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  • Find all substrings of a string - StringIndexOutOfBoundsException

    - by nazar_art
    I created class Word. Word has a constructor that takes a string argument and one method getSubstrings which returns a String containing all substring of word, sorted by length. For example, if the user provides the input "rum", the method returns a string that will print like this: r u m ru um rum I want to concatenate the substrings in a String, separating them with a newline ("\n"). Then return the string. Code: public class Word { String word; public Word(String word) { this.word = word; } /** * Gets all the substrings of this Word. * @return all substrings of this Word separated by newline */ public String getSubstrings() { String str = ""; int i, j; for (i = 0; i < word.length(); i++) { for (j = 0; j < word.length(); j++) { str = word.substring(i, i + j); str += "\n"; } } return str; } But it throws exception: java.lang.StringIndexOutOfBoundsException: String index out of range: -1 at java.lang.String.substring(String.java:1911) I stuck at this point. Maybe, you have other suggestions according this method signature public String getSubstrings(). How to solve this issue?

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  • LZMA for Delphi

    - by SaCi
    I got a LZMA library on 7-zip site, but that didn't worked. I'm not using files, just stream. And for some why the library on 7-zip site just write the header on the stream but not compress the stream. Some one faced the some problem ? Have some example ? Know other LZMA library for Delphi ? Tks

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  • How to find two most distant points?

    - by depesz
    This is a question that I was asked on a job interview some time ago. And I still can't figure out sensible answer. Question is: you are given set of points (x,y). Find 2 most distant points. Distant from each other. For example, for points: (0,0), (1,1), (-8, 5) - the most distant are: (1,1) and (-8,5) because the distance between them is larger from both (0,0)-(1,1) and (0,0)-(-8,5). The obvious approach is to calculate all distances between all points, and find maximum. The problem is that it is O(n^2), which makes it prohibitively expensive for large datasets. There is approach with first tracking points that are on the boundary, and then calculating distances for them, on the premise that there will be less points on boundary than "inside", but it's still expensive, and will fail in worst case scenario. Tried to search the web, but didn't find any sensible answer - although this might be simply my lack of search skills.

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  • Any specifications/docs around optimization of Google Apps Script, to avoid timeouts and "hangs"

    - by BruceM
    From my experience so far, it seems that if you write a script that makes lots of expensive calls close together, the functionality just "hangs", or you get inconsistent responses, and have to refresh the browser because sheets stop updating etc. Are there any docs or specs that clarify this, as releasing an app fr real-world use is not possible if users can expect it to work most of the time, and produce random results every now and then...

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  • How to determine visibility in 2D

    - by Jack
    Hello, I'm developing an AI sandbox and I would like to calculate what every living entity can see. The rule is to simply hide what's behind the edges of the shapes from the point of view of the entity. The image clarifies everything: I need it either as an input to the artificial intelligence either graphically, to show it for a specific entity while it moves.. Any cool ideas?

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  • how to manage a "resource" array efficiently

    - by Haiyuan Zhang
    The senario of my question is that one need to use a fixed size of array to keep track of certain number of "objects" . The object here can be as simply as a integer or as complex as very fancy data structure. And "keep track" here means to allocate one object when other part of the app need one instance of object and recyle it for future allocation when one instance of the object is returned .Finally ,let me use c++ to put my problme in a more descriptive way . #define MAX 65535 /* 65535 just indicate that many items should be handled . performance demanding! */ typedef struct { int item ; }Item_t; Item_t items[MAX] ; class itemManager { private : /* up to you.... */ public : int get() ; /* get one index to a free Item_t in items */ bool put(int index) ; /* recyle one Item_t indicate by one index in items */ } how will you implement the two public functions of itemManager ? it's up to you to add any private member .

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  • Word Jumble Algorithm

    - by MasterMax1313
    Given a word jumble (i.e. ofbaor), what would be an approach to unscramble the letters to create a real word (i.e. foobar)? I could see this having a couple of approaches, and I think I know how I'd do it in .NET, but I curious to see what some other solutions look like (always happy to see if my solution is optimal or not). This isn't homework or anything like that, I just saw a word jumble in the local comics section of the paper (yes, good ol' fashioned newsprint), and the engineer in me started thinking.

