Answered: Given the following array, what is the…

Given the following array, what is the content of the array after Two levels of merge function calls applied in Merge Sort with function call MergeSort(array, 0, 5); array: 44 12 50 3 40 23

C++ Programming: From Problem Analysis to Program Design

8th Edition

ISBN:9781337102087

Author:D. S. Malik

Publisher:D. S. Malik

Chapter12: Points, Classes, Virtual Functions And Abstract Classes

Section: Chapter Questions

Problem 29SA

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The program is the same as shown at the end of the Merge sort section, with the following changes: Numbers are entered by a user in a separate helper function, ReadNums(), instead of defining a specific array in main(). The first number is how many integers to be sorted, and the rest are the integers. Output of the array has been moved to the helper function PrintNums(). An output has been added to MergeSort(), showing the indices that will be passed to the recursive function calls. Add code to the merge sort algorithm to count the number of comparisons performed. Add code at the end of main() that outputs "comparisons: " followed by the number of comparisons performed (Ex: "comparisons: 12") Hint: Use a global variable to count the comparisons. Note: Take special care to look at the output of each test to better understand the merge sort algorithm.
Below is how the arrays are represented ARRAY1[] = [1, 5, 6, 6, 9, 9, 9, 11, 11, 21] Here length of ARRAY1 is m. ARRAY2[] = [6, 6, 9, 11, 21, 21, 21] Here length of ARRAY2 is n. Array to be returned would be: ARRAY[] = [6, 9, 11, 21] ATTN : Further, please be reminded that you cannot use library functions to either sort and or perform the de-duplication operation. solve the problem in two ways In a separate implementation, code up a solution in such a way that your solution solves the problem with O(nlog(m)) time complexity 2 or O(mlog(n)) time complexity. Here log means to the base of 2. I’m sure you already know that the hint is to use Binary Search. In the form of sentences, as a comment in your code (at the bottom of your Solution2), you are required to suggest how can Solution2 be improved by leveraging the fact that both the arrays are already sorted. Suggest a solution so that your suggested solution can run linearly with O(m + n) time complexity. Your suggestion should be no…
An array with any number of elements is said to be repeating if any two or more of the elements are appearing in sequential an indexes after each other.Write a C++ function that accepts an integer array and returns 1 or “Repeating” if it is a repeating array, otherwise it returns 0 or “Not Repeating”. For Instance:2, 3, 5, 6, 7, 7, 7, 6, 8, 10 - Repeating2, 3, 4, 6, 4, 6, 4, 7, 8, 9 - Not repeating
this function c++ accept vector , replace this vector to array void merge_sort_4(vector<int> lst,int start,int end){if (start < end){int quarter1 = (start + end) / 4;int quarter2 = (start + end) / 2;int quarter3 = (end - quarter1 - 1); merge_sort_4(lst, start, quarter1);merge_sort_4(lst, quarter1 + 1, quarter2);merge_sort_4(lst, quarter2 + 1, quarter3);merge_sort_4(lst, quarter3 + 1, end); merge4(lst, start, quarter1, quarter2, quarter3, end);}}void merge4(vector<int> lst,int start,int q1,int q2,int q3,int end){vector<int> first_q_list;for(int i=start;i<q1+1;i++)first_q_list.push_back(lst[i]);vector<int> sec_q_list;for(int i=q1+1;i<q2+1;i++)sec_q_list.push_back(lst[i]);vector<int> third_q_list;for(int i=q2+1;i<q3+1;i++)third_q_list.push_back(lst[i]);vector<int> last_q_list;for(int i=q1+1;i<end+1;i++)last_q_list.push_back(lst[i]);int…
Java Merge Sort but make it read the data 12, 11, 13, 5, 6, 7 from a file not an array /* Java program for Merge Sort */ class MergeSort { // Merges two subarrays of arr[]. // First subarray is arr[l..m] // Second subarray is arr[m+1..r] void merge(int arr[], int l, int m, int r) { // Find sizes of two subarrays to be merged int n1 = m - l + 1; int n2 = r - m; /* Create temp arrays */ int L[] = new int[n1]; int R[] = new int[n2]; /*Copy data to temp arrays*/ for (int i = 0; i < n1; ++i) L[i] = arr[l + i]; for (int j = 0; j < n2; ++j) R[j] = arr[m + 1 + j]; /* Merge the temp arrays */ // Initial indexes of first and second subarrays int i = 0, j = 0; // Initial index of merged subarray array int k = l; while (i < n1 && j < n2) { if (L[i] <= R[j]) { arr[k] = L[i];…
