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  • 3 page longest path on a website

    - by Kazoom
    i have a log file which maintains source entry for each page.all the pages share the common file. source means from what page did user arrive on the target page. I want to find the most common 3 page path for all the pages on the website. Example log file: source Target 1 2 1 3 2 1 3 2 3 2 2 1 The most common 3 page path here was from 3 to 2 to 1.

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  • How to do run length encoding?

    - by Phoenix
    I have a long string for example it could be "aaaaaabbccc". Need to represent it as "a6b2c3". What's the best way to do this ? I could do this in linear time by comparing characters and incrementing counts and then replacing the counts in the array, using two indexes in one pass. Can you guys think of a better way than this? Are any of the encoding techniques going to work here ?

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  • Interview question: Check if one string is a rotation of other string.

    - by Webdev
    A friend of mine was asked the following question today at interview for the position of software developer. Given two string s1 and s2 how will you check if s1 is a rotated version of s2 ? Example: if s1 = "stackoverflow"; then the following are some of its rotated versions: "tackoverflows" "ackoverflowst" "overflowstack" where as "stackoverflwo" is not a rotated version. The answer he gave was: Take s2 and find the logest prefix that is a substring of s1, that will give you the point of rotation. Once you find that point, break s2 at that point to get s2a and s2b, then just check if concatenate(s2a,s2b) == s1 It looks like a good solution to me and my friend. But the interviewr though otherwise. He asked for a simpler solution. Please help me by telling how would you do this in Java/C/C++ ? Thanks in advance.

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  • read-only memory and heap memory

    - by benjamin button
    hi, AFAIK, string literals are stored in read only memory in case of C language. where is this actually present on the hardware. as per my knowledge heap is on RAM.correct me if i am wrong. how different is heap from read only memory? is it OS dependant?

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  • Understanding the concept of inheritance in Java

    - by Nirmal
    Hello All.... I am just refreshing the oops features of the java. So, I have a little confusion regarding inheritance concept. For that I have a following sample code : class Super{ int index = 5; public void printVal(){ System.out.println("Super"); } } class Sub extends Super{ int index = 2; public void printVal(){ System.out.println("Sub"); } } public class Runner { public static void main(String args[]){ Super sup = new Sub(); System.out.println(sup.index+","); sup.printVal(); } } Now above code is giving me output as : 5,Sub. Here, we are overriding printVal() method, so that is understandable that it is accessing child class method only. But I could not understand why it's accessing the value of x from Super class... Thanks in advance....

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  • How to generate a number in arbitrary range using random()={0..1} preserving uniformness and density?

    - by psihodelia
    Generate a random number in range [x..y] where x and y are any arbitrary floating point numbers. Use function random(), which returns a random floating point number in range [0..1] from P uniformly distributed numbers (call it "density"). Uniform distribution must be preserved and P must be scaled as well. I think, there is no easy solution for such problem. To simplify it a bit, I ask you how to generate a number in interval [-0.5 .. 0.5], then in [0 .. 2], then in [-2 .. 0], preserving uniformness and density? Thus, for [0 .. 2] it must generate a random number from P*2 uniformly distributed numbers. The obvious simple solution random() * (x - y) + y will generate not all possible numbers because of the lower density for all abs(x-y)>1.0 cases. Many possible values will be missed. Remember, that random() returns only a number from P possible numbers. Then, if you multiply such number by Q, it will give you only one of P possible values, scaled by Q, but you have to scale density P by Q as well.

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  • Check my anagram code from a job interview in the past.

    - by Michael Dorgan
    Had the following as an interview question a while ago and choked so bad on basic syntax that I failed to advance (once the adrenalin kicks in, coding goes out the window.) Given a list of string, return a list of sets of strings that are anagrams of the input set. i.e. "dog","god", "foo" should return {"dog","god"}. Afterward, I created the code on my own as a sanity check and it's been around now for a bit. I'd welcome input on it to see if I missed anything or if I could have done it much more efficiently. Take it as a chance to improve myself and learn other techniques: void Anagram::doWork(list input, list &output) { typedef list SortType; SortType sortedInput; // sort each string and pair it with the original for(list<string>::iterator i = input.begin(); i != input.end(); ++i) { string tempString(*i); std::sort(tempString.begin(), tempString.end()); sortedInput.push_back(make_pair(*i, tempString)); } // Now step through the new sorted list for(SortType::iterator i = sortedInput.begin(); i != sortedInput.end();) { set<string> newSet; // Assume (hope) we have a match and pre-add the first. newSet.insert(i->first); // Set the secondary iterator one past the outside to prevent // matching the original SortType::iterator j = i; ++j; while(j != sortedInput.end()) { if(i->second == j->second) { // If the string matches, add it to the set and remove it // so that future searches need not worry about it newSet.insert(j->first); j = sortedInput.erase(j); } else { // else, next element ++j; } } // If size is bigger than our original push, we have a match - save it to the output if(newSet.size() > 1) { output.push_back(newSet); } // erase this element and update the iterator i = sortedInput.erase(i); } }

