This is most common interview question in java now-a-days. There are many techniques to find duplicate elements in array in java like using Collections.frequency(). I am writing yet another solution which is much easier and fast.
Here an array of integers is having 10 integers and 1 and 8 are duplicate integers. You need to filter them out.
package com.howtodoinjava.interview;
import java.util.HashSet;
import java.util.Set;
public class DuplicateInArray
{
public static void main(String[] args)
{
int[] array = {1,1,2,3,4,5,6,7,8,8};
Set<Integer> set = new HashSet<Integer>();
for(int i = 0; i < array.length ; i++)
{
//If same integer is already present then add method will return FALSE
if(set.add(array[i]) == false)
{
System.out.println("Duplicate element found : " + array[i]);
}
}
}
}
Output:
Duplicate element found : 1
Duplicate element found : 8
In this solution, all you have to do is to iterate over array and all elements in a set one by one. If Set.add() method returns false then element is already present in set and thus it is duplicate.
Happy Learning !!
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Rahul
also count the duplicate no. occurances?
Pawan Sharma
display
The type Set is not generic; it cannot be parameterized with arguments
Durga Prasad
In interview i have faced one question.
How to find second repeated value in Array?
Ex: {1,2,3,4,7,8,4,6,9,5,8,0,1}
o/p: 8
What is the solution?
Priyadharshini
public static void main(String []args){
int[] array={2,3,5,2,5,6,9,6,4,7};
int count=0;
int newcount;
int[] duparray=new int[count];
for(int i=0;i<array.length;i++){
for(int j=i+1;j<array.length;j++){
if(array[i]==array[j]){
count++;
if(count==2){
System.out.println("The Second repeated value in an array is" +array[i]);
}
}
}
}
}
Priyadharshini
Ignore duparray and newcount
Cherry
Great analysis. I would recommend another post about the same question – http://blog.gainlo.co/index.php/2016/05/10/duplicate-elements-of-an-array/?utm_source=comment&utm_campaign=comment&utm_medium=11316
Monika
Hi I need
program for Duplicate array store in another array?
sashi
Hi,
Is this optimized one ? time complexity O(n)
public static void main(String[] args) {
// TODO Auto-generated method stub
int A[] = {4,2,6,5,3,8,6};
List ar = Arrays.asList(IntStream.of(A).boxed().toArray());
int dist[] = Arrays.stream(A).distinct().toArray();
int count;
int maxCount =0;
int leader = 0;
for(int k : dist){
count = Collections.frequency(ar, k);
if(count > maxCount ){
leader = k;
maxCount = count;
}
}
System.out.println(“Leader of array:”+leader);
}
ALEXANDRU
What if I want to determine the number of duplicates, the total of duplicates?
super211
To avoid printing many times, another solution for taking the strategy space for time:
int[] array={10,10,20,10,20,41,55,10,10,10,11,12,66};
Set set = new HashSet();
Set set2 = new HashSet();
for(int i=0; i<array.length; i++){
if(!set.add(array[i])){
if(set2.add(array[i]))
System.out.println(array[i]);
}
}
Anil
Thanks All 🙂
Rohit
Hello,
int[] arr={10,10,20,10,20,41,55,10,10,10,11,12,66};
Your program may fail for this array declaration it will print 5 times 10 s
so make it correct as below:
int[] arr={10,10,20,10,20,41,55,10,10,10,11,12,66};
Set set=new HashSet();
for(int i=0;i<arr.length;i++)
{
for(int j=i+1;j<arr.length;j++)
{
if(arr[i]==arr[j])
{
if(!set.contains(arr[i]))
{
set.add(arr[i]);
System.out.println("Duplicate Element:"+arr[i]);
}
}
}//end inner for;
}//end outer for
Lokesh Gupta
Very clever. Thanks for sharing with others. Much appreciated.
Coder 4 Life
Super bro
SARAL SAXENA
public class dupliacteselection { public static void main(String[] args) { int a[]={1,2,3,3,4}; Set<Integer> s=new HashSet<Integer>(); for(int i=0;i<a.length;i++) { if(!s.add(a[i])) System.out.println("at index"+ i+" "+a[i]+"is duplicate"); } for(int i:s) { System.out.println(i); } } }SARAL SAXENA
@Lokesh perfect one just to add another similiar approach for this is ..
public class dupliacteselection {
public static void main(String[] args) {
int a[]={1,2,3,3,4};
Set s=new HashSet();
for(int i=0;i<a.length;i++)
{
if(!s.add(a[i]))
System.out.println("at index"+ i+" "+a[i]+"is duplicate");
}
for(int i:s)
{
System.out.println(i);
}
}
}
James Ting
nice algorithm, thanks
Vivek Kumar
import java.util.HashSet;
import java.util.Set;
public class DuplicatItems {
public static void main(String[] args) {
int items[] = {1,2,3,2,5,7,6,4,5,5,9};
Set set = new HashSet();
for (int item:items){
if(set.contains(item))
System.out.println(item +” is duplicate item”);
set.add(item);
}
System.out.println(“No. of duplicate items: “+(items.length-set.size()));
}
}
Lokesh Gupta
Thanks Vivek for sharing your answer. It’s almost same, with additional call to
set.contains()method.Vivek Kumar
Hi Lokesh,
How to find the total number of times a value is repeated in a HashMap containing duplicate values ?
Shashi
May be you can use Map, key as element in array and frequency as value.
Something like
if(map.contains(key)){
map.get(key)+1;
}else{
map.put(key, new Integer(1));
}
Guest
Thanks Lokesh 🙂 this works for me
vinodh
Thanks for this.
they asked me this recently