iven an array of integers, return an array with all duplicates removed. public class RemoveDuplicatesFromArray{ public static int[] solution(int[] nums){ // ↓↓↓↓ your code goes here ↓↓↓↓ return new int[0]; } }
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Given an array of integers, return an array with all duplicates removed.
public class RemoveDuplicatesFromArray{
public static int[] solution(int[] nums){
// ↓↓↓↓ your code goes here ↓↓↓↓
return new int[0];
}
}
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- (a) Write a method public static void insert(int[] a, int n, int x) that inserts x in order among the first n elements of a, assuming these elements are arranged in ascending order. Do NOT use arraylists. x is the last element in a. n does not include x. (b) Using the insert method from Part (a), write a recursive implementation of Insertion Sort.Given an ArrayList of integers, return an ArrayList where all the odds are before the evens in the order that they occur. import java.util.ArrayList;public class EvenAfterOdds{public static ArrayList<Integer> solution(ArrayList<Integer> nums){// ↓↓↓↓ your code goes here ↓↓↓↓return new ArrayList<>();}}Write a Java program to find the median of all elements in an array of integers. The median is the middle value in an ordered set of numbers. If the number of elements is even, the median is the average of the two middle values. Crease a class name Task5. There should be a main method and a “findMedian” method to find out the median of an input array.
- Write method findZero, as started below. findZero should return the index of the first element of array A that contains the value zero, starting from position 0 up through position pos. If no element of A from position 0 to position pos contains the value zero, then findZero should return –1. For example:(in java) Array (A) Position (pos) Value returned by findZero(A, pos) 1 0 2 5 6 0 -1 1 0 2 5 6 1 1 1 0 2 5 6 2 1 1 0 2 0 6 2 1 1 0 2 0 6 4 1 1 2 3 4 5 4 -1 Complete method findZero below. Assume that it is called only with values that satisfy its assumptions as stated below. // Description: Returns smallest index k // such that (0 <= k <= pos) and (A[k] == 0) // Assumptions: 0 <= pos < A.length public int findZero(int[] A, int pos) {Write a java method called SearchKey that return the indices of the first and the last occurences of an element k in an array.The function should Return [-1,1] in case the element was not foundFOR JAVA Write a Java method that takes two 2 dimensional int arrays (a and b) and a 2 dimensional boolean array (c) where all three arrays have the same size. Your method should return an array (result) such that, if a partic- ular element in c is true, then the corresponding (same indexed) element in result must be the multiplication of the the corresponding elements of a and b. If the element in c is false, then the the corresponding element in result must be the negative of multiplication of the the corresponding element in a and b. For example ifa = {{1,2,3},{4,5}},b = {{1,2,1},{0,2}} andc = {{true,false,true},{false,true}}, then the result should be result={{1, -4, 3}{0, 10}}.
- For this problem I have a java code: import java.util.Arrays; import java.util.Scanner; public class MergeSortWeightedSumInversion { public static long merge(int arr[], int l, int mid, int r) { long weight = 0; int n1 = mid-l+1; int n2 = r-mid; int[] a = new int[n1]; int[] b = new int[n2]; for(int i = 0; i < n1; i++) { a[i] = arr[l+i]; } for(int i = 0; i < n2; i++) { b[i] = arr[mid+i+1]; } int i = 0, j = 0, k = l; while(i<n1 && j<n2) { if(a[i] <= b[j]) { arr[k] = a[i]; k++; i++; } else { weight += (a[i] + b[j]); arr[k] = b[j]; k++; j++; } } while(i<n1) { arr[k] = a[i]; k++; i++; } while(j<n2) { arr[k] = b[j]; k++; j++; } return weight; } public static long mergeSort(int arr[], int l, int r) { long weight = 0; if(l < r) { int mid = (l+r)/2; weight += mergeSort(arr, l, mid); weight += mergeSort(arr,mid+1,r); weight += merge(arr,l,mid,r); } return weight; } public static void main(String[] args) { Scanner scan = new Scanner(System.in); int n; n =…Given an array with n objects colored red, white or blue, sort them so that objects of the same color are adjacent, with the colors in the order red, white and blue. Here, we will use the integers 0, 1, and 2 to represent the color red, white, and blue respectively. Please make the solution at least as efficient as O(nlog2n). public class Sort { public void sortColors(int[] nums) { // your code goes here } }FOR JAVA Write a method that takes an integer array and returns the difference between the sum of even indexed elements and the sum of odd indexed elements in that array.
- Given a non-negative integer N, the task is to find the Nth row of Pascal’s Triangle. import java.util.ArrayList;public class NthRowPascal{public static ArrayList<Integer> solution(int nth){// ↓↓↓↓ your code goes here ↓↓↓↓return new ArrayList<>();}}Given an array of numbers, and an integer, find the last index that that integer appears in the array. If the number is not found, return -1. import java.util.ArrayList;public class LastIndexFound{public static int solution(ArrayList<Integer> nums, int numToFind){// ↓↓↓↓ your code goes here ↓↓↓↓return 0;}}code: import java.util.Random;import java.util.ArrayList;import java.util.Scanner;public class WeekendOrders {/*** * This method takes as input the scanner instance created from main and* asks the user for input on the number of orders placed on each of the * days on the long weekend (Saturday, Sunday, and Monday). * Put the number of orders for each day in an array of fixed size* * * @param scnr (an instance of Scanner)* @return an array of int which includes the number of orders* placed on each day, with one day per index of the array*/public static int[] getUserInput(Scanner scnr){// DO NOT MODIFY THIS LINEint[] userInput = new int[3];// TODO: Prompt user for input on number of orders on each day// and store in userInputreturn userInput; }/*** * This method appends all the new orders for the given day onto the * arraylist. It first generates a random int between 0-2 (inclusive).* If the int is 0: it adds "Burger" to the prevOrders list* If the int is 1: it adds "Fries" to the…