Anagram | Check whether two strings are Anagram of each other

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

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2 жыл бұрын

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Пікірлер: 33

@coolguy2066 Жыл бұрын

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

@NZIT1 Жыл бұрын

Very informative tutorial about anagram. Thanks

@thebrothers6880 Жыл бұрын

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

@ShivamSingh-gc6pm Жыл бұрын

Thankyou sir 😊 Very helpful....

@user-jo3gs1km6d Ай бұрын

Thank you so much for this tutorial .🤗

@SilentAngelD Жыл бұрын

this program is showing error while executing🙄

@nawwintphyu1873 3 ай бұрын

Thank you!

@TurboThrottle200Rider 3 ай бұрын

best explaination sir very easy and simply helpful

@AmitYadav-nz5if Жыл бұрын

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.

@RahulKumar-sx8hm Жыл бұрын

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

@rachamallidorasrivignesh4499 4 ай бұрын

Tq so much brother❤❤❤❤

@whyu8012 2 жыл бұрын

Super bro

@knowledgegetter 8 ай бұрын

Best one!

@districte7177 9 ай бұрын

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

@ankurrunthala Жыл бұрын

Gajab

@saibaba3692 2 жыл бұрын

Nice

@KatikaAmrutha-vd2xe 3 ай бұрын

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

@amitamishra8210 Жыл бұрын

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

@thahirThavvagunta 10 ай бұрын

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

@theintrovert894 9 ай бұрын

ya he should go with Frequency array which will give t(n) around O(n)

@faizalyakub9380 Жыл бұрын

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)

@proxy5061 Жыл бұрын

By built in method you can use like first;;;; String[] strngarr = strng.split(" "); Loop that words now , inside and check strng.indexOf("day") If user defined means Split and loop as shown above Inside loop Check for every sequence occurence of string day ,,, like For (){ Str[i] == d str[i+1] == a str[i+2] == y This is just plane we have to enhance little bit u will get this inguess

@BharathKumar-nf6xt 11 ай бұрын

Why cant we use equal() method bro..

@pshpshh2845 Жыл бұрын

bro i love you greetings from germany

@BTechComputerScience Жыл бұрын

So nice to have students from other countries ❤

@prasadpatil4016 8 ай бұрын

Best

@Hitu619 5 ай бұрын

Bro,please give logic in mathematical approach of coding

@BTechComputerScience 5 ай бұрын

Hi, coding is nothing but talking too computer, it's like asking the computer to do some task for u... so there is nothing like mathematical approach of coding. For some problems u need to understand maths to tell the computer how to solve it. Not all coding problem requires maths... coding requires logical thinking and problem solving. Hope u can now strengthen those things and grow😊

@Hitu619 5 ай бұрын

@@BTechComputerScience kk bro

@rupasrithati8960 Жыл бұрын

Why char to chararray

@shiny.5019 Жыл бұрын

Strings doesn't have a built-in sort method but charArray does!! so we convert the string into charArray and apply the methods

@AmitYadav-nz5if Жыл бұрын

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"); } } }