Frequently Asked Java Program 19: How To Find Duplicate Elements in Array

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Topic : How To Find Duplicate Elements in Array
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15 сен 2024

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Комментарии : 59

@bhushanmali2598 Год назад

Finally i found this in a very perfect way of explanation. Thank you ❤️

@sdetpavan Год назад

Welcome

@divyad9144 2 года назад

In the first for loop i should be less than arr.length()-1, bcz when i is at index 4, then j will be i+1 (i.e) 5 which will lead to ArrayIndexOutofBound exception

@Testology-PradeepPandey 2 года назад

Hahaha...asking or telling

@msgudaduramath4980 Год назад

It is not nessesary to compare the last element with others because it is already compared by other elements

@samK12_18 2 года назад

2 mistakes: 1. i loop should be from 0 to array length-1. 2. if condition inside j loop is incorrect. for(int i=0; i

@bhagathgudivaka1520 2 года назад

@samK12_18 2 года назад

@@bhagathgudivaka1520 Yes because if i goes to 5, then j becomes 6 and it will return array index out of bounds error.

@bhagathgudivaka1520 2 года назад

@@samK12_18 if j becomes 6 then inner loop will be stopped and Comes to outer loop again

@sarathr4495 Год назад

Hash set way is life saver sir 👍

@sdetpavan Год назад

My pleasure

@srinivasrao-lw6xk 3 года назад

I wrote this program . Bt when I take two duplicate(java,c) in the set . The out is showing like duplicate element is : java and next line duplicate element is c.

@gauri992 3 года назад

Ur the best Sir. thanku

@sdetpavan 2 года назад

Welcome

@TheShubhamVsnv Год назад

?? how can fix this if i have three same element ?? arr[]={"java","c","c++","python","java","java"}; output : java java java

@World90yes Год назад

Same error

@World90yes Год назад

Same error

@thesurajpatade 2 года назад

Using HashSet approach, if we have{java, java, java} it will print three times java which is not right, any solution to this?

@csRHarish 2 года назад

How it allowing duplicate values to u?

@kanithisatishkumar7036 3 года назад

sir, please provide code for multiple duplicates.

@Johnsahu-m7k 10 месяцев назад

great sir. plz upload latest trending logical programs java in automation interview.

@BharathKumarEnishetty Год назад

we can directly write as if(flag) , it would look good right , though it is fine , thanks

@sridhar6358 3 года назад

The given array is {"java","C","C++","python","java"} the first iteration java is compared with c c++ python and java but with iterations as it goes forward the second iteration c is compared with c++ python and java the third iteration c++ is compared with python and java the fourth iteration python is compared with java

@dsantoshini 2 года назад

yes

@keshavrawal2746 2 года назад

it works as you said bcoz in first iteration "java" is already compared with all the elements

@Shwethatp Год назад

sir along with the finding the duplicate elements i want to print how many times it's repeated...can you help me with this program

@MDALAM-yh3pf Год назад

first java is small j , last 2 java is capital J; how you prove multiple entry give same result?

@jarvisfriday7452 4 года назад

suppose if we have multiple duplicates like {"java","python","ruby","c#","python","java"} ??? how to print those duplicates separately?

@kanithisatishkumar7036 3 года назад

if you got this code plzz post it ?

@baggumani 2 года назад

@@kanithisatishkumar7036 Iterator i=langs.iterator(); while(i.hasNext()) { System.out.println(i.next()); } Use Iterator to print all the values in Hashset.

@sarves2482 6 месяцев назад

@@baggumani iterator returns all the values in the String including duplicates, not only the duplicates

@snehalwakode5291 Год назад

Thank u sir..

