finding common elements in two integer arrays java

Question

The code returns 0 and the common numbers more than once. I want it to return an array with the common numbers once! So how do I return an array with numbers that are common to both arrays. I want to return {2,7,4} - something like this. I keep getting out of bounds exceptions when I try to return an array. Thanks, Barry

public class Test {
    public int findCommonElement(int[] a, int[] b){
        int counter=0;
        int temp= 0;
        int tempCounter = 0;
        for(int i=0; i<a.length; i++){
            temp=a[i];
            tempCounter=0;
            for(int j=0; j<b.length; j++){
                if (temp==b[j]){
                    tempCounter++;  
                }

            }

            if (tempCounter == 1) {
                temp = a[i];

                counter++;

                System.out.println(temp);

            }

        }

        return 0;
    }

    public static void main(String []args){
        int myArray[] = {2,2,7,7,2,1,5,4,5,1,1};
        int myArray2[] = {2,3,4,7,10};


        Test hello = new Test ();
        System.out.println(hello.findCommonElement(myArray, myArray2));

    }
}

Multiple problems here. First, your function returns 0. Second, if you want to return more than one number you must change findCommonElement to return an Array or List. — mdewit
– mdewit, Commented Nov 16, 2016 at 13:11
Also, the counter variable in findCommonElement currently does nothing. Why does the if(tempCounter == 1) statement only check whether the tempCounter is 1? Should it not be if(tempCounter > 0)? — mdewit
– mdewit, Commented Nov 16, 2016 at 13:17
Please indent you code properly — I have trouble reading it. — Anonymous
– Anonymous, Commented Nov 16, 2016 at 13:36

nail fei · Accepted Answer · 2016-11-16 13:18:40Z

7

an alternative solution for findCommonElement method

public int[] findCommonElement(int[] a, int[] b){
    List<Integer> array = new LinkedList<Integer>();
    Set<Integer> set = new HashSet<Integer>();
    for(int ele:a){
        set.add(ele);
    }

    for(int ele:b){
        if(set.contains(ele)){
            array.add(ele);
        }
    }

    int[] arr = new int[array.size()];
    for(int i = 0; i < array.size();i++){
        arr[i] = array.get(i);
    }
    return arr;
}

answered Nov 16, 2016 at 13:18

nail fei

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1 Comment

deFreitas Over a year ago

I think that solution is much better because has less complexity, is it n+n+n that is much better than nxn

talshahar · Accepted Answer · 2017-10-16 04:38:34Z

1

here is an O(m+n) solution:

static ArrayList<Integer> commonElements(int[] array1, int[] array2) {
    int p1 = 0;
    int p2 = 0;
    ArrayList<Integer> common = new ArrayList<Integer>();

    while(true) {
        if (array1[p1] == array2[p2]) {
            common.add(array1[p1]);
        }
        if (p1 == array1.length - 1 || p2 == array2.length - 1) break;
        if (array1[p1 + 1] < array2[p2 + 1]) {
            p1++;
        } else {
            p2++;
        }
    }
    return common;
}

answered Oct 16, 2017 at 4:38

talshahar

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3 Comments

Akshay Over a year ago

Are you considering both the arrays to be pre-sorted.

Ashish Agrawal Yodlee Over a year ago

this is working only for sorted array. int []array1= {2,3,5,6,7,9,11,22}; int []array2= {22,2,4,5,6,8,10,12}; ArrayList<Integer> list = commonElements(array1, array2); System.out.println(list); Output : [2, 5, 6]

Ayush Kumar Over a year ago

I've made some improvements in this and posted it. Covers the edge cases. stackoverflow.com/a/67289502/6438896

Mrudav Shukla · Accepted Answer · 2016-11-16 15:02:17Z

0

Basically the number of common element from the two elements will be dynamic. Hence if you try to put common elements into an array then it won't be possible as you need to declare size of this array (which in this case will be dynamic).

