CareerCup

/**
 * 
 */


import java.util.ArrayList;
import java.util.List;

/**
 * @author singhnirajara
 *
 */
public class SameElement {

	/**
	 * @param args
	 */
	@SuppressWarnings({ "rawtypes", "unchecked" })
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		
		List x = new ArrayList();
		List y = new ArrayList();
		x.add(1L);
		x.add("Niraj");
		x.add(123.52f);
		x.add(123);
		
		y.add(1L);
		y.add('C');
		y.add(123.52f);
		y.add(123d);
		 
		for(Object t:getSameObjects(x,y)){
			System.out.println(t);
		}
		
		
	}
	

	private static <T extends Object> List<T> getSameObjects(List<T> x, List<T> y){
		List<T> xy = new ArrayList<T>();
		for(T t:x){
				if(y.contains(t))xy.add(t);
		 }
			return xy;
	
	}

}

using contains method we can get the same object from list 1

public static <T> Collection <T> intersect (Collection <? extends T> a, Collection <? extends T> b)
	{
	    Collection <T> result = new ArrayList <T> ();

	    for (T t: a)
	    {
	        if (b.contains (t)) result.add (t);
	    }

	    return result;
	}

Use this code find the common elements.

Hash table can solve that in linear time.

use retainAll method of collection

use retainAll method of collection

/// Collection<Object> collection = new ArrayList<Object>();
collection.add("A");
collection.add("B");
collection.add("C");
collection.add("D");

Collection<Object> collection2 = new ArrayList<Object>();
collection2.add("A");
collection2.add("B");
collection2.add("C");
collection2.add("E");

collection.retainAll(collection2);

for (Object str : collection) {
System.out.println(str);
} ///

Collection<Object> collection = new ArrayList<Object>();
		collection.add("A");
		collection.add("B");
		collection.add("C");
		collection.add("D");
		
		Collection<Object> collection2 = new ArrayList<Object>();
		collection2.add("A");
		collection2.add("B");
		collection2.add("C");
		collection2.add("E");
		
		collection.retainAll(collection2);
			
		for (Object str : collection) {
			System.out.println(str);
		}

Won't the solutions containing retainAll() or contains() have time complexity O(n^2). Why not use HashMap. Use a HashMap to store the first List and then while taking second List as input, just compare in HashMap whether element is already there or not.
Time complexity: O(n)