I NEED THIS IN JAVA

Merge-sort and quicksort, are two advanced sorting algorithms which can obtain an average running time complexity of O(n log n). Usually, the actual sorting time depends on the way the elements arranged themselves (We assume the sorting is in an ascending order in this question). Counting the inversions in an array is a way to measure the order. An inversion is a pair of elements (a_i, a_j) such that i a_j. For example, the array {2, 4, 1, 3, 5} has three inversions (2, 1), (4, 1), (4, 3). For a sorted array, the inversion number is 0. (a) Implement a method with time complexity of O(n^2) to count how many inversions there are for a given array. Then print out the number of inversions for array {47, 24, 83, 78, 36, 17, 96, 55}. (b) Through using the divide and conquer technique, you can improve the time complexity of counting inversions to O(n log n). Implement a method to count inversions by modifying the code of merge-sort. Then print out the number of inversions for array {96, 83, 78, 55, 47, 36, 24, 17}. (c) In Quicksort, the element, chosen as the pivot, can be any element in an array. Pick the first element as the pivot. For array {47, 24, 83, 78, 36, 17, 96, 55}, write the numbers in the array after each partitioning step. Taking the first, partitioning step for example, your answer should be 47 (as the pivot), {36, 24, 17, 47, 78, 83, 96, 55} (as the array after the partitioning step). (d) In the real world, our unsorted data is not necessarily in a random order. The data could be almost sorted (i.e., its inversion number is small). In this situation, using neither the first element nor the last element, as the pivot in Quicksort is a good choice. Because the two parts divided by the pivot element is quite uneven, this will make the Quicksort algorithm close to the worst case (O(n^2) in complexity). To avoid this situation, we can choose the pivot more wisely by picking a random element in the array. Implement a quicksort method by using a random element in the array as a pivot