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• Determining the order of a list of numbers (possibly without sorting)

  - by Victor Liu

  I have an array of unique integers (e.g. val[i]), in arbitrary order, and I would like to populate another array (ord[i]) with the the sorted indexes of the integers. In other words, val[ord[i]] is in sorted order for increasing i. Right now, I just fill in ord with 0, ..., N, then sort it based on the value array, but I am wondering if we can be more efficient about it since ord is not populated to begin with. This is more of a question out of curiousity; I don't really care about the extra overhead from having to prepopulate a list and then sort it (it's small, I use insertion sort). This may be a silly question with an obvious answer, but I couldn't find anything online.

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• question on loop in java

  - by davit-datuashvili

  can anybody tell me what is mistake here? while ((char t==(char) System.in.read())!='0') ?

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• Picking apples off a tree

  - by John Retallack

  I have the following problem: I am given a tree with N apples, for each apple I am given it's weight and height. I can pick apples up to a given height H, each time I pick an apple the height of every apple is increased with U. I have to find out the maximum weight of apples I can pick. 1 = N = 100000 0 < {H, U, apples' weight and height, maximum weight} < 231 Example: N=4 H=100 U=10 height weight 82 30 91 10 93 5 94 15 The answer is 45: first pick the apple with the weight of 15 then the one with the weight of 30. Could someone help me approach this problem?

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• Generate encoding String according to creation order.

  - by Tony

  I need to generate encoding String for each item I inserted into the database. for example: x00001 for the first item x00002 for the sencond item x00003 for the third item The way I chose to do this is counting the rows. Before I insert the third item, I count against the database, I know there're already 2 rows, so the next encoding is ended with 3. But there is a problem. If I delete the second item, the forth item will not be the x00004,but x00003. I can add additional columns to table, to store the next encoding, I don't know if there's other better solutions ?

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• please help please

  - by davit-datuashvili

  http://stackoverflow.com/questions/2836934/how-to-compare-two-char-variables please show me a few steps how to start to write this code please

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• help implementing 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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• maximum of given function

  - by davit-datuashvili

  first of all i am doing programs in java language this code is merely taken from web site i have not question about divide and conqurer but about function and it's argument here is code of ternary search def ternarySearch(f, left, right, absolutePrecision): #left and right are the current bounds; the maximum is between them if (right - left) < absolutePrecision: return (left + right)/2 leftThird = (2*left + right)/3 rightThird = (left + 2*right)/3 if f(leftThird) < f(rightThird): return ternarySearch(f, leftThird, right, absolutePrecision) return ternarySearch(f, left, rightThird, absolutePrecision) i am not asking once again how implement it in java i am asking for example how define function?for example let y=x^+3 yes we can determine it as public static int y(int x){ return x*x+3; } but here return ternarySearch(f, leftThird, right, absolutePrecision) function f does not have argument and how do such?please help me

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• What is the best way to find the digit at n position in a decimal number?

  - by Elijah

  Background I'm working on a symmetric rounding class and I find that I'm stuck with regards to how to best find the number at position x that I will be rounding. I'm sure there is an efficient mathematical way to find the single digit and return it without having to resort to string parsing. Problem Suppose, I have the following (C#) psuedo-code: var position = 3; var value = 102.43587m; // I want this no ? (that is 5) protected static int FindNDigit(decimal value, int position) { // This snippet is what I am searching for } Also, it is worth noting that if my value is a whole number, I will need to return a zero for the result of FindNDigit. Does anyone have any hints on how I should approach this problem? Is this something that is blaringly obvious that I'm missing?

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• How to detect if a certain range resides (partly) within an other range?