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  • RTNETLINK answers: File exists... maybe because assigned a new mac adress

    - by steven
    I got a "RTNETLINK answers: File exists Failed to bring up eth0:1" on "ifup eth0:1". I suspect it happens because i assigned a new mac adress in my VM's network adapter. Can you tell me how to fix the issue? My configuration looks like this: # The loopback network interface auto lo iface lo inet loopback # The primary network interface auto eth0 allow-hotplug eth0 iface eth0 inet static address 192.168.1.80 netmask 255.255.255.0 gateway 192.168.1.1 dns-nameservers 192.168.1.1 # Alias being connected to 192.168.10.x Network auto eth0:1 allow-hotplug eth0:1 iface eth0:1 inet static address 192.168.10.83 netmask 255.255.255.0 gateway 192.168.10.10 dns-nameservers 192.168.10.1 Why do I get "RTNETLINK answer: File exists.." suddenly? I worked with this configuration before without problems. All i did in the past is to renew the adapters mac adress. At the moment I am connected to the 192.168.10.x Network and if I do /etc/init.d/networking stop /etc/init.d/networking start then i got "RTNETLINK [...] falied to bring up eth0:1" but the strage thing is that i am able to connect to 192.168.10.83 via ssh from my host machine. But I cannot reach the internet from the debian client. I hope it is clear what my problem is, now. update if i change my /etc/network/interfaces like this then "ifup eth0" fails, too with the same error! # The loopback network interface auto lo iface lo inet loopback # The primary network interface auto eth0 allow-hotplug eth0 iface eth0 inet static address 192.168.10.83 netmask 255.255.255.0 gateway 192.168.10.10 dns-nameservers 192.168.10.1 with verbose option enabled i got: Configuring interfache eth0=eth0 (inet) run-parts --verbose /etc/network/if-pre-up.d ip addr add 192.168.10.83/255.255.255.0 broadcast 192.168.10.255 dev eth0 label eth0 RTNETLINK answers: File exists Failed to bring up eth0. same if i type this manually: ip addr add 192.168.10.83/255.255.255.0 broadcast 192.168.10.255 dev eth0 label eth0

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  • How to find largest common sub-tree in the given two binary search trees?

    - by Bhushan
    Two BSTs (Binary Search Trees) are given. How to find largest common sub-tree in the given two binary trees? EDIT 1: Here is what I have thought: Let, r1 = current node of 1st tree r2 = current node of 2nd tree There are some of the cases I think we need to consider: Case 1 : r1.data < r2.data 2 subproblems to solve: first, check r1 and r2.left second, check r1.right and r2 Case 2 : r1.data > r2.data 2 subproblems to solve: - first, check r1.left and r2 - second, check r1 and r2.right Case 3 : r1.data == r2.data Again, 2 cases to consider here: (a) current node is part of largest common BST compute common subtree size rooted at r1 and r2 (b)current node is NOT part of largest common BST 2 subproblems to solve: first, solve r1.left and r2.left second, solve r1.right and r2.right I can think of the cases we need to check, but I am not able to code it, as of now. And it is NOT a homework problem. Does it look like?

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  • Given a string of red and blue balls, find min number of swaps to club the colors together

    - by efficiencyIsBliss
    We are given a string of the form: RBBR, where R - red and B - blue. We need to find the minimum number of swaps required in order to club the colors together. In the above case that answer would be 1 to get RRBB or BBRR. I feel like an algorithm to sort a partially sorted array would be useful here since a simple sort would give us the number of swaps, but we want the minimum number of swaps. Any ideas? This is allegedly a Microsoft interview question according to this.

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