Isomorphic Strings | Simple Thought Process | Leetcode 205| LinkedIn | codestorywithMIK

[@techtalk528]

bestest explanation of question and the code. I mean seriously I don't comment on any youtuber video, I learn from striver and other youtubers video as well but still didn't find this good explanation. huge respect to you buddy. just subscribed. thanks a lot brother......

[@gui-codes]

nice. I also failed the same test case and then figured out the problem.

[@rohitrajput2097]

what an amazing expaliation

[@NeerajKumar-cj6ky]

I subsrcibed your channel 8 month ago and the energy of teaching remains the same.

[@oqant0424]

Solved it on my own Came here to learn optimisation Thanks :)

[@shloksuman8164]

even i tried using map, and mapped the frequency of both the strings. Then iterated over a string to to check any mismatch in frequency. passed 43/45 testcases. got stuck at the preserving order point , which was i guess the crux of the problem. thanks for solution

[@EB-ot8uu]

kaafi tricky Problem statement tha. thanks for making it easy

[@_FruitBasket]

thank you for existing

[@shashiraj6226]

Sir can you tell me which editor do you use while explaining? It is really very clear explanation. Thank you.

[@avendersharma8532]

Apple notes hain ipad mein inbuilt hota hain

[@soumyadwipsom5014]

Thank you sir for sharing your important time with us ❤

[@Mohit-fe5hx]

november se ye question ko 6 attempt ho gye the wrong submission k..lfinally aaj clear hua

[@aritrakar2492]

thank you bhaiyaa , corner cases dekhke hi samaj gaya kaha code phassa tha mera .

[@wearevacationuncoverers]

nice one

[@abhijitroy1958]

can we use the find function in such a way that when we are searching for the first string element we can also check if the associated elements with it also exists or not

[@gaganjasuja1602]

Can you give a solution using bit manipulation??

[@adityaraj-zm7zk]

@ codestorywithMIK bhaiya ye orthogonal greedy algo ka question kaha se kare and pattern of this question

[@tutuimam3381]

Thanks a lot

[@gauravbanerjee2898]

Thanks a lot bhaiya ❤❤

[@ugcwithaddi]

Thank you ☺️

[@Nofaltuguy1]

i have a doubt will there a problem if length of s and t are different????

[@avtarchandra2407]

You are lovely mik ....thanks

[@thaman701]

Java code guyz... class Solution { public boolean isIsomorphic(String s, String t) { int n=s.length(); int m=t.length(); if(n!=m) return false; HashMap mp=new HashMap(); for(int i=0;i

[@gui-codes]

thanks

[@bipulmukherjee422]

Thanks

[@subhajitdey135]

Another approach but same space and time complexity :) bool f(string s,string t){ int size1=s.size(),size2=t.size(); if(s==t) return true; unordered_mapmp; unordered_maptaken; for(int i=0;i

[@layathal7460]

sir, here we are taking 2 additional maps right? So how the space complexity is still 0(1) ?

[@akshay3841]

With 2 additional maps, its O(2 x 1(for limited characters)) so = 0(1).

[@wearevacationuncoverers]

Because the constraints says that there will be only valid ascii characters in the string. We have only 256 ASCII characters. so map size will never exceed 256. Hence constant.

[@gui-codes]

s and t consist of any valid ascii character.

[@harshavshah7929]

Bro pls Longest substring with atleast k repeating character ka ek video bana do

[@mohdafzal4017]

Why 256 as size ( when both capital and small letters are their ). i could understand 60(65-124). Please tell me

[@codestorywithMIK]

In ASCII, for example, there are 128 characters, but if you include extended ASCII, there are 256 characters. This includes both uppercase and lowercase letters, digits, punctuation marks, and control characters. So, to ensure that you can handle both lowercase and uppercase letters, you might use an array size of 256 to cover all possibilities.

[@souravjoshi2293]

I was able to solve this 🙂

[@tushartyagi4965]

🔥🔥🔥🔥🔥

[@kungagyaltsenlachungpa8915]

s = "paper", t = "title", 'e' is alreday mapped with ''l' and and 'r' is mapped with 'e'? 2 different character mapped with same character?