  - by Tom

  Lets say I've got two squares and I know their positions, a red and blue square: redTopX; redTopY; redBotX; redBotY; blueTopX; blueTopY; blueBotX; blueBotY; Now, I want to check if square blue resides (partly) within (or around) square red. This can happen in a lot of situations, as you can see in this image I created to illustrate my situation better: Note that there's always only one blue and one red square, I just added multiple so I didn't have to redraw 18 times. My original logic was simple, I'd check all corners of square blue and see if any of them are inside square red: if ( ((redTopX >= blueTopX) && (redTopY >= blueTopY) && (redTopX <= blueBotX) && (redTopY <= blueBotY)) || //top left ((redBotX >= blueTopX) && (redTopY >= blueTopY) && (redBotX <= blueBotX) && (redTopY <= blueBotY)) || //top right ((redTopX >= blueTopX) && (redBotY >= blueTopY) && (redTopX <= blueBotX) && (redBotY <= blueBotY)) || //bottom left ((redBotX >= blueTopX) && (redBotY >= blueTopY) && (redBotX <= blueBotX) && (redBotY <= blueBotY)) //bottom right ) { //blue resides in red } Unfortunately, there are a couple of flaws in this logic. For example, what if red surrounds blue (like in situation 1)? I thought this would be pretty easy but am having trouble coming up with a good way of covering all these situations.. can anyone help me out here? Regards, Tom

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• Get `n` random values between 2 numbers having average `x`

  - by Somnath Muluk

  I want to get n random numbers(e.g n=16)(whole numbers) between 1 to 5(including both) so that average is x. x can be any value between (1, 1.5, 2, 2.5, 3, 3.5, 4, 4.5, 5). I am using PHP. e.g. Suppose I have average x= 3. Then required 16 whole numbers between 1 to 5(including both). like (1,5,3,3,3,3,2,4,2,4,1,5,1,5,3,3) Update: if x=3.5 means average of 16 numbers should be between 3.5 to 4. and if x=4 means average of 16 numbers should be between 4 to 4.5 and if x=5 means all numbers are 5

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• How to find validity of a string of parentheses, curly brackets and square brackets?

  - by Rajendra

  I recently came in contact with this interesting problem. You are given a string containing just the characters '(', ')', '{', '}', '[' and ']', for example, "[{()}]", you need to write a function which will check validity of such an input string, function may be like this: bool isValid(char* s); these brackets have to close in the correct order, for example "()" and "()[]{}" are all valid but "(]", "([)]" and "{{{{" are not! I came out with following O(n) time and O(n) space complexity solution, which works fine: Maintain a stack of characters. Whenever you find opening braces '(', '{' OR '[' push it on the stack. Whenever you find closing braces ')', '}' OR ']' , check if top of stack is corresponding opening bracket, if yes, then pop the stack, else break the loop and return false. Repeat steps 2 - 3 until end of the string. This works, but can we optimize it for space, may be constant extra space, I understand that time complexity cannot be less than O(n) as we have to look at every character. So my question is can we solve this problem in O(1) space?

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• Parsing a comma-separated list

  - by alex

  I have a comma-separated list of values, for example: strins s = "param1=true;param2=4;param3=2.0f;param4=sometext;"; I need a functions: public bool ExtractBool(string parameterName, string @params); public int ExtractInt(string parameterName, string @params); public float ExtractFloat(string parameterName, string @params); public string ExtractString(string parameterName, string @params); Is there a special functions in .net that can help me with csl ? PS: parameter names are equal within a list.

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• Finding cities close to one another using longitude and latitude

  - by Jamie

  Each user in my db is associated to a city (with it's longitude and latitude) How would I go about finding out which cities are close to one another? i.e. in England, Cambridge is fairly close to London. So If I have a user who lives in Cambridge. Users close to them would be users living in close surrounding cities, such as London, Hertford etc. Any ideas how I could go about this? And also, how would I define what is close? i.e. in the UK close would be much closer than if it were in the US as the US is far more spread out. Ideas and suggestions. Also, do you know any services that provide this sort of functionality? Thanks

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• Formula for popularity? (based on "like it", "comments", "views")

  - by paullb

  I have some pages on a website and I have to create an ordering based on "popularity"/"activity" The parameters that I have to use are: views to the page comments made on the page (there is a form at the bottom where uses can make comments) clicks made to the "like it" icon Are there any standards for what a formula for popularity would be? (if not opinions are good too) (initially I thought of views + 10*comments + 10*likeit)

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• in-place permutation of a array follows this rule

  - by Mgccl

  Suppose there is an array, we want to find everything in the odd index, and move it to the end. Everything in the even index move it to the beginning. The relative order of all odd index items and all even index items are preserved. Suppose the values of the array, a[i] = i, n is even. Then we have. 0,1,2,3,4,5,...,n-1 after the operation 0,2,4,6,...,n-2,1,3,5,7,...,n-1 Can this be done in-place and in O(n) time?

