Isomorphic Strings | LeetCode problem 205

Isomorphic Strings
Leetcode problem number 205
Solution in JAVA
JAVA interview programming playlist:
• Interview Programming ...
Linkedin: / amrita-bala-a2b79460
#java #interviewquestions #leetcode

[@GurjeetSingh-zr7vj]

It is an partial algorithm , The code didn't check condition weather the used char is again used or not, if it is used we do not say the strings are isomorphic. For Resolving - we have to use hashset which contains only those char who used only once. code - class Solution { public boolean isIsomorphic(String s, String t) { HashMap map = new HashMap(); HashSet check = new HashSet(); for(int i=0;i

[@nitishkumar-mj9tp]

@GurjeetSingh-zr7vj ,Your code is small ,but it takes 8-10 minutes for me to understand. Thanks

[@saratdas8886]

the indentation and the brackets opening and closing are not taken care of , and also this code is partially complete , this doesn't pass all test cases

[@nitishkumar-mj9tp]

To the point video, very nice for entry-level experience people(1-2 yr), they don't want everything from basics which takes more time.

[@androiddeveloperlpu301]

For those code is not running write the condition of checking length of both the strings is not equal then return false

[@amitsgada]

Please zoom out the page little bit cannot see the above code lines when explaining or writing code. Keep the zoom in such a way that whole code logic is visible everytime

[@TechnosageLearning]

Sure..Will take care next time

[@MsArn3]

Please soom out, we cant barly see an intire line of code

[@amitchaurasia592]

Can you check the below one abcs pqrb

[@WHAGDEVARAKONDAVAISHNAVI]

thank you mam.....for your explanation...................

[@tufguygaming]

This is the Code beats 100% users in C++ class Solution { public: bool isIsomorphic(string s, string t) { if(s.size()!=t.size()){ return false; } int hash[256]={0}; // mapping of characters from string s int isMapped[256]={0}; // ensures that mapping goes to which of the word in t for(int i=0;i

[@adityamishra1868]

explanation is good but they way of writting code is worst, u are placing cbraces before adding if or else

[@roopkumar2803]

very well explained ma'am

[@shindeprem228]

class Solution { public boolean isIsomorphic(String s, String t) { if(s.length()!=t.length()) return false; HashMap hm = new HashMap(); for(int i=0 ; i

[@rishikanth4519]

🙏

[@kathakalisaha9735]

thanks

[@shindeprem228]

@@rishikanth4519 👍

[@shindeprem228]

@@kathakalisaha9735 👍

[@hirahasan]

check your solution with "asdvd" and "kkk"

[@TechnosageLearning]

It is returning false since these strings are not isomorphic strings

[@suhasdevang9332]

​@@TechnosageLearninginstead coding in local ide do code in leetcode so that your code not running successfully for all the testcases

[@aapkikhushi4689]

Code ka last mai image dal do atleast

[@TechnosageLearning]

Git ka url dia hai..Usme sare codes hai..You can try it out..

[@androiddeveloperlpu301]

Map m=new HashMap(); int n=s.length(); if(s.length()!=t.length()){ return false; } for(int i=0;i

[@sachinpatil1010]

The Above solution seems to be not passing for all the tests.... Please refer below solution , It works s = "foo" t="bar" 1) sToT['f']= b and tToS['b'] = f 2) sToT['o']= a and tToS['a'] = o 3) sToT['o']= a and tChar = r public bool IsIsomorphic(string s, string t) { char[] sTOT = new char[256]; char[] tTOS = new char[256]; for (int i = 0;i< s.Length;i++) { char sChar = s[i]; char tChar = t[i]; if (sTOT[sChar] ==0 && tTOS[tChar] == 0) { sTOT[sChar] = tChar; tTOS[tChar] = sChar; }else if(sTOT[sChar] != tChar) { return false; } } return true; }

[@mhirroaceroyportillo875]

I’m afraid this solution will fail given s = abdc t = abab

[@TechnosageLearning]

It would return false...since these two strings are not isomorphic .."No two characters may map to same character" But in this case..a-->a and d-->a so it should return false only..

[@suhasdevang9332]

​@@TechnosageLearning but your code not running for all the testcase

[@pkshriram2530]

you are too fast mam, couldn't follow it 🥲

[@TechnosageLearning]

Will try to slow the pace in upcoming videos