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  • Help me understand Inorder Traversal without using recursion

    - by vito
    I am able to understand preorder traversal without using recursion, but I'm having a hard time with inorder traversal. I just don't seem to get it, perhaps, because I haven't understood the inner working of recursion. This is what I've tried so far: def traverseInorder(node): lifo = Lifo() lifo.push(node) while True: if node is None: break if node.left is not None: lifo.push(node.left) node = node.left continue prev = node while True: if node is None: break print node.value prev = node node = lifo.pop() node = prev if node.right is not None: lifo.push(node.right) node = node.right else: break The inner while-loop just doesn't feel right. Also, some of the elements are getting printed twice; may be I can solve this by checking if that node has been printed before, but that requires another variable, which, again, doesn't feel right. Where am I going wrong? I haven't tried postorder traversal, but I guess it's similar and I will face the same conceptual blockage there, too. Thanks for your time! P.S.: Definitions of Lifo and Node: class Node: def __init__(self, value, left=None, right=None): self.value = value self.left = left self.right = right class Lifo: def __init__(self): self.lifo = () def push(self, data): self.lifo = (data, self.lifo) def pop(self): if len(self.lifo) == 0: return None ret, self.lifo = self.lifo return ret

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  • Time complexity of a certain program

    - by HokageSama
    In a discussion with my friend i am not able to predict correct and tight time complexity of a program. Program is as below. /* This Function reads input array "input" and update array "output" in such a way that B[i] = index value of nearest greater value from A[i], A[i+1] ... A[n], for all i belongs to [1, n] Time Complexity: ?? **/ void createNearestRightSidedLargestArr(int* input, int size, int* output){ if(!input || size < 1) return; //last element of output will always be zero, since no element is present on its right. output[size-1] = -1; int curr = size - 2; int trav; while(curr >= 0){ if(input[curr] < input[curr + 1]){ output[curr] = curr + 1; curr--; continue; } trav = curr + 1; while( input[ output [trav] ] < input[curr] && output [trav] != -1) trav = output[trav]; output[curr--] = output[trav]; } } I said time complexity is O(n^2) but my friend insists that this is not correct. What is the actual time complexity?

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  • Iterative Cartesian Product in Java

    - by akappa
    Hi, I want to compute the cartesian product of an arbitrary number of nonempty sets in Java. I've wrote that iterative code... public static <T> List<Set<T>> cartesianProduct(List<Set<T>> list) { List<Iterator<T>> iterators = new ArrayList<Iterator<T>>(list.size()); List<T> elements = new ArrayList<T>(list.size()); List<Set<T>> toRet = new ArrayList<Set<T>>(); for (int i = 0; i < list.size(); i++) { iterators.add(list.get(i).iterator()); elements.add(iterators.get(i).next()); } for (int j = 1; j >= 0;) { toRet.add(Sets.newHashSet(elements)); for (j = iterators.size()-1; j >= 0 && !iterators.get(j).hasNext(); j--) { iterators.set(j, list.get(j).iterator()); elements.set(j, iterators.get(j).next()); } elements.set(Math.abs(j), iterators.get(Math.abs(j)).next()); } return toRet; } ...but I found it rather inelegant. Someone has a better, still iterative solution? A solution that uses some wonderful functional-like approach? Otherwise... suggestion about how to improve it? Errors? Thanks :)

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  • Capturing time intervals when somebody was online? How would you impement this feature?

