Determine if two strings/phrases are valid Anagrams | Study Algorithms

[@anirudhv0062]

In the Optimized code, I don't think we need to check for spaces. Since anagrams are supposed to be single words, if there are spaces the string, it will become a phrase instead of a word. Also if we check in the beginning if the strings are of equal length, there is no need for 2 for loops of bucket arrays. The code can be simplified to : class Solution { public boolean isAnagram(String s, String t) { if (s.length() != t.length()) return false; int counts = new int[26]; for (int i=0; i

[@SatishSingh-eb5tq]

Why didn't you used HashMap to fill the values first with key and counter and then to decrement. Why to go for A-Z map?

[@lalanabiridi8358]

instead of bucket array we can use hashtable too right??? by removing spaces

[@jritzeku]

I took this approach: -remove all empty spaces first -early exit cond; if length dont equal -In first loop evaluated first string populated hash map with occurrence of each char. -In second loop, evaluated second string and decremented the count from map. If the count was at 1, i removed it from map. -Ultimately, at the end if the map is empty (length of 0), we found anagram.

[@mai_aasim]

Thanks for helping

[@eduardoignacioroblessosa6349]

brilliant as always

[@nageshnimmala7366]

Can i get by using compareTo metheod?

[@nageshnimmala7366]

One more request pls suggest me a book where I can understand advanced java and data structures and algorithms easily

[@nikoo28]

If you check the link the video description, I do mention some resources :) Hope that helps

[@nageshnimmala7366]

@@nikoo28 Yaa that's ok but i searched for that but it's very advanced one

[@nageshnimmala7366]

Pls recommend me some good books to learn servelets, jsp, microservices

[@nikoo28]

Have you tried the headfirst series? They are really easy to follow, have a lot of visuals and you will make projects too as you read along. Highly recommended :)

[@pradeepella]

package javaprac; public class JavaPrac { public boolean anagram(String str1,String str2){ str1=str1.toLowerCase(); str2=str2.toLowerCase(); str1=str1.replace(" ", ""); str2=str2.replace(" ", ""); int[] arr=new int[26]; for(int i=0;i

[@mr_funtor]

You looped 3 times. Shouldn't this be n^3?

[@nikoo28]

3 loops do not mean O(n^3)...3 nested loops will cause the time complexity to be n^3. 3 separate loops mean O(3 * n) = O(n)

[@mr_funtor]

@@nikoo28 Alright. Thanks