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  • Adapting pseudocode to java implementation for finding the longest word in a trie

    - by user1766888
    Referring to this question I asked: How to find the longest word in a trie? I'm having trouble implementing the pseudocode given in the answer. findLongest(trie): //first do a BFS and find the "last node" queue <- [] queue.add(trie.root) last <- nil map <- empty map while (not queue.empty()): curr <- queue.pop() for each son of curr: queue.add(son) map.put(son,curr) //marking curr as the parent of son last <- curr //in here, last indicate the leaf of the longest word //Now, go up the trie and find the actual path/string curr <- last str = "" while (curr != nil): str = curr + str //we go from end to start curr = map.get(curr) return str This is what I have for my method public static String longestWord (DTN d) { Queue<DTN> holding = new ArrayQueue<DTN>(); holding.add(d); DTN last = null; Map<DTN,DTN> test = new ArrayMap<DTN,DTN>(); DTN curr; while (!holding.isEmpty()) { curr = holding.remove(); for (Map.Entry<String, DTN> e : curr.children.entries()) { holding.add(curr.children.get(e)); test.put(curr.children.get(e), curr); } last = curr; } curr = last; String str = ""; while (curr != null) { str = curr + str; curr = test.get(curr); } return str; } I'm getting a NullPointerException at: for (Map.Entry<String, DTN> e : curr.children.entries()) How can I find and fix the cause of the NullPointerException of the method so that it returns the longest word in a trie?

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  • question about in -place sort

    - by davit-datuashvili
    for example we have following array char data[]=new char[]{'A','S','O','R','T','I','N','G','E','X','A','M','P','L','E'}; and index array int a[]=new int[]{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14}: void insitu(char data[],int a[],N){ for (int i=0;i<N;i++) { char v=data[i]; int j,int k; for (k=i;a[k]!=i;k=a[j];a[j]=j) { j=k;data[k]=data[a[k]; } data[k]=v; a[k]=k; } i have question what is initialize value of j? when i run this code it asks me to initialize j and what should do? }

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  • Distributing players to tables

    - by IVlad
    Consider N = 4k players, k tables and a number of clans such that each member can belong to one clan. A clan can contain at most k players. We want to organize 3 rounds of a game such that, for each table that seats exactly 4 players, no 2 players sitting there are part of the same clan, and, for the later rounds, no 2 players sitting there have sat at the same table before. All players play all rounds. How can we do this efficiently if N can be about ~80 large? I thought of this: for each table T: repeat until 4 players have been seated at T: pick a random player X that is not currently seated anywhere if X has not sat at the same table as anyone currently at T AND X is not from the same clan as anyone currently at T seat X at T break I am not sure if this will always finish or if it can get stuck even if there is a valid assignment. Even if this works, is there a better way to do it?

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  • question about siftdown operation on heap

    - by davit-datuashvili
    i have following pseudo code which execute siftdown operation on heap array suppose is x void siftdown(int n) pre heap(2,n) && n>=0 post heap(1,n) i=1; loop /*invariant heap(1,n) except perhaps between i and it's (0,1,or 2) children*/ c=2*i; if (c>n) break; // c is left child of i if (c+1)<=n /* c+1 is rigth child of i if (x[c+1]<x[c]) c++ /* c is lesser child of i if (x[i]<=x[c]) break; swap(c,i) i=c; i have wrote following code is it correct? public class siftdown{ public static void main(String[]args){ int c; int n=9; int a[]=new int[]{19,100,17,2,7,3,36,1,25}; int i=1; while (i<n){ c=2*i; if (c>n) break; //c is the left child of i if (c+1<=n) //c+1 ir rigth child of i if (a[c+1]<a[c]) c++; if (a[i]<=a[c]) break; int t=a[c]; a[c]=a[i]; a[i]=t; i=c; } for (int j=0;j<a.length;j++){ System.out.println(a[j]); } } } // result is 19 2 17 1 7 3 36 100 25

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  • F# insert/remove item from list

    - by Timothy
    How should I go about removing a given element from a list? As an example, say I have list ['A'; 'B'; 'C'; 'D'; 'E'] and want to remove the element at index 2 to produce the list ['A'; 'B'; 'D'; 'E']? I've already written the following code which accomplishes the task, but it seems rather inefficient to traverse the start of the list when I already know the index. let remove lst i = let rec remove lst lst' = match lst with | [] -> lst' | h::t -> if List.length lst = i then lst' @ t else remove t (lst' @ [h]) remove lst [] let myList = ['A'; 'B'; 'C'; 'D'; 'E'] let newList = remove myList 2 Alternatively, how should I insert an element at a given position? My code is similar to the above approach and most likely inefficient as well. let insert lst i x = let rec insert lst lst' = match lst with | [] -> lst' | h::t -> if List.length lst = i then lst' @ [x] @ lst else insert t (lst' @ [h]) insert lst [] let myList = ['A'; 'B'; 'D'; 'E'] let newList = insert myList 2 'C'