    - by Kirzilla
    Hello, Our aim is to build timelines saying about periods of time when user was online. (It really doesn't matter what user we are talking about and where he was online) To get information about onliners we can call API method, someservice.com/api/?call=whoIsOnline whoIsOnline method will give us a list of users currently online. But there is no API method to get information about who IS NOT online. So, we should build our timelines using information we got from whoIsOnline. Of course there will be a measurement error (we can't track information in realtime). Let's suppose that we will call whoIsOnline method every 2 minutes (yes, we will run our script by cron every 2 minutes). For example, calling whoIsOnline at 08:00 will return Peter_id Michal_id Andy_id calling whoIsOnline at 08:02 will return Michael_id Andy_id George_id As you can see, Peter has gone offline, but we have new onliner - George. Available instruments are Db(MySQL) / text files / key-value storage (Redis/memcache); feel free to choose any of them (or even all of them). So, we have to get information like this George_id was online... 12 May: 08:02-08:30, 12:40-12:46, 20:14-22:36 11 May: 09:10-12:30, 21:45-23:00 10 May: was not online And now question... How would you store information to implement such timelines? How would you query/calculate information about periods of time when user was online? Additional information.. You cannot update information about offline users, only users who are "currently" online. Solution should be flexible: timeline information could be represented relating to any timezone. We should keep information only for last 7 days. Every user seen online is automatically getting his own identifier in our database. Uff.. it was really hard for me to write it because my English is pretty bad, but I hope my question will be clear for you. Thank you.

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  • submatrix from a matrix

    - by Grv
    A matrix is of size n*n and it consists only 0 and 1 find the largest submatrix that consists of 1's only eg 10010 11100 11001 11110 largest sub matrix will be of 3*2 from row 2 to row 4 please answer with best space and time complexity

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  • Is there any simple way to test two PNGs for equality?

    - by Mason Wheeler
    I've got a bunch of PNG images, and I'm looking for a way to identify duplicates. By duplicates I mean, specifically, two PNG files whose uncompressed image data are identical, not necessarily whose files are identical. This means I can't do something simple like compare CRC hash values. I figure this can actually be done reliably since PNGs use lossless compression, but I'm worried about speed. I know I can winnow things down a little by testing for equal dimensions first, but when it comes time to actually compare the images against each other, is there any way to do it reasonably efficiently? (ie. faster than the "double-for-loop checking pixel values against each other" brute-force method?)

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  • Merging k sorted linked lists - analysis

    - by Kotti
    Hi! I am thinking about different solutions for one problem. Assume we have K sorted linked lists and we are merging them into one. All these lists together have N elements. The well known solution is to use priority queue and pop / push first elements from every lists and I can understand why it takes O(N log K) time. But let's take a look at another approach. Suppose we have some MERGE_LISTS(LIST1, LIST2) procedure, that merges two sorted lists and it would take O(T1 + T2) time, where T1 and T2 stand for LIST1 and LIST2 sizes. What we do now generally means pairing these lists and merging them pair-by-pair (if the number is odd, last list, for example, could be ignored at first steps). This generally means we have to make the following "tree" of merge operations: N1, N2, N3... stand for LIST1, LIST2, LIST3 sizes O(N1 + N2) + O(N3 + N4) + O(N5 + N6) + ... O(N1 + N2 + N3 + N4) + O(N5 + N6 + N7 + N8) + ... O(N1 + N2 + N3 + N4 + .... + NK) It looks obvious that there will be log(K) of these rows, each of them implementing O(N) operations, so time for MERGE(LIST1, LIST2, ... , LISTK) operation would actually equal O(N log K). My friend told me (two days ago) it would take O(K N) time. So, the question is - did I f%ck up somewhere or is he actually wrong about this? And if I am right, why doesn't this 'divide&conquer' approach can't be used instead of priority queue approach?

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  • what is order notation f(n)=O(g(n))?

    - by Lopa
    2 questions: question 1: under what circumstances would this[O(f(n))=O(k.f(n))] be the most appropriate form of time-complexity analysis? question 2: working from mathematical definition of O notation, show that O(f(n))=O(k.f(n)), for positive constant k.? My view: For the first one I think it is average case and worst case form of time-complexity. am i right? and what else do i write in that? for the second one I think we need to define the function mathematically, so is the answer something like because the multiplication by a constant just corresponds to a readjustment of value of the arbitrary constant 'k' in definition of O.

