How to print common elements in 2 arrays without sorted output in Java

Question

Given two arrays, find the common elements in them.

Example: [1,45,33,23,22,45,233,21], [5,23,45,0,9,23,1,9] => Output: [1,45, 23]

import java.io.*;
import java.util.*;

class Mycode {
    public static void main(String args[]) {
        int a[] = {1, 45, 33, 23, 22, 45, 233, 21};
        int b[] = {5, 23, 45, 0, 9, 45, 1, 9};

        Mycode test = new Mycode();

        test.testNumber(a, b);
    }

    void testNumber(int c[], int d[]) {

        System.out.println(Arrays.toString(c));
        System.out.println(Arrays.toString(d));
        Set<Integer> hset = new HashSet<Integer>();

        for (int i = 0; i < c.length; i++) {
            for (int j = 0; j < d.length; j++) {
                if (c[i] == d[j]) {
                    System.out.println(c[i]);
                    hset.add(c[i]);
                }
            }
        }

        System.out.println(hset);
    }
}

Actual Output: [1, 45, 33, 23, 22, 4, 233, 21] [5, 23, 45, 0, 9, 5, 1, 9] => [1, 23, 45]

Possible duplicate of [How do I get the intersection between two arrays as a new array?](https://stackoverflow.com/questions/13270491/how-do-i-get-the-intersection-between-two-arrays-as-a-new-array) — icarumbas, Oct 27 '17 at 19:12
@icarumbas I wouldn't say that this is a duplicate because it involves preserving the order of the output elements — Juan Carlos Mendoza, Oct 27 '17 at 19:28

Juan Carlos Mendoza · Accepted Answer · 2017-10-27T21:20:49.233

HashSet makes no guarantees of the preservation of the insertion order as indicated in the JavaDoc:

It makes no guarantees as to the iteration order of the set; in particular, it does not guarantee that the order will remain constant over time.

So I would prefer using a LinkedHashSet. This Set implementation guarantees the preservation of the insertion order. From the JavaDocs:

This linked list defines the iteration ordering, which is the order in which elements were inserted into the set (insertion-order)

void testNumber(int c[], int d[]) {
    System.out.println(Arrays.toString(c));
    System.out.println(Arrays.toString(d));

    Set<Integer> hset = new LinkedHashSet<>();

    for (int i = 0; i < c.length; i++) {
        for (int j = 0; j < d.length; j++) {
            if (c[i] == d[j]) {
                hset.add(c[i]);
            }
        }
    }
    System.out.println(hset);
}

Output:

[1, 45, 23]

Thanks Juan Carlos. I really appreciate – Vjay Oct 28 '17 at 04:37 — Vjay, Oct 28 '17 at 04:37

score 0 · Answer 2 · answered Oct 27 '17 at 19:08

0

You can use a HashMap Loop through the first array and add each element with the value false. Loop through the second and if it's in the hashMap then it's common.

answered Oct 27 '17 at 19:08

tiborK

385
1
6

HashSet maybe ? – icarumbas Oct 27 '17 at 19:19
HashSet, HashMap, TreeMap. In this case it's all the same since getting or adding and element from/ to the map has O(1) time complexity. – tiborK Oct 27 '17 at 20:16
HashMap and TreeMap hold two elements Key and Value. What do you think he should use for second element? – icarumbas Oct 27 '17 at 20:20
My initial thought was to use HashMap, but you are right. If he needs only the common elements than HashSet is the best. – tiborK Oct 27 '17 at 21:12

score 0 · Answer 3 · answered Oct 27 '17 at 19:31

What about something like below? This will be more neat since you don't need to have two for loops, and you don't need to sort it before hand.

public class Mycode {
    public static void main(String args[]) {
        Integer a[] = {1, 45, 33, 23, 22, 45, 233, 21};
        Integer b[] = {5, 23, 45, 0, 9, 45, 1, 9};

        Mycode test = new Mycode();

        System.out.println(test.testNumber(a, b));
    }

    public  <T> Set<T>  testNumber(T [] arra1, T [] arr2){
        Set<T> set1 = new HashSet<T>();
        Set<T> interset = new HashSet<T>();

        Collections.addAll(set1, arra1);
        for(T t: arr2){
            if(set1.contains(t))
                interset.add(t);
        }
        return interset;
    }

}

score 0 · Answer 4 · answered Oct 27 '17 at 20:02

0

List<Integer> list = new ArrayList<>(Arrays.asList(array1));
list.retainAll(Arrays.asList(array2));
System.out.println(list);

answered Oct 27 '17 at 20:02

daniu

14,137
4
32
53

score 0 · Answer 5 · answered Jun 19 '20 at 10:52

Solution with time complexity of order N

import java.io.*;
import java.util.*;

public class Main {

    public static void main(String[] args) throws Exception {
        // write your code here

        Scanner scn = new Scanner(System.in);
        int n1 = scn.nextInt();
        int[] arr1 = new int[n1];
        for (int i = 0; i < n1; i++) {
            arr1[i] = scn.nextInt();
        }
        int n2 = scn.nextInt();
        int[] arr2 = new int[n2];
        for (int i = 0; i < n2; i++) {
            arr2[i] = scn.nextInt();
        }
        HashMap < Integer, Integer > freqMap1 = new HashMap < > ();

        for (int i = 0; i < n1; i++) {
            int ch = arr1[i];

            if (freqMap1.containsKey(ch)) {
                int f = freqMap1.get(ch);
                freqMap1.put(ch, f + 1);
            } else {
                freqMap1.put(ch, 1);
            }
        }
        for (int i = 0; i < n2; i++) {
            int ch = arr2[i];

            if (freqMap1.containsKey(ch)) {
                System.out.println(ch);
                freqMap1.remove(ch);
            }
        }

    }

}

How to print common elements in 2 arrays without sorted output in Java

5 Answers5