Consider using list. I've tried to keep the logic as simple as possible along with comprehensive variable names.

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

    public class Test {

    public static void main(String[] args) {

        int myArray[] = { 2, 2, 7, 7, 2, 1, 5, 4, 5, 1, 1 };
        int myArray2[] = { 2, 3, 4, 7, 10 };

        Test hello = new Test();
        System.out.println(hello.findCommonElement(myArray, myArray2));
    }
    /**
     * 
     * @param a
     * @param b
     * @return commonElements
     */
    public List<Integer> findCommonElement(int[] a, int[] b) {

        List<Integer> commonElements = new ArrayList<Integer>();

        for(int i = 0; i < a.length ;i++) {
            for(int j = 0; j< b.length ; j++) {
                    if(a[i] == b[j]) {  
                    //Check if the list already contains the common element
                        if(!commonElements.contains(a[i])) {
                            //add the common element into the list
                            commonElements.add(a[i]);
                        }
                    }
            }
        }
        return commonElements;
    }
}

edited Nov 16, 2016 at 15:02

answered Nov 16, 2016 at 14:00

Mrudav Shukla

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4 Comments

Barry Reeves Over a year ago

Thanks nice simple solution!

Mrudav Shukla Over a year ago

Thanks Barry! Keep exploring!👍🏻

Akki Over a year ago

@deFreitas I can think of O(n) solution, where I am use a hashmap for one array and while traversing other i just do map.containsKey() everytime.... I am trying to think if there is any better way to do that...other than creating O(n) space

Ayush Kumar Over a year ago

This would be a very costly approach with O(n*n) complexity. Check out my solution with O(n) complexity and without the use of HashSets; thereby saving memory. stackoverflow.com/a/67289502/6438896

Anonymous · Accepted Answer · 2016-11-16 13:57:26Z

0

I see the following issues with your code:

I want it to return an array with the common numbers once!

So you need to declare that your method returns an array. Add square brackets:

public int[] findCommonElement(int[] a, int[] b) {

Inside your method you must also keep track of all common elements found so far. You may use a new array or more conveniently an ArrayList or even more conveniently a HashSet (since the set automatically eliminates duplicates so you get each common number only once). I think you meant the counter variable to keep track of the number of elements in the new array, only the array is not there yet.

You check:

if (tempCounter == 1) {

This is not right if the number occurs more than once in b. Instead do

if (tempCounter > 0) {

As I said, you need a way to filter away duplicates from a so you don’t get [2, 2, 7, 7, 2, 4] but only [2, 7, 4]. You may use a set I as I mentioned, or you may use ArrayList.contains() or introduce another loop to check whether the number is already in your array of common numbers. Just don’t add it again if it is.

Finally, to print the contents of an array, use Arrays.toString():

    System.out.println(Arrays.toString(hello.findCommonElement(myArray, myArray2)));

answered Nov 16, 2016 at 13:57

Anonymous

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1 Comment

Barry Reeves Over a year ago

Thanks for the info!

Manjunath Jakkandi · Accepted Answer · 2018-09-11 08:42:25Z

0

    int arr1[] = {1,2,5,7,89,3};
    int arr2[] = {1,45,87,34,3};

    for(int i=0;i<arr1.length;i++) {
        for(int j=0;j<arr2.length;j++) {
            if(arr1[i] == arr2[j]) {
                System.out.print(arr1[i] +" ");
            }
        }
    }

answered Sep 11, 2018 at 8:42

Manjunath Jakkandi

1

1 Comment

Thom Over a year ago

While this code may solve the problem, a few words of explanation will help current and future readers understand your answer even better.

