Find common elements in two ArrayLists in Java

Prerequisite: ArrayList in Java

Given two ArrayLists, the task is to print all common elements in both the ArrayLists in Java .

Examples:

Input: List1 = ["Hii", "Geeks", "for", "Geeks"], 
       List2 = ["Hii", "Geeks", "Gaurav"]
Output: [Hii, Geeks, Geeks]

Input: List1 = ["a", "b", "c", "d", "e", "f"], 
       List2 = ["b", "d", "e", "h", "g", "c"]
Output:[b, c, d, e]

1. Using Collections.retainAll() method

   Syntax:

   Collections1.retainAll(Collections2)
   
   This method keeps only the common elements
   of both Collection in Collection1.
   

   Approach:

   1. Get the two ArrayLists.
   2. Find the common elements in both the Lists using Collection.retainAll() method. This method keeps only the common elements of both Collection in Collection1.
   3. The List 1 now contains the common elements only.

   Below is the implementation of the above approach:
   Program: By modifying the contents of List1.

   // Java Program to find common elements
   // in two ArrayLists
   // Using retainAll() method
   
   // import ArrayList package
   import java.util.ArrayList;
   
   public class GFG {
   
       // main method
       public static void main(String[] args)
       {
   
           // create ArrayList list1
           ArrayList<String>
               list1 = new ArrayList<String>();
   
           // Add values in ArrayList
           list1.add("Hii");
           list1.add("Geeks");
           list1.add("for");
           list1.add("Geeks");
   
           // print list 1
           System.out.println("List1: "
                              + list1);
   
           // Create ArrayList list2
           ArrayList<String>
               list2 = new ArrayList<String>();
   
           // Add values in ArrayList
           list2.add("Hii");
           list2.add("Geeks");
           list2.add("Gaurav");
   
           // print list 2
           System.out.println("List2: "
                              + list2);
   
           // Find the common elements
           list1.retainAll(list2);
   
           // print list 1
           System.out.println("Common elements: "
                              + list1);
       }
   }
   Output:

   List1: [Hii, Geeks, for, Geeks]
   List2: [Hii, Geeks, Gaurav]
   Common elements: [Hii, Geeks, Geeks]
   

   Program 2: By retaining the contents of List1.

   // Java Program to find common elements
   // in two ArrayLists
   // Using retainAll() method
   
   // import ArrayList package
   import java.util.ArrayList;
   
   public class GFG {
   
       // main method
       public static void main(String[] args)
       {
   
           // create ArrayList list1
           ArrayList<String>
               list1 = new ArrayList<String>();
   
           // Add values in ArrayList
           list1.add("Hii");
           list1.add("Geeks");
           list1.add("for");
           list1.add("Geeks");
   
           // print list 1
           System.out.println("List1: "
                              + list1);
   
           // Create ArrayList list2
           ArrayList<String>
               list2 = new ArrayList<String>();
   
           // Add values in ArrayList
           list2.add("Hii");
           list2.add("Geeks");
           list2.add("Gaurav");
   
           // print list 2
           System.out.println("List2: "
                              + list2);
   
           // Create ArrayList list3
           ArrayList<String>
               list3 = new ArrayList<String>(list1);
   
           // Store the comparison output
           // in ArrayList list3
           list3.retainAll(list2);
   
           // print list 3
           System.out.println("Common elements: "
                              + list3);
       }
   }
   Output:

   List1: [Hii, Geeks, for, Geeks]
   List2: [Hii, Geeks, Gaurav]
   Common elements: [Hii, Geeks, Geeks]
   

2. Using Stream filter

   Syntax:

   list1.stream()
       .filter(list2::contains)
       .collect(Collectors
       .toList()));
   
   This method returns element if found in second list.
   

   Approach:

   1. First create two ArrayList and add values of list.
   2. Convert the ArrayList to Stream using stream() method.
   3. Set the filter condition to be distinct using contains() method.
   4. Collect the filtered values as List using collect() method. This list will be return common element in both list.
   5. Print list3

   Below is the implementation of the above approach:

   Program:

   // Java Program to find common elements
   // in two ArrayLists
   // Using  Stream filter method
   
   // import ArrayList package
   import java.util.*;
   import java.util.stream.*;
   
   public class GFG {
   
       // main method
       public static void main(String[] args)
       {
   
           // create ArrayList list1
           ArrayList<String>
               list1 = new ArrayList<String>();
   
           // Add values in ArrayList
           list1.add("Hii");
           list1.add("Geeks");
           list1.add("for");
           list1.add("Geeks");
   
           // print list 1
           System.out.println("List1: "
                              + list1);
   
           // Create ArrayList list2
           ArrayList<String>
               list2 = new ArrayList<String>();
   
           // Add values in ArrayList
           list2.add("Hii");
           list2.add("Geeks");
           list2.add("Gaurav");
   
           // print list 2
           System.out.println("List2: "
                              + list2);
   
           // Find common elements
           System.out.print("Common elements: ");
           System.out.println(list1.stream()
                                  .filter(list2::contains)
                                  .collect(Collectors
                                               .toList()));
       }
   }
   Output:

   List1: [Hii, Geeks, for, Geeks]
   List2: [Hii, Geeks, Gaurav]
   Common elements: [Hii, Geeks, Geeks]
   

3. Naive approach:
   1. First create two ArrayList and add values of list.
   2. Create a temporary ArrayList to contain common elements.
   3. Iterate through the list1 and check if that element is present in the list2 using ArrayList.contains() method.
   4. If found, add it to the list3
   5. Print the common elements from list3

   Below is the implementation of the above approach:

   // Java Program to find common elements
   // in two ArrayLists
   // Using  Stream filter method
   
   // import ArrayList package
   import java.util.ArrayList;
   
   public class GFG {
   
       // main method
       public static void main(String[] args)
       {
   
           // create ArrayList list1
           ArrayList<String>
               list1 = new ArrayList<String>();
   
           // Add values in ArrayList
           list1.add("Hii");
           list1.add("Geeks");
           list1.add("for");
           list1.add("Geeks");
   
           // print list 1
           System.out.println("List1: "
                              + list1);
   
           // Create ArrayList list2
           ArrayList<String>
               list2 = new ArrayList<String>();
   
           // Add values in ArrayList
           list2.add("Hii");
           list2.add("Geeks");
           list2.add("Gaurav");
   
           // print list 2
           System.out.println("List2: "
                              + list2);
   
           // Create ArrayList list3
           ArrayList<String>
               list3 = new ArrayList<String>();
   
           // Find common elements
           // while iterating through list1
           for (String temp : list1) {
   
               // Check if the element is
               // present in list2 or not
               if (list2.contains(temp)) {
   
                   // Since present, add it to list3
                   list3.add(temp);
               }
           }
   
           // print common elements from list 3
           System.out.println("Common elements: "
                              + list3);
       }
   }
   Output:

   List1: [Hii, Geeks, for, Geeks]
   List2: [Hii, Geeks, Gaurav]
   Common elements: [Hii, Geeks, Geeks]