@sdetpavan Год назад

Welcome

@silvere4830 2 года назад

Please, provide an example with Java Streams

@indiantourism975 Год назад

you're explained clearly and easily I understood how to find duplicate elements thank you so much, Sir. But the count of duplicate elements did not get the correct value when using the HashSet approach and I've given my coding below public class DuplicatesArrays { public static void main(String[] args) { Boolean flag = false; Boolean dntdup = false; int count = 0; int count1 = 0; String [] ar = {"l","m","v","m","n","y","m"}; System.out.println(ar.length); // using for loop for (int i = 0; i < ar.length; i++) { for (int j = i+1; j < ar.length; j++) { if (ar[i]==ar[j]) { count = count+1; System.out.println("found duplicate :" + ar[i]); flag = false; } } } System.out.println("count duplicate elements by for loop: "+count); if (flag == true) { System.out.println("not found duplicate:"); } //using hashset HashSet h = new HashSet(); for (String l : ar) { if (h.add(l)==false) { count1 = count1+1; System.out.println("found duplicate :" + l); } } System.out.println("count duplicate elements by hashset :" + count1); if (flag == true) { System.out.println("not found duplicate:"); } } }

@Forty3Stories Год назад

In hashset the first time the element is found count is zero . So while printing count u should print count + 1

@tarunsharma9974 2 года назад

How can I get the count of this duplicate element

@omprakash7040 2 года назад

@tarunsharma9974 2 года назад

@@omprakash7040 it didn't work like that, can you please share the code

@gurunathkulkarni-ry8yv Год назад

@@tarunsharma9974 First intinalize count as 0 after printing the duplicate elements increment the count.

@as8212 4 года назад

Can some body explain; on first logic first element is compared with elements after that, so say 3 rd element is compared with 4th element, it is not compared with first or second, Is that logic right???

@Mm-zt1gk 4 года назад

No need coz duplicate off first nd second already searched

@renjithdhoni7 4 года назад

Thats not needed when you search,i=1. And j=3 ,the same thing need to be searched in j=3 and i=1

@bharathkoneru4008 4 года назад

Not needed because i=0, j=1 (arr[0]== arr[1]) comparison is already done to check for duplicates. So, there is no need of comparing again with the former i.e; i=1, j=0 arr[1] == arr[0]

@harish3787 3 года назад

Take the upper for loop once and check what it is printing ok Then check what the lower for loop checking Now put i = let's say 4 And then u put 2nd for loop you will see it will give values upto 4 ok so it always checks values hoto that point

@mohammedajazquadri7869 4 года назад

Sir your mail is there?? I have some doubts??

@sureshsubramaniyan3201 2 года назад

suppose we have multiple duplicates like below and want to the print the output only once like duplicate elements = java,c++,python. Could you please guide me. String arr[] = {"java","python","c++","java","c","c++","devops","java","c++","python","python"};

@maharudra947 Год назад

what about multiple duplicates

@sridhar6358 3 года назад

The first logic wont work well as it does not compare with the previous values

@faisalsakib9586 4 года назад

How to remove duplicate duplicate values from array ??

@theflutterboi 3 года назад

use Set

@suganthym1438 3 года назад

Create a new array and add the elements only when Not duplicate. Or use a java collection type that ignores duplicates

@suiiiiii5798 2 месяца назад

❤️

@Bulldoges90 3 года назад

the methods are time complexity are really bad. the 1st one is O(N^2). you can do it in O(N) with 1 loop and check for 3 or more duplicates/ the code:(same for String) class FindDuplicate { // Function to print duplicates void printRepeating(int arr[], int size) { int i; System.out.println("The repeating elements are : "); for (i = 0; i < size; i++) { int j = Math.abs(arr[i]); if (arr[j] >= 0) arr[j] = -arr[j]; else System.out.print(j + " "); } } // Driver code public static void main(String[] args) { FindDuplicate duplicate = new FindDuplicate(); int arr[] = { 1, 2, 3, 1, 3, 6, 6 }; int arr_size = arr.length; duplicate.printRepeating(arr, arr_size); } } for String: String[] users = "User1,User2,User1,User,User".split(","); Set uniquUsers = new HashSet(); for (int i = 0; i < users.length; i++) { if (!uniquUsers.add(users[i])) users[i] = "Duplicate"; // here I am assigning Duplicate instead if find duplicate // you can assign as null or whatever you want to do with duplicates. } System.out.println(Arrays.toString(users));