Which one of the following function prototype can possible do the following? A function with one reference parameter is a pointer to an integer. The function allocates a dynamic array of n integers, making the pointer point to this new array. It then fills the array with 0 through n - 1. Group of answer choices void exercise(int*& p, size_t n); void exercise(const int*& p, size_t n); void exercise(const int p, size_t n); void exercise(int& p, size_t n); void exercise( int p, size_t n);
Please help me with this: using js create an array of 30 random numbers that range between 1and 100. And yet again, write a function that will receive a number from the userand determine if that number exists in the array or not. But this time, start bySORTING your input list. After a sort, the list in problem 1 is as follows:[2, 2, 3, 5, 12, 14, 14, 15, 23, 36, 39, 41, 44, 44, 45, 48,49, 50, 52, 52, 59, 71, 81, 82, 88, 89, 89, 93, 96, 97] Approach: Implement a method called findC(x, A, i, j), where x is the number we arelooking for in array A, the first index of the array is i and the last index is j. We wantto determine whether x exists in A anywhere between index i and index j. Your firstcall to this method will therefore look like this: findC(x, A, 0, A.length-1). In the body of your function, compare x with the item that is in the middle of thearray, as you did before. As before, call the middle of index of the array mid. But thistime, if x<=a[mid], recursively call your…
Using following initialization of array allStudent[5], A)Create a sorted linked list structure using array allStudent[5] by writing a function named createSortLL() using field studNr with a head node headNum. B)Using linked list structure obtained at step (A) ,list only students whose grades are greater than average grade of 5 students by writing list() function with a head node headNum. struct studentInfo { int studNr; char name[12]; int grade; }; struct node // NODE STRUCTURE { struct studentInfo data; struct node *link; }; int main() { structure studentInfo allStudent[5]={ 47790, “MUHSIN”, 77, 63582, ”AYSE”, 85, 67434, “OSMAN”, 100, 31229, ”ZEKI”, 65,…
In the Mergesort, the idea is to merge two arrays or list of two numbers, where each of those arrays are sorted. The merged array becomes fully sorted. The following code snippet makes the process of merge happen. Assume an array A={4, 11, 2, 9, 6}, where we want to split that array and then merge it. If we call the function "merge(A, 0, 2, 4)", how would the array A look like after the merge operation?* 4, 11, 2, 9, 6 2, 4, 6, 9, 11 4, 11, 9, 2, 6 4, 11, 6, 9, 2 11, 9, 6, 4, 2 4, 9, 6, 11, 2
Write a program to compute the exact value of the number of array accesses usedby top-down mergesort and by bottom-up mergesort. Use your program to plot the values for N from 1 to 512, and to compare the exact values with the upper bound 6N lg N
(a) In what cases should Merge Sort be used? In what cases is it not efficient? Explain.(b) The following is an implementation of merge and mergeSort - Making sure to take note of the added print statement in merge, what would the following call in main print to the screen? int[] a = {4, 3, 2, 1};int[] temp = new int[a.length];mergeSort(a, 0, a.length-1, temp);
2.) Revise the function quickSort so that it always chooses the last item in the array as the pivot. Add a counter to the function partition that counts the number of comparisons that are made. Compare the behavior of the revised function with the original one (pivot should be selected using sortFirstMiddleLast algorithm), using arrays of various sizes. At what size array does the difference in the number of comparisons become significant? For which pivot selection strategy does the difference in the number of comparisons become significant? Compare your analysis with the actual running times and counter as a function of the input size n = 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 8192, 16384 <time.h> and clock() function. For comparison, you need to create two tables for execution times and a counter for both algorithms