[@HossainAhmedSiam-ot5jr]

(e->L ) making pair in hash1. (L->e) making pair in hash2. now (r->e) making pair in hash1 and (e->r) in hash2. are you noticed that, "e" not appear in hash2 before, where you adding now . "e" as a key existing in different hash hash1 hash2 e-->L L-->e r-->e e-->r

[@gui-codes]

@@HossainAhmedSiam-ot5jr thanks bro

[@kushagrapandey8898]

10:22 kya hum ek hi map me value ke through key ko ni check kr skte? Jaise key se value check kr rhe? 🤔

[@gui-codes]

No, because we need something to distinguish that the character I see in the map belongs to s or t

[@heathens2867]

Yes we can. We just need two if cases

[@iamnottech8918]

using 1 map class Solution { public: bool isIsomorphic(string s, string t) { unordered_map mpp; // length same given int i = 0; while (i < s.length()) { if (mpp.find(s[i]) == mpp.end()) // not there { for (const auto& pair : mpp) { // as actual mapp is {key ,val} pair if (pair.second == t[i]) { return false; } } mpp[s[i]] = t[i]; } else if (mpp.find(s[i]) != mpp.end()) // there { // if already presmt and now we got different val if (mpp[s[i]] != t[i]) // value was not as expected then not isomorphic return false; } i++; } return true; } };

[@as2002ajaysingh]

if some one want a simple solution with not many condition are - class Solution { public: bool isIsomorphic(string s, string t) { vectormark(257,false); vectorc(257,' '); for(int i = 0;i

[@FifthArima]

man, this question confused the hell out of me.

[@dayashankarlakhotia4943]

class Solution { public void create(char[]arr){ char start='a'; char[]mp=new char[256]; for(int i=0;i

[@learningbuddy8575]

What if string s= "abc" and string t="bas"??

[@gui-codes]

a -> b b -> a c -> s No same character in s is matched with different characters in t. So it's true.

[@dhruvrawat7023]

class Solution { public: bool isIsomorphic(string s, string t) { int sn = s.size(); int tn = t.size(); if(sn != tn){ return false; } unordered_map mp; vector ta(128, false); for(int i = 0; i < sn; i++){ char sc = s[i]; char tc = t[i]; if(mp.find(sc) == mp.end()){ if(ta[tc] == true){ return false; } mp[sc] = tc; ta[tc] = true; } else{ if(mp[sc] != tc){ return false; } } } return true; } };

[@shraban8508]

You haven’t uploaded this map code in github repo rather you have added the same arraylist code twice. Pls modify.

[@codestorywithMIK]

Done ❤️

[@Algorithmswithsubham]

sir yee approach me kya galati haii mapmp1,mp2; int n=s.size(); vectorv(256,1); for(int i=0;i

[@devmadaan5146]

kyaa haal h bhai ( aap software engineer ban jaun na ...paise chhapo daba ke) khair fir yt mat choddna (LG jayenge fur mere)

[@helpmore9389]

@codestorywithMIK #codestorywithMIK Thank you bhaiya to make difficult question, easy. Would you please make the video for this question kzbin.info/www/bejne/aHq2f3Z6f6qap7M As the solutions available for this is very less, and also complicated. Hope to get the solution for this question soon. Thank you so much for giving time to read this and think about it.

[@_FruitBasket]

bhai yeh link toh issi video ka hai

[@gui-codes]

link is of this video only

[@yashagarwal9784]

Important Substitue: I think we can Replace Map2 with unordered_set. It can work. Here is the updated code: class Solution { public: bool isIsomorphic(string s, string t) { int n = s.size(); unordered_map mp; unordered_set mapped; for (int i = 0; i < n; i++) { if (mp.find(s[i]) == mp.end()) { if (mapped.find(t[i]) != mapped.end()) return false; mp[s[i]] = t[i]; mapped.insert(t[i]); } else { if (mp[s[i]] != t[i]) return false; } } return true; } };