Node>(); queue.offer(root); while(!queue.isEmpty()){ Node node = queue.poll(); visit(node); if (node.left != null) queue.offer(node.left); if (node.right != null) queue.offer(node.right); } } Is it possible to write a recursive function to do the same? At first, I thought this would be easy, so I came out with this: void breadthFirstRecursive(){ Queue<Node> q = new LinkedList<Node>(); breadthFirst(root, q); } void breadthFirst(Node node, Queue<Node> q){ if (node == null) return; q.offer(node); Node n = q.poll(); visit(n); if (n.left != null) breadthFirst(n.left, q); if (n.right != null) breadthFirst(n.right, q); } Then I found it doesn't work. It is actually does the same thing as this: void preOrder(Node node) { if (node == null) return; visit(node); preOrder(node.left); preOrder(node.right); } Has any one thought about this before?

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