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  • Print number series in java

    - by user1898282
    I have to print the series shown below in java: ***1*** **2*2** *3*3*3* 4*4*4*4 My current implementation is: public static void printSeries(int number,int numberOfCharsinEachLine){ String s="*"; for(int i=1;i<=number;i++){ int countOfs=(numberOfCharsinEachLine-(i)-(i-1))/2; if(countOfs<0){ System.out.println("Can't be done"); break; } for(int j=0;j<countOfs;j++){ System.out.print(s); } System.out.print(i); for(int k=1;k<i;k++){ System.out.print(s); System.out.print(i); } for(int j=0;j<countOfs;j++){ System.out.print(s); } System.out.println(); } } But there are lot of for loops, so I'm wondering whether this can be done in a better way or not?

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  • Trie vs B+ tree

    - by Fakrudeen
    How does Trie and B+ tree compare for indexing lexicographically sorted strings [on the order some billions]? It should support range queries as well. From perf. as well as implementation complexity point of view.

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  • Reduce text length to fit cell width in a smart manner

    - by Andrei Ciobanu
    Hello, I am in project where we are building a simple web calendar using Java EE technologies. We define a table where every row is an employee, and every column represents an hour interval. The table width and column widths are adjustable. In every cell we have a text retrieved from a database, indicating what the employee is doing / should do in that time interval. The problem is that sometimes the text in cells is getting bigger than the actual cell. My task is to make the text more "readable" by reducing it's length in a "smart way" so that it can fit in the cell more "gracefully". For example if initially in a cell I have: "Writing documents", after the resize I should retrieve: "Wrtng. dcmnts" or "Writ. docum." so that the text can fit well. Is there a smart way to do it ? Or removing vocals / split the string in two is enough ?

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  • how to find out if a shape is passable

    - by jd
    I have a complex polygon (possibly concave) and a few of its edges marked as entry/exit points. there is a possibility that inside this polygon may lie one or more blockades of arbitrary shape. what approaches could I use to determine whether a path of certain width exists between a pair of entry/exit edges? having read through the question it looks like a homework type - it is not. I just wish to have a at least a few leads I could pursue, as this is new to me.

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  • Interview question: C program to sort a binary array in O(n)

    - by Zacky112
    I've comeup with the following program to do it, but it does not seem to work and goes into infinite loop. Its working is similar to quicksort. int main() { int arr[] = {1,1,0,1,0,0,0,1,0,1,0,1,0,1,0,1,0,1}; int N = 18; int *front, *last; front = arr; last = arr + N; while(front <= last) { while( (front < last) && (*front == 0) ) front++; while( (front < last) && (*last == 1) ) last--; if( front < last) { int temp = *front; *front = *last; *last = temp; front ++; last--; } } for(int i=0;i<N;i++) printf("%d ",arr[i]); return 0; }

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  • please help me to implement such kind of sort

    - by davit-datuashvili
    hi i need to write sucj kind of sorting maybe it is similary to radix sort and also (this is not homework because i created it problem myself and please if anobody can help me) problem is like this suppose i have array int x[]=new int[]{4,5,3,2,1}; let write it into binary form 5 -0101 4- 0100 3-0011 2-0010 1-0001 i want to sort this elements by using bitwise operatos or check each bit and if less exchange it can anybody help me for example take 5 and 4 check first rightmost bit 0==0 so continue in the 1 index also 1==1 next the same 0=0 and last one 10 it means that first element is greater then second so exchange it please help me

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  • One Database Field to Hold Survey Answer

    - by yar
    Skipping the question of whether this is bad design (and the question of why I would want/need to do this), I'm just wondering if my 'math' is right... I have n things that need to be put in order (n is always less than 5): thing1 thing2 thing3 ... and I'd like to store these results in a single integer for thing. My initial thought is that (obviously the code will not look like this): thing = thing1 * 1 + thing2 * 2 + thing3 * 4 + thing5 * 8 will always give me a unique value, and that any value for thing will always translate back to its values for thing1...thingn. Is this correct?

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  • Find all A^x in a given range

    - by Austin Henley
    I need to find all monomials in the form AX that when evaluated falls within a range from m to n. It is safe to say that the base A is greater than 1, the power X is greater than 2, and only integers need to be used. For example, in the range 50 to 100, the solutions would be: 2^6 3^4 4^3 My first attempt to solve this was to brute force all combinations of A and X that make "sense." However this becomes too slow when used for very large numbers in a big range since these solutions are used in part of much more intensive processing. Here is the code: def monoSearch(min, max): base = 2 power = 3 while 1: while base**power < max: if base**power > min: print "Found " + repr(base) + "^" + repr(power) + " = " + repr(base**power) power = power + 1 base = base + 1 power = 3 if base**power > max: break I could remove one base**power by saving the value in a temporary variable but I don't think that would make a drastic effect. I also wondered if using logarithms would be better or if there was a closed form expression for this. I am open to any optimizations or alternatives to finding the solutions.

