CodeGym/Help with Java Tasks/removeAllNumbersGreaterThan10 must work with set of any s...

Level 15

Bangalore

30.09.2018
2103views
8comments

removeAllNumbersGreaterThan10 must work with set of any size not satisfying this condition.

Question about the task

Greater than 10? You're not a good fit for us

Java Syntax, Level 8, Lesson 8

Under discussion

Now what's wrong in my code

package com.codegym.task.task08.task0814; import java.util.HashSet; import java.util.Set; import java.util.Iterator; /* Greater than 10? You're not a good fit for us */ public class Solution { public static HashSet<Integer> createSet() { // write your code here HashSet<Integer> set = new HashSet<Integer>(); set.add(1); set.add(2); set.add(3); set.add(4); set.add(5); set.add(6); set.add(7); set.add(8); set.add(9); set.add(10); set.add(11); set.add(12); set.add(13); set.add(14); set.add(15); set.add(16); set.add(17); set.add(18); set.add(19); set.add(20); return set; } public static HashSet<Integer> removeAllNumbersGreaterThan10(HashSet<Integer> set) { // write your code here for(Iterator<Integer> iterator = set.iterator(); iterator.hasNext();){ Integer n = iterator.next(); if(n > 10){ set.remove(n); } } return set; } public static void main(String[] args) { } }

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Sebastian

Level 18 , Malmoe, Sweden

17 April 2019, 12:10

I can think of two things: 1. use a for loop to add items to the set, ex:

for (int i = 0; i < 20; i++){
          setArr.add(i);
      }

2. And an easy way of removing the numbers would be to use the removeIf, ex:

setArr.removeIf(value -> value > 10);

to remove the numbers above 10 and then return it.

hidden #10447638

Level 22

12 June 2019, 16:05

Sebastian, would you care to explain how the removeIf() method works? I just can't understand how the statement works (already tries to read about it) but it seems to be the simplest solution for this task.

Alek

Level 23 , Warsaw, Poland

30 March 2019, 18:10

beside options sugdested in other comments, you can try also more elegant solution using the: set.removeIf(); method :)

satya vath

Level 9 , Mumbai, India

9 December 2018, 10:42

//use this
package com.codegym.task.task08.task0814;

import java.util.HashSet;
import java.util.Set;
import java.util.Iterator;
/*
Greater than 10? You're not a good fit for us

*/

public class Solution {
    public static HashSet<Integer> createSet() {
        // write your code here
        HashSet<Integer> set=new HashSet<Integer>();

       set.add(1);
        set.add(2);
        set.add(3);
        set.add(4);
        set.add(5);
        set.add(11);
        set.add(12);
        set.add(13);
        set.add(14);
        set.add(15);
        set.add(6);
        set.add(7);
        set.add(8);
        set.add(9);
        set.add(10);
        set.add(16);
        set.add(17);
        set.add(18);
        set.add(19);
        set.add(20);
      //  removeAllNumbersGreaterThan10(set);
        return set;

    }

    public static HashSet<Integer> removeAllNumbersGreaterThan10(HashSet<Integer> set) {
        // write your code here
      //  HashSet<Integer> setA = Solution.createSet();

        for (Iterator<Integer> iterator = set.iterator(); iterator.hasNext();){
            Integer r =  iterator.next();
            if(r>10)
                iterator.remove();
        }

        return set;

    }

    public static void main(String[] args) {

    }
}

Ankush Rajput

Level 25 , Bhopal, India

30 September 2018, 12:38

Instead of

set.remove(n);

Do -

iterator.remove();

It will remove the element iterator is pointing at. Here the problem is that you are removing element from set but not changing the iterator position.

Tayyab Mubeen

Level 16 , Lahore, Pakistan

12 October 2018, 14:20

why we use iterator.remove instead of set.remove(n);

Ankush Rajput

Level 25 , Bhopal, India

12 October 2018, 14:58

set, map, arraylist etc do not support concurrent modification. That means you cannot change them while some other task is running over them. Consider above example. When an iterator is going over the set, if you remove or add any items to set, concurrentModificationException will be thrown. For analogy - You can think of it as dealing with two different objects and if you change one without knowledge of other, there will be a problem. Two overcome this problem, deal with only one object i.e Iterator. Now you are iterating and removing both using iterator object only. if you are not iterating over collections using iterator, then remove method of that particular collection will work fine. But if you are accessing the collection using an iterator, then you should use iterator.remove().

Przemek

Level 9 , Poznan, Poland

8 December 2019, 17:12

you can replace:

public static HashSet<Integer> removeAllNumbersGreaterThan10(HashSet<Integer> set) {
    for (Iterator<Integer> iterator = set.iterator(); iterator.hasNext(); ) {
        Integer next = iterator.next();
        if (next > 10) {
            iterator.remove();
        }
    }
    return set;

public static HashSet<Integer> removeAllNumbersGreaterThan10(HashSet<Integer> set) {
      set.removeIf(next -> next > 10);
      return set;

it was the same