Implement the following algorithms in Java:  A) A variant of QUICKSORT which returns without sorting subarrays with fewer than k elements and then uses INSERTION-SORT to sort the entire nearly-sorted array (slide 24). B) A variant of QUICKSORT using the median-of-three partitioning scheme. Slide 24: •Cutoff to INSERTION-SORT (as in MERGE-SORT). Alternatively: −When calling QUICKSORT on a subarray with fewer than k elements, return without sorting the subarray −After the top-level call to QUICKSORT returns, run INSERTION-SORT on the entire array to finish the sorting process −Taking advantage of the fast running time of INSERTION-SORT when its input is “nearly” sorted •Tail call optimisation convert the code so that it makes only one recursive call −Usually good compilers do that for us • Iterative version with the help of an auxiliary stack

Task - 2: Write a java program (AnyTypeMergeSort.java) to implement the Merge Sort algorithm to sort the array containing any type of elements (suppose strings). Your program will have the following method signature: ● ● public void mergeSort(T[] A, int lowerBound, int upperBound) public void merge(T[] A, int lowerBound, int midPoint, int upperBound) Following is the Merge Sort Algorithm MERGE-SORT (A, p,r) 1 if p≥r 2 3 4 5 6 // Merge A[p:q] and A[q + 1:r] into A[p:r]. 7 MERGE(A, p, q, r) 4 Where p and r represent lower (starting index) and upper (ending index) bounds of array A respectively. The Merge Sort calls the following Merge Algorithm that accepts q as the mid of the array along with p and r: MERGE(A, p, q, r) 1 n₂ = q-p+1 2 nR=r-9 3 let L[0:n 5 6 7 8 9 10 11 return ANZEIGEN q = [(p+r)/2] MERGE-SORT(A, p, q) MERGE-SORT (A, q + 1,r) 22 25 26 27 i = 0 j = 0 k = p 12 13 14 15 16 17 18 19 // Having gone through one of L and R entirely, copy the // remainder of the other to the end…

Write the state of the elements of the vector below after each of the first 4 passes of the outermost loop of the insertion sort algorithm. (After the first pass, 2 elements should be sorted. After the second pass, 3 elements should be sorted. And so on.) // index 0 1 2 3 4 5 6 7 vector<int> numbers{29, 17, 3, 94, 46, 8, -4, 12}; insertionSort(numbers); NOTE: Write your answer inside curly braces with numbers separated by commas {29, 17, 3, 94, 46, 8, -4, 12}   after pass 1   after pass 2   after pass 3   after pass 4

in java

Quicksort is a powerful divide-and-conquer sorting algorithm that can be described in just four lines ofpseudocode. The key to Quicksort is the PARTITION(A, p, r) procedure, which inputs elementsptorof array A,and chooses the final element x = A[r] as the pivot element. The output is an array where all elementsto the left ofxare less thanx, and all elements to the right of x are greater than x. In this question, we will use the Lomuto Partition Method from class and assume that the pivot isalwaysthe last (right-most) element of the input array. Question: Let A be an array withn= 2k−1 elements, where k is some positive integer. Determine a formula (in terms of n) for the minimum possible number of total comparisons required by Quicksort, as well as a formula for the maximum possible number of total comparisons required by Quicksort. Use your formulas to show that the running time of Quicksort is O(nlogn) in the best case and O(n2) in the worst case.

Write the full Java code for NaturalMergeSorter.java

A. Can you draw the animation of Merge sort: given an unsorted array of size 10, show the divide andconquer techniqueb. What is Running time for Merge Sort: O(nlogn) such that n is the size of thearray: Reference code attached

Let A = [n, n − 1, n − 2, . . . , 3, 2, 1] be an array where the first n positive integers are listed in decreasing order. Determine whether Heapsort or Quicksort sorts this array faster.For this question, assume the Quicksort pivot is always the right-most element.

Java Quick Sort but make it read the data 10, 7, 8, 9, 1, 5 from a file not an array // Java implementation of QuickSort import java.io.*;    class GFG {        // A utility function to swap two elements     static void swap(int[] arr, int i, int j)     {         int temp = arr[i];         arr[i] = arr[j];         arr[j] = temp;     }        /* This function takes last element as pivot, places        the pivot element at its correct position in sorted        array, and places all smaller (smaller than pivot)        to left of pivot and all greater elements to right        of pivot */     static int partition(int[] arr, int low, int high)     {            // pivot         int pivot = arr[high];            // Index of smaller element and         // indicates the right position         // of pivot found so far         int i = (low - 1);            for (int j = low; j <= high - 1; j++) {                // If current element is smaller             // than the pivot             if…