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  • recursively implementing 'minimum number of coins' in python

    - by user5198
    This problem is same as asked in here. Given a list of coins, their values (c1, c2, c3, ... cj, ...), and the total sum i. Find the minimum number of coins the sum of which is i (we can use as many coins of one type as we want), or report that it's not possible to select coins in such a way that they sum up to S. I"m just introduced to dynamic programming yesterday and I tried to make a code for it. # Optimal substructure: C[i] = 1 + min_j(C[i-cj]) cdict = {} def C(i, coins): if i <= 0: return 0 if i in cdict: return cdict[i] else: answer = 1 + min([C(i - cj, coins) for cj in coins]) cdict[i] = answer return answer Here, C[i] is the optimal solution for amount of money 'i'. And available coins are {c1, c2, ... , cj, ...} for the program, I've increased the recursion limit to avoid maximum recursion depth exceeded error. But, this program gives the right answer only someones and when a solution is not possible, it doesn't indicate that. What is wrong with my code and how to correct it?

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  • big O notation algorithm

    - by niggersak
    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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  • Mathematical attack on the Digital Signature Algorithm

    - by drelihan
    Does anybody know the mathematics behind an attack on DSA where modulus p has p-1 made up of only small factors. In reality, this would not happen as the key generator would guarantee that this is not so. There is much information on the web on generating good input paramters for DSA so that it is hard to crack but no information on how you find X if modulus p has p-1 made up of only small factors.

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  • How to use Haar wavelet to detect LINES on an image?

    - by Ole Jak
    So I have Image like this I want to get something like this (I hevent drawn all lines I want but I hope you can get my idea) I want to use SURF ( (Speeded Up Robust Features) is a robust image descriptor, first presented by Herbert Bay et al. in 2006 ) or something that is based on sums of 2D Haar wavelet responses and makes an efficient use of integral images for finding all straight lines on image. I want to get relative to picture pixel coords start and end points of lines. So on this picture to find all lines between tiles and thouse 2 black lines on top. Is there any such Code Example (with lines search capability) to start from? I love C and C++ but any other readable code will probably work for me=)

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  • Suggestions for duplicate file finder algorithm (using C)

