
Concept explainers
********************Answer question below:*****************
Write Java program.
Write an
We assume that:
- List elements are inetger
- Input list is already sorted
Sample example:
INPUT:
1 1 2 2 3 3 4 4 5 5 5 6 7 67
OUTPUT:
8
- DriverMain.java:
(Do not change anyparts of the given code, only add more code)
import java.util.*;
import java.util.stream.Collectors;
import java.lang.*;
import java.io.*;
class DriverMain {
public static void main(String args[]) {
Scanner input = new Scanner(System.in);
String str = input.nextLine();
input.close();
int[] arr = Arrays.stream(str.substring(0, str.length()).split("\\s"))
.map(String::trim).mapToInt(Integer::parseInt).toArray();
List<Integer> list = Arrays.stream(arr).boxed().collect(Collectors.toList());
Collections.sort(list);
System.out.println(ProblemSolution.removeDuplicate(list));
}
}
- solution.java:
import java.util.*;
import java.util.stream.Collectors;
import java.lang.*;
import java.io.*;
class ProblemSolution {
public static int removeDuplicate(List<Integer> list) {
}
}

Trending nowThis is a popular solution!
Step by stepSolved in 2 steps with 1 images

- Question 66 Computer Science in javaarrow_forwardHow can I apply this python code wherein I have an input n. Then create a list from 2, n and remove the multiples of n. Example: Input: 10 Output: 2, 3, 4, 5, 6, 7, 8, 9 def createList(n): #Base Case/s #TODO: Add conditions here for your base case/s #if <condition> : #return <value> #Recursive Case/s #TODO: Add conditions here for your recursive case/s #else: #return <operation and recursive call> #remove the line after this once you've completed all the TODO for this function return [] def removeMultiples(x, arr): #Base Case/s #TODO: Add conditions here for your base case/s #if <condition> : #return <value> #Recursive Case/s #TODO: Add conditions here for your recursive case/s #else: #return <operation and recursive call> #remove the line after this once you've completed all the TODO for this function return []arrow_forwardA sequential search of a sorted list can halt when the target is less than a given element in the list. Modify the program to stop when the target becomes less than the current value being compared. In a sorted list, this would indicate that the target is not in the list and searching the remaining values is unnecessary. ----------------------------------------------------------------------------------- """ File: search.py Project 11.1 """ def sequentialSearch(target, lyst): """Returns the position of the target item if found, or -1 otherwise. The lyst is assumed to be sorted in ascending order.""" position = 0 while position < len(lyst): if target == lyst[position]: return position position += 1 return -1 def main(): """Tests with three lists.""" print(sequentialSearch(3, [0, 1, 2, 3, 4])) print(sequentialSearch(3, [0, 1, 2])) # Should stop at second position. print(sequentialSearch(3, [0, 4, 5, 6])) if…arrow_forward
- Database System ConceptsComputer ScienceISBN:9780078022159Author:Abraham Silberschatz Professor, Henry F. Korth, S. SudarshanPublisher:McGraw-Hill EducationStarting Out with Python (4th Edition)Computer ScienceISBN:9780134444321Author:Tony GaddisPublisher:PEARSONDigital Fundamentals (11th Edition)Computer ScienceISBN:9780132737968Author:Thomas L. FloydPublisher:PEARSON
- C How to Program (8th Edition)Computer ScienceISBN:9780133976892Author:Paul J. Deitel, Harvey DeitelPublisher:PEARSONDatabase Systems: Design, Implementation, & Manag...Computer ScienceISBN:9781337627900Author:Carlos Coronel, Steven MorrisPublisher:Cengage LearningProgrammable Logic ControllersComputer ScienceISBN:9780073373843Author:Frank D. PetruzellaPublisher:McGraw-Hill Education





