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  • + 0 comments

    if("###".equals(it.next())) break;

  • + 0 comments

    Here is Java Iterator solution - https://programmingoneonone.com/hackerrank-java-iterator-problem-solution.html

  • + 0 comments
    static Iterator func(ArrayList mylist){
          Iterator it=mylist.iterator();
          while(it.hasNext()){
             Object element = it.next();
             if(element instanceof String)
                if(element.equals("###"))
    			break;
    		 }
         return it;
          
       }
    
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    public class Solution {
        @SuppressWarnings({ "unchecked", "rawtypes" })
        public static void main(String[] args) {
            Scanner scanner = new Scanner(System.in);
            int numLenght = scanner.nextInt();
            int strLenght = scanner.nextInt();
            ArrayList list = new ArrayList();
            for (int i = 0; i < numLenght; i++) {
                list.add(scanner.nextInt());
            }
            list.add("###");
            for (int i = 0; i < strLenght; i++) {
                list.add(scanner.next());
            }
            list.stream().filter(s -> s instanceof String && ((String) s).matches("[a-zA-Z]+")).forEach(System.out::println);
            scanner.close();
        }
    }
    
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