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  • Graph search problem with route restrictions

    - by Darcara
    I want to calculate the most profitable route and I think this is a type of traveling salesman problem. I have a set of nodes that I can visit and a function to calculate cost for traveling between nodes and points for reaching the nodes. The goal is to reach a fixed known score while minimizing the cost. This cost and rewards are not fixed and depend on the nodes visited before. The starting node is fixed. There are some restrictions on how nodes can be visited. Some simplified examples include: Node B can only be visited after A After node C has been visited, D or E can be visited. Visiting at least one is required, visiting both is permissible. Z can only be visited after at least 5 other nodes have been visited Once 50 nodes have been visited, the nodes A-M will no longer reward points Certain nodes can (and probably must) be visited multiple times Currently I can think of only two ways to solve this: a) Genetic Algorithms, with the fitness function calculating the cost/benefit of the generated route b) Dijkstra search through the graph, since the starting node is fixed, although the large number of nodes will probably make that not feasible memory wise. Are there any other ways to determine the best route through the graph? It doesn't need to be perfect, an approximated path is perfectly fine, as long as it's error acceptable. Would TSP-solvers be an option here?

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  • Finding whether a point lies inside a rectangle or not

    - by avd
    The rectangle can be oriented in any way...need not be axis aligned. Now I want to find whether a point lies inside the rectangle or not. One method I could think of was to rotate the rectangle and point coordinates to make the rectangle axis aligned and then by simply testing the coordinates of point whether they lies within that of rectangle's or not. The above method requires rotation and hence floating point operations. Is there any other efficient way to do this??

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  • What is the most efficient method to find x contiguous values of y in an array?

    - by Alec
    Running my app through callgrind revealed that this line dwarfed everything else by a factor of about 10,000. I'm probably going to redesign around it, but it got me wondering; Is there a better way to do it? Here's what I'm doing at the moment: int i = 1; while ( ( (*(buffer++) == 0xffffffff && ++i) || (i = 1) ) && i < desiredLength + 1 && buffer < bufferEnd ); It's looking for the offset of the first chunk of desiredLength 0xffffffff values in a 32 bit unsigned int array. It's significantly faster than any implementations I could come up with involving an inner loop. But it's still too damn slow.

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  • Constructing a tree using Python

    - by stealthspy
    I am trying to implement a unranked boolean retrieval. For this, I need to construct a tree and perform a DFS to retrieve documents. I have the leaf nodes but I am having difficulty to construct the tree. Eg: query = OR ( AND (maria sharapova) tennis) Result: OR | | AND tennis | | maria sharapova I traverse the tree using DFS and calculate the boolean equivalent of certain document ids to identify the required document from the corpus. Can someone help me with the design of this using python? I have parsed the query and retrieved the leaf nodes for now.

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  • How to store and collect data for mining such information as most viewed for last 24 hours, last 7 d

    - by Kirzilla
    Hello, Let's imagine that we have high traffic project (a tube site) which should provide sorting using this options (NOT IN REAL TIME). Number of videos is about 200K and all information about videos is stored in MySQL. Number of daily video views is about 1.5KK. As instruments we have Hard Disk Drive (text files), MySQL, Redis. Views top viewed top viewed last 24 hours top viewed last 7 days top viewed last 30 days top rated last 365 days How should I store such information? The first idea is to log all visits to text files (single file per hour, for example visits_20080101_00.log). At the beginning of each hour calculate views per video for previous hour and insert this information into MySQL. Then recalculate totals (for last 24 hours) and update statistics in tables. At the beginning of every day we have to do the same but recalculate for last 7 days, last 30 days, last 365 days. This method seems to be very poor for me because we have to store information about last 365 days for each video to make correct calculations. Is there any other good methods? Probably, we have to choose another instruments for this? Thank you.

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  • How can we find second maximum from array efficiently?

    - by Xinus
    Is it possible to find the second maximum number from an array of integers by traversing the array only once? As an example, I have a array of five integers from which I want to find second maximum number. Here is an attempt I gave in the interview: #define MIN -1 int main() { int max=MIN,second_max=MIN; int arr[6]={0,1,2,3,4,5}; for(int i=0;i<5;i++){ cout<<"::"<<arr[i]; } for(int i=0;i<5;i++){ if(arr[i]>max){ second_max=max; max=arr[i]; } } cout<<endl<<"Second Max:"<<second_max; int i; cin>>i; return 0; } The interviewer, however, came up with the test case int arr[6]={5,4,3,2,1,0};, which prevents it from going to the if condition the second time. I said to the interviewer that the only way would be to parse the array two times (two for loops). Does anybody have a better solution?

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