QuickJava

An integer array with unique elements has the following property – elements initially are in increasing order till a point after which they start to decrease. Implement a function to find the index of the maximum element in the array in less than linear time, i.e., O(n). Example: Input: array = {1, 3, 5, 7, 9, 8, 6, 4} Output: max index = 4 Max Index : 4.  Max Element value is : 9

In this programming problem and the next you'll code up the greedy algorithms from lecture for minimizing the weighted sum of completion times.. This file describes a set of jobs with positive and integral weights and lengths. It has the format [number_of_jobs] [job_1_weight] [job_1_length] [job_2_weight] [job_2_length] ... For example, the third line of the file is "74 59", indicating that the second job has weight 74 and length 59. You should NOT assume that edge weights or lengths are distinct. Your task in this problem is to run the greedy algorithm that schedules jobs in decreasing order of the difference (weight - length). This algorithm is not always optimal. IMPORTANT: if two jobs have equal difference (weight - length), you should schedule the job with higher weight first. Beware: if you break ties in a different way, you are likely to get the wrong answer. You should report the sum of weighted completion times of the resulting schedule --- a posi...

once  i had an interview with a tech startup;  there were few programming problems; i am sharing one from that interview. Interview Location :Bangalore Company : A product based start up that works on search and big data. Round : 2 nd Level: semi difficult Program & Algo written on : white board Problem statement:   Given an array like A B C A A Z K I A Z Sort   the array based on frequency of character, if frequency is same sort them based on ASCII. {a=4,b=1,c=1,z=2,k=1,i=1} Out put : A A A A Z Z B C I K What i did  during that interview i am writing here: Algo To solve the problem:              1. Find their frequency  using hashtable.       2.  Prepare a class with {data : frequency } and put them in a list.       3. S...

Already we have a nice article about about problems with double check locking. we will see two nice quick workarounds  and its pros and cons . Double-checked locking: Clever, but broken and Double-Checked Locking is Broken . before getting into solution, classic singleton uses the following signature for get instance:      public static  Resource getResource() {     if (resource == null)       resource = new Resource();     return resource;   } so the above code will only create the new resource object if  no instance of resource is found. This mechanism is  called lazy loading more preciously need based loading we only load resources if and only if it is required. Benefit associated with this approach is performance  improvement, if resource creation is a heavy process and we do not want it to be carried when class is loading t  then this classic approach does make sense...

Simple simulation of  muli-threading : few threads will read data from different sources and reduce them to get maximum from all the sources, An example of double check locking. lets make a  prototype of above problem. we will not go and solve it for generic case, we will simply create two thread thread1 and thread2;  this two threads will read data from two different sources and will will use a shared variable to get maximum of two data sets.  we will not go into details about what is singleton and double check locking, may be we see them in future post. in a sentence singleton is a design pattern that enable us to have only one     instantiation  of an object. that means we can't create  object of that class using new when and when ever we need to have object. So, our implementation logic goes as follows       1. start both the threads               find the max in each ...