Java Program To Find First Non Repeated Character | Java | Ashok IT

[@ashokit]

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[@Ravikumar-gj6qw]

Ur way of explaining or teaching is excellent 👌😉bro superb 👌👍👏, nobody is there ...

[@rptech6902]

public class practicee {
public static void main(String args[])
{
String s="AABCDBE";
int count=0;
char[] k =s.toCharArray();
for (int i = 0; i < k.length; i++) {
for (int j = 0; j < k.length; j++) {
if (k[i]==k[j]) {
count++;
}

}
if (count==1) {
System.out.println(k[i]);
break;
}
count=0;
}
}
}
this is my approach

[@YashSingh-xs5yj]

Collection approach is the best ❤

[@ashokit]

Thankyou so much.. 🔥 Please follow us in WhatsApp channel : bit.ly/49wCWje

[@dipikasharma748]

Shouldn't the map be LinkedHashMap so that it maintains the insertion order to correctly identify first non-repeated character?

[@mishra1576]

Yes it should be😊

[@rajeevkalangi2472]

Thanks Bro. Your explanation is super cool.

[@ajayghode3602]

Thank you Sir.

[@ashokit]

Thankyou so much..

[@jagadeeshbaskaran4881]

Excellent . Thanks sir

[@jatinsharma3792]

In hashMap the insertion order is not maintain, how can you be sure it will give your the First Non Repeated Char boss ????

[@ashokit]

please contact our admin team : [phone]

[@TechieToHelpYou]

After finding frequency for every element then take another loop iterate thourgh loop index from 0 to string.length
Then for each character check in the map wheather that character has 1 or not

[@DanishAli-ql1es]

Very nice explanation

[@ashokit]

Thanks for liking

[@kameshwaran6689]

Really superb

[@Ravikumar-gj6qw]

Tq Ashok bro

[@ram8705]

My approach :
String unique = "AABDCE";
for (int i = 0; i < unique.length(); i++) {
for (int j = i+1; j < unique.length(); j++) {
if(unique.charAt(i) != unique.charAt(j)) {
System.out.println("1st non repeating char is : "+unique.charAt(j));
return;
}
}
}

[@ashokit]

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[@eshwarmuruganti9758]

It won't work for all the cases it will print all the characters from given string expect

[@kkiran-zk8ed]

You can use j=i+1

[@alokkumar-pl6ds]

same doubt

[@krishkhemani96]

I don't think j=i+1 would work because in that case first iteration A and A would be compared in second iteration A and b would be compared and thus compiler would declare A as the first non repeating character.I am also not sure.

[@garvitrangi3263]

@@krishkhemani96 if we take j = i+1 then it will consider the last duplicate character as the first non-repeating character for eg It will consider 'A' of index 1 as a non-repeating character.

[@moumajhi23]

using this collection approach it won't give correct output , if the non repeated occurs at the first position , for example input = "kaabcdbe" output should be k but it would give c.

[@ganeshbodduaddasuryanitinb7811]

Using index Of method?

[@DILEEPKUMAR-gp7ic]

It is coming last non repeated char we need first non repeated characters

[@DILEEPKUMAR-gp7ic]

Could u correct the code thro hashmap

[@showdownx8276]

Cant use HashMap here.

[@ashokit]

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[@hchemlal]

Your coding still needs some polishing, first approach you should have used j=i+1 and the second approach you should have used LinkedHashMap.

[@anbuv.g.k495]

Yeass bro and the hashmap does not maintain insertion order right? How he got the answer bro..!

[@hchemlal]

@@anbuv.g.k495 It will work 90% of the time, but in production setting you'll face reality.

[@krishkhemani96]

I don't think j=i+1 would work because in that case first iteration A and A would be compared in second iteration A and b would be compared and thus compiler would declare A as the first non repeating character.I am also not sure.