Anagram | Check whether two strings are Anagram of each other | Java

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[coolguy]

It is the best anagram video I found on the internet. So clear. Great handwriting too.

[thebrothers]

helped me alot confused about this problem and code given by my teacher is very complex thanks brother keep doing irt

[AmitYadav-nzif]

please note that last line you wrote for printing anagram will print every now and then until loop becomes 0..so you can use if loop to print properly.

[NZIT]

Very informative tutorial about anagram. Thanks

[MamtaJaiswal-oi]

Thank you so much for this tutorial .🤗

[Motoraddictz_mehkma]

best explaination sir very easy and simply helpful

[RahulKumar-sxhm]

bro vedio is super plz do on more vedios on strings like removing duplicate elements in the strings etc

[districte]

You explained it really well tbh. Next time please add the time complexity.

[thahirThavvagunta]

Nice explanation 👏. I want to add one point, here the time complexity for the code you wrote is very high
O(n) converting string to array
2xO(nlog(n)) for sorting
O(n)

Overall i feel nlog(n) as time complexity

[Shadow-pjuv]

helpful video thanks we can also use hash Map to reduce the TC

[pshpshh]

bro i love you
greetings from germany

[faizalyakub]

Brother can you help me in the below question.
Suppose a string id given "Today is sunday" and interviewer asked me to find day occures howmany times and at what position.(Hint: position must be of only "d" like 2, 12 & without using built in methods)

[amitamishra]

but after loop it will always print anagram string so what is the logic behind this

[SilentAngelD]

this program is showing error while executing🙄

[Hitu]

Bro, please give logic in mathematical approach of coding

[KatikaAmrutha-vdxe]

For me Arrays.sort(c1);
Arrays.sort(c2);
Showing error

[AmitYadav-nzif]

Here i the corrected code from my end
import java.util.Arrays;
public class Anagram_or_not
{
public static void main(String[] args)
{
String str1 = "SILENT";
String str2 = "LISTEN";
char ch1[] = str1.toCharArray();
char ch2[]= str2.toCharArray();
boolean status = false;
Arrays.sort(ch1);
Arrays.sort(ch2);
if (ch1.length!= ch2.length)
{
System.out.println("Not an Anagram Because length is not same");

}


for (int i=0;i< ch1.length;i++)
{
if (ch1[i]!=ch2[i])
{
System.out.println("Not An Anagram");
status=false;
break;
}

}

if (status!=true)
{
System.out.println("It is An Anagram");
}
}
}