Find duplicate element occur more than two times from an array in java

I want to find out the duplicate element and there index number from an array. I write down a code for that. It works well but only fail to generate exact output when number of duplicate element more than 2. I read the value from a file and then build an array and then searching duplicate element from this array.

import java.io.File;
import java.util.Arrays;
import java.util.Scanner;

public class T1 {
public static void main(String args) throws Exception{
Scanner x=new Scanner(new File("C:\Duplicate_array.txt"));
int duplicate_data=new int[9];
int i1=0;
while(x.hasNext()){
int a=x.nextInt();
duplicate_data[i1]=a;
i1++;
}
System.out.println(Arrays.toString(duplicate_data));
for (int i = 0; i < duplicate_data.length-1; i++) {
for (int j = i+1; j < duplicate_data.length; j++) {
if ((duplicate_data[i] == duplicate_data[j]) && (i != j)) {
System.out.println("Duplicate Element : "+duplicate_data[j]);
System.out.println("Index of that duplicate element : "+j);
}
}
}
}
}


Here is my output:

[5, 6, 1, 6, 9, 5, 2, 1, 5]
Duplicate Element : 5
Index of that duplicate element : 5
Duplicate Element : 5
Index of that duplicate element : 8
Duplicate Element : 6
Index of that duplicate element : 3
Duplicate Element : 1
Index of that duplicate element : 7
Duplicate Element : 5
Index of that duplicate element : 8


Error at last line.It already find 5 at the beginning at position no: 8. But at the end of the program it again searching for 5 and give the position no. This last search is unnecessary. How to get rid of this last search?

Set

4 Answers
4

i starts at 0 (value 5) and j starts at the end of the array(value 5) it outputs the the position of duplicate right. However when I is at the end of array and j starts at the end it will do the same thing, to solve this you can make a copy of the array and when traversing the array delete the dulpicate items.

The simplest way that I can think of is to create a set that holds all the indexes that have already been found. Also, the check (i != j) is redundant since j=i+1.

import java.io.File;
import java.util.Arrays;
import java.util.Scanner;
import java.util.Set;

public class T1 {
public static void main(String args) throws Exception{
Scanner x=new Scanner(new File("C:\Duplicate_array.txt"));
int duplicate_data=new int[9];
Set<Integer> found = new HashSet<>();
int i1=0;
while(x.hasNext()){
int a=x.nextInt();
duplicate_data[i1]=a;
i1++;
}
System.out.println(Arrays.toString(duplicate_data));
for (int i = 0; i < duplicate_data.length-1; i++) {
for (int j = i+1; j < duplicate_data.length; j++) {
if ((duplicate_data[i] == duplicate_data[j]) && (i != j) && found.add(j)) {
System.out.println("Duplicate Element : "+duplicate_data[j]);
System.out.println("Index of that duplicate element : "+j);
}
}
}
}
}


It is searching for the same duplicates again because you are not storing previously found duplicates by any means. So, You have to use a data structure to store previously found duplicates and not searching for them again.
This takes us to a better solution for finding duplicates which is using a hashset from the beginning because it's O(n) instead of O(n^2)

import java.io.File;
import java.util.Arrays;
import java.util.Scanner;

public class T1 {
public static void main(String args) throws Exception {
Scanner x=new Scanner(new File("C:\Duplicate_array.txt"));
Set<Integer> set = new HashSet<Integer>();
int index = 0;
while(x.hasNext()){
int nextNumber = x.nextInt();
if (set.contains(nextNumber)) {
System.out.println("Duplicate Element : " + nextNumber);
System.out.println("Index of that duplicate element : "+index);
} else
set.add(i);
}
}
}


(i != j) is not necessary in your if statement, since j is always ahead of i by 1, but that's not your issue.

(i != j)

j

i

You could try using a duplicate array flag to know when you've already found a duplicate.

import java.util.Arrays;

public class StackOverflow {
public static void main(String args) throws Exception {
int duplicate_data = {5,6,1,6,9,5,2,1,5};
boolean duplicate = new boolean[duplicate_data.length];

System.out.println(Arrays.toString(duplicate_data));
for (int i = 0; i < duplicate_data.length - 1; i++) {
for (int j = i + 1; j < duplicate_data.length; j++) {
// Make sure you haven't flagged this as a duplicate already
if (!duplicate[j] && duplicate_data[i] == duplicate_data[j]) {
duplicate[j] = true;
System.out.println("Duplicate Element : " + duplicate_data[j]);
System.out.println("Index of that duplicate element : " + j);
}
}
}
}
}


Result:

[5, 6, 1, 6, 9, 5, 2, 1, 5]
Duplicate Element : 5
Index of that duplicate element : 5
Duplicate Element : 5
Index of that duplicate element : 8
Duplicate Element : 6
Index of that duplicate element : 3
Duplicate Element : 1
Index of that duplicate element : 7


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.