An unordered array search time complexity is which of the following

This problem has been solved! You'll get a detailed solution from a subject matter expert that helps you learn core concepts. See Answer See Answer See Answer done loading Question:Question 2(20 points) a) Analyze the running time complexity of the following function which finds the kth smallest integer in an unordered array. (15 points) int selectkth (int a[], int k, int n) \{ int i,j, mini, tmp; for (i=0;i Question 2 ( 20 points) a) Analyze the running time complexity of the following function which finds the kth smallest integer in an unordered array. (15 points) int selectkth (int a[], int k, int n) \{ int i , j, mini, tmp; for ( i = 0 ; i < k ; i + + ) { mini = i ; for ( j = i + 1 ; j < n ; j + + ) if ( a [ j ] < a [ mini ]) mini = j ; tmp = a [ i ] ; a [i] = a [ mini ] ; a[mini] = tmp; return a[k-1]; b)If we assume that the input array is always sorted, how can we rewrite the above function to return the kth smallest element of the sorted array? what would be the running time complexity in that case? (5 points) Previous question Next question

3.1 Elementary Symbol Tables Symbol table. The primary purpose of a symbol table is to associate a value with a key. The client can insert key–value pairs into the symbol table with the expectation of later being able to search for the value associated with a given key. API. Here is the API. We consider several design choices for our implementations to make our code consistent, compact, and useful. • Generics. We consider the methods without specifying the types of keys and values being processed, using generics. • Duplicate keys. Only one value is associated with each key (no duplicate keys in a table). When a client puts a key-value pair into a table already containing that key (and an associated value), the new value replaces the old one. These conventions define the associative array abstraction, where you can think of a symbol table as being just like an array, where keys are indices and values are array entries. • Null values. No key can be associated with the value null. This convention is directly tied to our specification in the API that get() should return null for keys not in the table. This convention has two (intended) consequences: First, we can test whether or not the symbol table defines a value associated with a given key by testing whether get() returns null. Second, we can use the operation of calling put() with null as its second (value) argument to implement deletion. • Deletion. Deletion in symbol tables generally involves one of two strategies: lazy ...

Table of Contents • • • • Introduction The Java programming language created by James Gosling in the early 1990s is one of the most popular programming languages. No wonder Java is the second most demanded programming language with around 70K job postings in January 2020, as per Indeed. Over 90% of the Fortune 500 companies still rely on Java for their development projects, and globally there are over 8 million Java developers. So if you get an interview call for the Java developer role and are looking for a quick guide before your interview, then you have come to the right place. The whole blog consists of 100 interview questions divided into three parts: Beginner, Intermediate and Advanced. This article is Part 3 of the Java Interview Questions and Answer Series covering advanced-level Java Interview questions. Go through all the questions to enhance your chances of performing well in the interviews. Advanced Java Interview Questions Q1) What is Encapsulation? Ans 1) Encapsulation in Java is a mechanism of wrapping the data(i.e., variables) and code acting on the data(methods) together as a single unit. The variables of an encapsulated class are hidden from other classes and can be accessed only using the methods of the current class. It is also known as Data hiding. Data hiding can be achieved by making all the instance variables private; The Java bean class is an example of a fully encapsulated class. Consider the following example of an encapsulated class in Java. cla...

Time Complexity of Binary Search Algorithm: Best Case Time Complexity of Binary Search Algorithm: O(1) Best case is when the element is at the middle index of the array. It takes only one comparison to find the target element. So the best case complexity is O(1). Average Case Time Complexity of Binary Search Algorithm: O(log N) Consider array arr[] of length N and element X to be found. There can be two cases: • Case1: Element is present in the array • Case2: Element is not present in the array. There are N Case1 and 1 Case2. So total number of cases = N+1. Now notice the following: • An element at index N/2 can be found in 1 comparison • Elements at index N/4 and 3N/4 can be found in 2 comparisons. • Elements at indices N/8, 3N/8, 5N/8 and 7N/8 can be found in 3 comparisons and so on. Based on this we can conclude that elements that require: • 1 comparison = 1 • 2 comparisons = 2 • 3 comparisons = 4 • x comparisons = 2 x-1 where x belongs to the range [1, logN] because maximum comparisons = maximum time N can be halved = maximum comparisons to reach 1st element = logN. So, total comparisons = 1*(elements requiring 1 comparisons) + 2*(elements requiring 2 comparisons) + . . . + logN*(elements requiring logN comparisons) = 1*1 + 2*2 + 3*4 + . . . + logN * (2 logN-1) = 2 logN * (logN – 1) + 1 = N * (logN – 1) + 1 Total number of cases = N+1. Therefore, the average complexity = ( N*(logN – 1) + 1)/N+1 = N*logN / (N+1) + 1/(N+1). Here the dominant term is N*logN/(N+1) which is...

I just bumped on to this question today and was trying for a solution that is better than O(N) but could not come up with one. Searched through SO but couldn't find this question. Is there any solution better than O(n) or is it a problem that cannot be solved better than that? My initial thought was Binary Search but again for that you need to sort it which is again >n. I also thought of applying quicksort for just the half of the array to which the search element might belong but again we are making n comparisons initially and discarding the other half only later. Am I getting this right or am I looking at the solution in a wrong direction? I was trying for a solution in c++ and no javascript's IndexOf() or C# Array.find() or LINQ's. Usually, we check one element of array in one iteration... which takes n iterations to completely loop through the array... therefore, worst case time complexity becomes O(n). for(int i=0;i

Closed 5 years ago. All tutorials on algorithms show the complexity for the linear search in the unsorted array in the average case as N/2. I understand that the average case means the items in the list are randomly distributed. Can anyone show how I would arrive at N/2 if I have items randomly distributed? Or does it come out of randomly shuffling array bazillion times and recording the number of operations? First assume input is uniformly distributed. More precisely it is $\frac$$ $\begingroup$ @Maximus Notion of Average or expected number of steps has meaning only in context of some probability distribution. In other words, "array should be randomly distributed" means that input comes in a random way, i.e. if you know that input consists of positive integers between 1 and $n$ then randomly distributed means that any permutation on that integers is possible. In my answer I assumed that any permutation has equal probability, known as uniform distribution. $\endgroup$ $\begingroup$ @Maximus For example if know that your input is always $1,2,3, 4,5,6,7,8,9,10$, then you don't need to search over all array. You just need to check if $x 10$, if it is true then $x$ is not found, otherwise found. You just need only one comparison. This is an example of "lack" of randomness. $\endgroup$