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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 to select number of lines from large text files?

  - by MiNdFrEaK

  I was wondering how to select number of lines from a certain text file. As an example: I have a text file containing the following lines: branch 27 : rect id 23400 rect: -115.475609 -115.474907 31.393650 31.411301 branch 28 : rect id 23398 rect: -115.474907 -115.472282 31.411301 31.417351 branch 29 : rect id 23396 rect: -115.472282 -115.468033 31.417351 31.427151 branch 30 : rect id 23394 rect: -115.468033 -115.458733 31.427151 31.438181 Non-Leaf Node: level=1 count=31 address=53 branch 0 : rect id 42 rect: -115.768539 -106.251556 31.425039 31.717550 branch 1 : rect id 50 rect: -109.559479 -106.009361 31.296721 31.775299 branch 2 : rect id 51 rect: -110.937401 -106.226143 31.285870 31.771971 branch 3 : rect id 54 rect: -109.584412 -106.069092 31.285240 31.775230 branch 4 : rect id 56 rect: -109.570961 -106.000954 31.296721 31.780769 branch 5 : rect id 58 rect: -115.806213 -106.366188 31.400450 31.687519 branch 6 : rect id 59 rect: -113.173859 -106.244057 31.297440 31.627750 branch 7 : rect id 60 rect: -115.811478 -106.278252 31.400450 31.679470 branch 8 : rect id 61 rect: -109.953888 -106.020111 31.325319 31.775270 branch 9 : rect id 64 rect: -113.070969 -106.015968 31.331841 31.704750 branch 10 : rect id 68 rect: -113.065689 -107.034576 31.326300 31.770809 branch 11 : rect id 71 rect: -112.333344 -106.059860 31.284081 31.662920 branch 12 : rect id 73 rect: -115.071083 -106.309677 31.267879 31.466850 branch 13 : rect id 74 rect: -116.094414 -106.286308 31.236290 31.424770 branch 14 : rect id 75 rect: -115.423264 -106.286308 31.229691 31.415510 branch 15 : rect id 76 rect: -116.111656 -106.313110 31.259390 31.478300 branch 16 : rect id 77 rect: -116.247467 -106.309677 31.240231 31.451799 branch 17 : rect id 78 rect: -116.170792 -106.094543 31.156429 31.391781 branch 18 : rect id 79 rect: -116.225723 -106.292709 31.239960 31.442850 branch 19 : rect id 80 rect: -116.268013 -105.769913 31.157240 31.378111 branch 20 : rect id 82 rect: -116.215424 -105.827202 31.198441 31.383421 branch 21 : rect id 83 rect: -116.095734 -105.826439 31.197460 31.373819 branch 22 : rect id 84 rect: -115.423264 -105.815018 31.182640 31.368891 branch 23 : rect id 85 rect: -116.221527 -105.776512 31.160931 31.389830 branch 24 : rect id 86 rect: -116.203369 -106.473831 31.168350 31.367611 branch 25 : rect id 87 rect: -115.727631 -106.501587 31.189100 31.395941 branch 26 : rect id 88 rect: -116.237289 -105.790756 31.164780 31.358959 branch 27 : rect id 89 rect: -115.791344 -105.990044 31.072620 31.349529 branch 28 : rect id 90 rect: -115.736847 -106.495079 31.187969 31.376900 branch 29 : rect id 91 rect: -115.721710 -106.000130 31.160351 31.354601 branch 30 : rect id 92 rect: -115.792236 -106.000793 31.166620 31.378811 Leaf Node: level=0 count=21 address=42 branch 0 : rect id 18312 rect: -106.412270 -106.401367 31.704750 31.717550 branch 1 : rect id 18288 rect: -106.278252 -106.253387 31.520321 31.548361 I just want those lines which are in between Non-Leaf Node level=1 to Leaf Node Level=0 and also there are a lot of segments like this and I need them all.

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• C - How to implement Set data structure?