    - by Andrei Ciobanu
    Hello, I wanted to write a program that test if two files are duplicates (have exactly the same content). First I test if the files have the same sizes, and if they have i start to compare their contents. My first idea, was to "split" the files into fixed size blocks, then start a thread for every block, fseek to startup character of every block and continue the comparisons in parallel. When a comparison from a thread fails, the other working threads are canceled, and the program exits out of the thread spawning loop. The code looks like this: dupf.h #ifndef __NM__DUPF__H__ #define __NM__DUPF__H__ #define NUM_THREADS 15 #define BLOCK_SIZE 8192 /* Thread argument structure */ struct thread_arg_s { const char *name_f1; /* First file name */ const char *name_f2; /* Second file name */ int cursor; /* Where to seek in the file */ }; typedef struct thread_arg_s thread_arg; /** * 'arg' is of type thread_arg. * Checks if the specified file blocks are * duplicates. */ void *check_block_dup(void *arg); /** * Checks if two files are duplicates */ int check_dup(const char *name_f1, const char *name_f2); /** * Returns a valid pointer to a file. * If the file (given by the path/name 'fname') cannot be opened * in 'mode', the program is interrupted an error message is shown. **/ FILE *safe_fopen(const char *name, const char *mode); #endif dupf.c #include <errno.h> #include <pthread.h> #include <stdio.h> #include <stdlib.h> #include <string.h> #include <sys/types.h> #include <sys/stat.h> #include <unistd.h> #include "dupf.h" FILE *safe_fopen(const char *fname, const char *mode) { FILE *f = NULL; f = fopen(fname, mode); if (f == NULL) { char emsg[255]; sprintf(emsg, "FOPEN() %s\t", fname); perror(emsg); exit(-1); } return (f); } void *check_block_dup(void *arg) { const char *name_f1 = NULL, *name_f2 = NULL; /* File names */ FILE *f1 = NULL, *f2 = NULL; /* Streams */ int cursor = 0; /* Reading cursor */ char buff_f1[BLOCK_SIZE], buff_f2[BLOCK_SIZE]; /* Character buffers */ int rchars_1, rchars_2; /* Readed characters */ /* Initializing variables from 'arg' */ name_f1 = ((thread_arg*)arg)->name_f1; name_f2 = ((thread_arg*)arg)->name_f2; cursor = ((thread_arg*)arg)->cursor; /* Opening files */ f1 = safe_fopen(name_f1, "r"); f2 = safe_fopen(name_f2, "r"); /* Setup cursor in files */ fseek(f1, cursor, SEEK_SET); fseek(f2, cursor, SEEK_SET); /* Initialize buffers */ rchars_1 = fread(buff_f1, 1, BLOCK_SIZE, f1); rchars_2 = fread(buff_f2, 1, BLOCK_SIZE, f2); if (rchars_1 != rchars_2) { /* fread failed to read the same portion. * program cannot continue */ perror("ERROR WHEN READING BLOCK"); exit(-1); } while (rchars_1-->0) { if (buff_f1[rchars_1] != buff_f2[rchars_1]) { /* Different characters */ fclose(f1); fclose(f2); pthread_exit("notdup"); } } /* Close streams */ fclose(f1); fclose(f2); pthread_exit("dup"); } int check_dup(const char *name_f1, const char *name_f2) { int num_blocks = 0; /* Number of 'blocks' to check */ int num_tsp = 0; /* Number of threads spawns */ int tsp_iter = 0; /* Iterator for threads spawns */ pthread_t *tsp_threads = NULL; thread_arg *tsp_threads_args = NULL; int tsp_threads_iter = 0; int thread_c_res = 0; /* Thread creation result */ int thread_j_res = 0; /* Thread join res */ int loop_res = 0; /* Function result */ int cursor; struct stat buf_f1; struct stat buf_f2; if (name_f1 == NULL || name_f2 == NULL) { /* Invalid input parameters */ perror("INVALID FNAMES\t"); return (-1); } if (stat(name_f1, &buf_f1) != 0 || stat(name_f2, &buf_f2) != 0) { /* Stat fails */ char emsg[255]; sprintf(emsg, "STAT() ERROR: %s %s\t", name_f1, name_f2); perror(emsg); return (-1); } if (buf_f1.st_size != buf_f2.st_size) { /* File have different sizes */ return (1); } /* Files have the same size, function exec. is continued */ num_blocks = (buf_f1.st_size / BLOCK_SIZE) + 1; num_tsp = (num_blocks / NUM_THREADS) + 1; cursor = 0; for (tsp_iter = 0; tsp_iter < num_tsp; tsp_iter++) { loop_res = 0; /* Create threads array for this spawn */ tsp_threads = malloc(NUM_THREADS * sizeof(*tsp_threads)); if (tsp_threads == NULL) { perror("TSP_THREADS ALLOC FAILURE\t"); return (-1); } /* Create arguments for every thread in the current spawn */ tsp_threads_args = malloc(NUM_THREADS * sizeof(*tsp_threads_args)); if (tsp_threads_args == NULL) { perror("TSP THREADS ARGS ALLOCA FAILURE\t"); return (-1); } /* Initialize arguments and create threads */ for (tsp_threads_iter = 0; tsp_threads_iter < NUM_THREADS; tsp_threads_iter++) { if (cursor >= buf_f1.st_size) { break; } tsp_threads_args[tsp_threads_iter].name_f1 = name_f1; tsp_threads_args[tsp_threads_iter].name_f2 = name_f2; tsp_threads_args[tsp_threads_iter].cursor = cursor; thread_c_res = pthread_create( &tsp_threads[tsp_threads_iter], NULL, check_block_dup, (void*)&tsp_threads_args[tsp_threads_iter]); if (thread_c_res != 0) { perror("THREAD CREATION FAILURE"); return (-1); } cursor+=BLOCK_SIZE; } /* Join last threads and get their status */ while (tsp_threads_iter-->0) { void *thread_res = NULL; thread_j_res = pthread_join(tsp_threads[tsp_threads_iter], &thread_res); if (thread_j_res != 0) { perror("THREAD JOIN FAILURE"); return (-1); } if (strcmp((char*)thread_res, "notdup")==0) { loop_res++; /* Closing other threads and exiting by condition * from loop. */ while (tsp_threads_iter-->0) { pthread_cancel(tsp_threads[tsp_threads_iter]); } } } free(tsp_threads); free(tsp_threads_args); if (loop_res > 0) { break; } } return (loop_res > 0) ? 1 : 0; } The function works fine (at least for what I've tested). Still, some guys from #C (freenode) suggested that the solution is overly complicated, and it may perform poorly because of parallel reading on hddisk. What I want to know: Is the threaded approach flawed by default ? Is fseek() so slow ? Is there a way to somehow map the files to memory and then compare them ?