Keshav Gera · Accepted Answer · 2019-08-20 13:00:54Z

0

Remove Complexity Using HashSet

Finding common elements in two integer arrays in java

import java.util.*;
public class Complexity
{
    public static void main(String args[])
    {
    int arr1[] = {2,2,7,7,2,1,5,4,5,1,1};
        int arr2[] = {2,3,4,7,10};

        HashSet<Integer> hashset= new HashSet<Integer>();

        for (int i : arr1){
            hashset.add(i);
        }
        for (int i : arr2) 
        {
            if (hashset.contains(i))
        {
            // found duplicate!   
                System.out.println("Common Elements --> " +i );
        }
       }
    }
}

edited Aug 20, 2019 at 13:00

answered Jun 24, 2019 at 7:09

Keshav Gera

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Comments

Ayush Kumar · Accepted Answer · 2021-04-27 19:29:10Z

The following is a simple O(n) solution that takes into consideration that arrays are sorted. If not, you can sort them. This is an improvement of the solution provided by @talshahar that also covers the last element being common(an edge case).

public List<Integer>  getCommon(int[]array1, int[] array2){
    int p1 = 0;
    int p2 = 0;
    ArrayList<Integer> common = new ArrayList<Integer>();
    while(p1<array1.length || p2<array2.length) {       
       if (array1[p1] == array2[p2]) {
           common.add(array1[p1]);
           p1++;p2++;
       }
      
       else if (array1[p1] < array2[p2]) {
           p1++;
       } else {
           p2++;
       }
   }
    return common;
}

Ayush Kumar · Accepted Answer · 2021-04-28 05:39:22Z

O(m+n) solution with the usage of Java Provided excellent Collection Data Structures. The key being .retainAll() function used in Hashset, which retains all the common elements:

It is worth mentioning that retainAll() works with any of the Collection class and internally calls contains() on it. Hence, O(m+n) will only be if the collection here is HashSet since it gives 0(1) lookup. If it is a linear one, like List, complexity will be 0(n^2)

public class common {
public static void main(String[] args) {

    Integer[] arr1 = new Integer[]{1,1,3,4,6,6,7,2,2};
    Integer[] arr2 = new Integer[]{1,1,3,2,4,8,9,5,6};
    List<Integer> alist = Arrays.asList(arr1);
    List<Integer> blist = Arrays.asList(arr2);

    HashSet<Integer> aset =  new HashSet<>(alist);
    aset.retainAll(blist);

    System.out.println(aset);
}

Bharathiraja · Accepted Answer · 2021-09-19 19:21:58Z

0

Time complexity o(n) i.e. using single for loop to get the common elements in an array.

    int[] arr1 = { 1, 2, 5, 5, 8, 9, 7, 10 };
    int[] arr2 = { 1, 0, 6, 5, 6, 4, 7, 0 };

    System.out.println("Array1 : " + Arrays.toString(arr1));
    System.out.println("Array2 : " + Arrays.toString(arr2));

    int i = 0;
    for (int j = 0; j < arr2.length; j++) {
        if (arr1[i] == (arr2[j])) {
            System.out.println("Common element is : " + (arr1[i]));
        }
        i++;
    }

Output:

Array1 : [1, 2, 5, 5, 8, 9, 7, 10]
Array2 : [1, 0, 6, 5, 6, 4, 7, 0]
Common element is : 1
Common element is : 5
Common element is : 7

answered Sep 19, 2021 at 19:21

Bharathiraja

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Comments

Ishan Lakshitha · Accepted Answer · 2018-10-15 15:36:54Z

-1

int x[] = {5, 3, 7, 2, 8}; int y[] = {6, 3, 8, 0, 2, 7, 4, 9};

    for (int i = 0; i < x.length; i++) {
        for (int j = 0; j < y.length; j++) {
            if (x[i] == y[j]) {
                System.out.println(x[i]);
            }
        }
    }

answered Oct 15, 2018 at 15:36

Ishan Lakshitha

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1 Comment

Acapulco Over a year ago

Welcome to Stack Overflow! It is recommended to add a brief explanation of your code, which goes a long way in helping others understand it.

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