  - by psihodelia

  Is there any tricky way to implement a set data structure (a collection of unique values) in C? All elements in a set will be of the same type and there is a huge RAM memory. As I know, for integers it can be done really fast'N'easy using value-indexed arrays. But I'd like to have a very general Set data type. And it would be nice if a set could include itself.

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• Can my tortoise vs. hare race be improved?

  - by FredOverflow

  Here is my code for detecting cycles in a linked list: do { hare = hare.next(); if (hare == back) return; hare = hare.next(); if (hare == back) return; tortoise = tortoise.next(); } while (tortoise != hare); throw new AssertionError("cyclic linkage"); Is there a way to get rid of the code duplication inside the loop? Am I right in assuming that I don't need a check after making the tortoise take a step forward? As I see it, the tortoise can never reach the end of the list before the hare (contrary to the fable). Any other ways to simplify/beautify this code?

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• How do I find the most complex convex polygon enclosing a set of points?

  - by Marius

  I have a list of (about 200-300) 2d points. I know needs to find the polygon that encloses all of them. The polygon has to be convex, and it should be as complex as possible (i.e. not a rectangular bounding box). It should find this in as low as possible time, but there are no restrictions on memory. You may answer in pseudocode or any language you want to use.

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• Is my program taking too much time to execute?

  - by Conrad C

  I wanted to solve a question from project euleur about finding the largest prime number of a big number. I run my code on a virtual machine on Visual studio 2012, and the code seems froze. When I step into the loop, the code works well, but when I execute it, the console is always there. It is as if the program is still running. Could it be that the program takes time to execute? My Code static void Main(string[] args) { long number = 5; for (long i = 1; i < 600851475143; i++) { if (i % 2 != 0 && i % 1 == 0 && i % i == 0) number = i; } }

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• Given a few strings, how many strings can be lexicographically least by modifying the alphabet?

  - by Jackson W

  Number of strings can be huge as in 30000. Given N strings, output which ones can be lexicographically least after modifying the english alphabet. e.g. acdbe...... for example if the strings were: omm moo mom ommnom "mom" is already lexicographically least with the original english alphabet. we can make the word "omm" least by switching "m" and "o" in the alphabet ("abcdefghijklonmpqrstuvwxyz"). the other ones you cant make lexicographically last, no matter what you do. any fast way to do this? I have no ways to approach this except try every single possible alphabet

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• Google Jam 2009. C. Welcome to Code Jam. Can't understand Dynamic programming

  - by vibneiro

  The original link of the problem is here: https://code.google.com/codejam/contest/90101/dashboard#s=p2&a=2 In simple words we need to find how many times the string S="welcome to code jam" appears as a sub-sequence of given string S, e.g. S="welcome to code jam" T="wweellccoommee to code qps jam" I know the theory but not good at DP in practice. Would you please explain step-by-step process to solve this DP problem on example and why it works?

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• Given a number N, find the number of ways to write it as a sum of two or more consecutive integers

  - by hilal

  Here is the problem (Given a number N, find the number of ways to write it as a sum of two or more consecutive integers) and example 15 = 7+8, 1+2+3+4+5, 4+5+6 I solved with math like that : a + (a + 1) + (a + 2) + (a + 3) + ... + (a + k) = N (k + 1)*a + (1 + 2 + 3 + ... + k) = N (k + 1)a + k(k+1)/2 = N (k + 1)*(2*a + k)/2 = N Then check that if N divisible by (k+1) and (2*a+k) then I can find answer in O(N) time Here is my question how can you solve this by dynamic-programming ? and what is the complexity (O) ? P.S : excuse me, if it is a duplicate question. I searched but I can find

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• Faster way to compare two sets of points in N-dimensional space?

  - by Amit

  List1 contains a high number (~7^10) of N-dimensional points (N <=10), List2 contains the same or fewer number of N-dimensional points (N <=10). My task is this: I want to check which point in List2 is closest (euclidean distance) to a point in List1 for every point in List1 and subsequently perform some operation on it. I have been doing it the simple- the nested loop way when I didn't have more than 50 points in List1, but with 7^10 points, this obviously takes up a lot of time. What is the fastest way to do this? Any concepts from Computational Geometry might help?

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