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  • How to detect a Triangle gesture with kinect?

    - by Akhilesh Mishra
    I am trying to implement a gesture recognition system which interprets the geometric gestures user makes and draws it on screen, I have some idea of how circle can be recognized, however i have no clue how to get started with triangle recognition. the data I have is X and Y coordinates of all points the gesture passed through. I get this data by tracking right hand. I found something online called Hough Transform , which is used for detecting lines but i am not sure whether it will work for discrete collection of points, Any ideas folks?

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  • Stack and queue operations on the same array.

    - by Passonate Learner
    Hi. I've been thinking about a program logic, but I cannot draw a conclusion to my problem. Here, I've implemented stack and queue operations to a fixed array. int A[1000]; int size=1000; int top; int front; int rear; bool StackIsEmpty() { return (top==0); } bool StackPush( int x ) { if ( top >= size ) return false; A[top++] = x; return true; } int StackTop( ) { return A[top-1]; } bool StackPop() { if ( top <= 0 ) return false; A[--top] = 0; return true; } bool QueueIsEmpty() { return (front==rear); } bool QueuePush( int x ) { if ( rear >= size ) return false; A[rear++] = x; return true; } int QueueFront( ) { return A[front]; } bool QueuePop() { if ( front >= rear ) return false; A[front++] = 0; return true; } It is presumed(or obvious) that the bottom of the stack and the front of the queue is pointing at the same location, and vice versa(top of the stack points the same location as rear of the queue). For example, integer 1 and 2 is inside an array in order of writing. And if I call StackPop(), the integer 2 will be popped out, and if I call QueuePop(), the integer 1 will be popped out. My problem is that I don't know what happens if I do both stack and queue operations on the same array. The example above is easy to work out, because there are only two values involved. But what if there are more than 2 values involved? For example, if I call StackPush(1); QueuePush(2); QueuePush(4); StackPop(); StackPush(5); QueuePop(); what values will be returned in the order of bottom(front) from the final array? I know that if I code a program, I would receive a quick answer. But the reason I'm asking this is because I want to hear a logical explanations from a human being, not a computer.

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  • Pong: How does the paddle know where the ball will hit?

    - by Roflcoptr
    After implementing Pacman and Snake I'm implementing the next very very classic game: Pong. The implementation is really simple, but I just have one little problem remaining. When one of the paddle (I'm not sure if it is called paddle) is controlled by the computer, I have trouble to position it at the correct position. The ball has a current position, a speed (which for now is constant) and a direction angle. So I could calculate the position where it will hit the side of the computer controlled paddle. And so Icould position the paddle right there. But however in the real game, there is a probability that the computer's paddle will miss the ball. How can I implement this probability? If I only use a probability of lets say 0.5 that the computer's paddle will hit the ball, the problem is solved, but I think it isn't that simple. From the original game I think the probability depends on the distance between the current paddle position and the position the ball will hit the border. Does anybody have any hints how exactly this is calculated?

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