# Trying Figure Get Code Work Question Download File Implementing Algorithm Given Unordered Q24891622

Been trying to figure out to get this code to work!Question:

Download this file in which you will be implementing analgorithm which, given an unordered List, takes only*O*(*n*) time to return it’s *n*thsmallest element. The easiest (and most natural) way to implementthis algorithm is to use recursion. To receive credit, **youmust implement the following (recursive) algorithm:**

Integer getItemInOrder(*al*, *left*, *right*, *sizeRanking*) { *pivotIndex* ⇐ partialOrder(*al*, *left*, *right*) // Partially order the elements between left and right // The pivot is in its final sorted position if *sizeRanking* = *pivotIndex* return the element in *al* at index *sizeRanking* // Solution found! else if *sizeRanking* < *pivotIndex* return getItemInOrder(*al*, *left*, *pivotIndex* – 1, *sizeRanking*) // The pivot was too large, so recurse with the smaller values else return getItemInOrder(*al*, *pivotIndex* + 1, *right*, *sizeRanking*) // The pivot was too small, so recurse with larger values end if

**Complete the upper getItemInOrder method so that itstarts the recursion (serves as the public method hiding therecursion)**.

Code Given:

package edu.buffalo.cse116;import java.util.ArrayList;public class OrderedItem { /** * Finds the item at the given ranking (from smallest to largest) in the array list WITHOUT sorting the data. This * process is best done using recursion. * * @param al The ArrayList of unordered data from which we will select the nth largest * @param sizeRanking Ranking of the item to be selected. When {@code sizeRanking} is 0, the method needs to return * the smallest item in {@code al}; if {@code sizeRanking} is {@code al.size() – 1}, the largest item is * returned. * @return The {@code sizeRanking}-th smallest item from the array list. */ public static int getItemInOrder(ArrayList<Integer> al, int sizeRanking) { // You can just assume that sizeRanking is between 0 and al.size()-1 (inclusive) } /** * Finds the item at the given ranking (from smallest to largest) in the array list WITHOUT sorting the data. It will * do this by only considering the ArrayList elements at the indices between left and right. Each recursion halves the * number of elements under consideration. * * @param al ArrayList of unordered data to be examined * @param left First index of data which is in consideration * @param right Last index of data which is in consideration * @param sizeRanking Ranking of the item to be selected. When {@code sizeRanking} is 0, the method needs to return * the smallest item in {@code al}; if {@code sizeRanking} is {@code al.size() – 1}, the largest item is * returned. * @return The {@code sizeRanking}-th smallest item from the array list. */ private static int getItemInOrder(ArrayList<Integer> al, int left, int right, int sizeRanking) { // You can just assume that sizeRanking is between 0 and al.size()-1 (inclusive) } /** * This is a helper method which partially orders the array list into left and right parts. The left part of the array * list will contain all of the values smaller than the element initially midway through the range between left and * right. The right part of the array list will contain all of the larger elements. * * @param al ArrayList of data which will be partitioned * @param left Leftmost index to be included in the partition * @param right Rightmost index to be included in the partition * @return The index at which the element originally at pivotIdx ends up. */ private static int partialOrder(ArrayList<Integer> al, int left, int right) { int pivotIdx = (left + right) / 2; int pivot = al.get(pivotIdx); al.set(pivotIdx, al.get(right)); al.set(right, pivot); int storeIdx = left; for (int i = left; i < right; i++ ) { if (al.get(i) < pivot) { int temp = al.get(storeIdx); al.set(storeIdx, al.get(i)); al.set(i, temp); storeIdx += 1; } } al.set(right, al.get(storeIdx)); al.set(storeIdx, pivot); return storeIdx; }}

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