what an explanation👏

I urge everyone to watch full ads in his videos and not skip it...This guy deserves this!

@codestorywithMIK

Жыл бұрын

That’s overwhelming. I want to thank you from the bottom of my heart. That’s very kind ❤️

@nagmakhan3165

Жыл бұрын

True

@souravjoshi2293

Жыл бұрын

Seriously man. I am really gonna do that for this guy. He deserves it.

@user-qo6vg5jb6y

Жыл бұрын

me using adblocker

Finally wait is over

Lovely Explanation...Next level Expalnation....maza hii aagya...

great explanation, huge respect man seriously 👍 keep posting such videos daily and also imp questions from imp topics and interview perspective

this channel is really underrated.... your explanations are crazy good.... thanks for such great contents ❤❤

Crisp and perfect. Thanks a lot for the recursive tree explanation and dry run

Learned alot of new things from this video, thankyou so much 🥺. You always made everything so ezz...

nice explanation

Wonderful explanation

Man you are just awesome. I just saw you crossed 2k subscribers. Congratulations and soon you will reach 1M trust me

Your explanation is really good bro . Keep bringing such videos

@codestorywithMIK

10 ай бұрын

Thank you so much 😇🙏

Mannn!!!🙇🏻‍♀️🙇🏻‍♀️ You're the best 🔥🔥🔥

@codestorywithMIK

3 ай бұрын

🙏🙏❤️❤️😇😇

Congratulations for 2K, Wish you many more in future.👏👏👏

you literally made it really easy !!

you are superbbbb.

God Level explanation

He is back 🔥

Best best bhai keep going👍

Thanks a lot

Bestest explanation😇

@codestorywithMIK

Жыл бұрын

Thank you Anshika 😇

Dope 🔥

lit

very helpful!! can you please make a vedio on constraints analysis .(how to take help from constraints in solving questions)

@codestorywithMIK

Жыл бұрын

Thanks Deepa. I have planned to make a video on that too soon. Thanks for watching 😇

Java Solution as per above Approach: class Solution { int n; List result; public boolean valid(String s) { if(s.charAt(0)=='0') return false; if(Integer.parseInt(s)>255)return false; return true; } public void solve(String s,int i,int parts,String cur) { if(parts>4) return; if(parts==4 && i==n) { //remove the dot result.add(cur.substring(0,cur.length()-1)); return; } if(i+1

Really thankyou bhaiya your explanation helped a lot at first I was not able to even approach the problem but after listening to you for few minutes I began to understand the problem and for that really thankyou bhaiya I will be always great full to you One thing wanted to ask you if this type of question came in online interview then what should we do face that type of situation ?

@codestorywithMIK

Жыл бұрын

Thanks a lot Uday During interviews they generally help also , hints are given if you need But it’s important that you atleast share your thoughts out loud

if (parts>4 ) return ; this condition should be added na otherwise it'll call recursion unnecessarily even when we crossed 4Parts

@codestorywithMIK

Жыл бұрын

Yep you can add it too. Good one 👍🏻

god 🛐🛐

You're awesome. But, Please post videos in English, so that it reaches to wider audience.

@codestorywithMIK

Жыл бұрын

Soon I will try to add subtitles. 😇 And make another video in english too

If possible end me Java ka code v ek baar dikha diya karo, baki you teach really good. Congratulations for 2k

@codestorywithMIK

Жыл бұрын

Sure ❤️

Nice approach & Solution If you don't mind can you make 1 video on explaining recursion? Like how generally Recursion are used in leetcode programs, how to think on problems solving from that perspective.. As watching the video I got the hint you will be using recursion but calling recursion 3 times with different parameters etc TBH its hard to trace the calls inside brain n think from that perspective I have started watching this channel start of 2023 I see you release video daily .. It will be tough for you to manage your day IF YOU CAN CREATE VIDEO Then only DO or else its whole fine no issues Thanks Mik 😊

@codestorywithMIK

Жыл бұрын

Hi Akshay, I am soon planning to create playlists on these topics also from scratch. I will just be ok break next month for 3 weeks for travelling, then i will be back again. And definitely i will create videos on what you want. Thanks for watching ❤️❤️❤️

@akshayzade9761

Жыл бұрын

​@@codestorywithMIK - Thanks a lot for your efforts and yaa enjoy your days very well We are not running out 😄Your channel has different uniqueness of "Story To Code" Which makes peeps like me to understand easy You are killing the leetcode discussion section.. Keep it up mate 👍

@codestorywithMIK

Жыл бұрын

Thanks a lot 😇❤️

Bro how do u figure out so many things to solve a prob. I am not able to do that 😢 between congrats for 2k bro u deserve more 🔥

@codestorywithMIK

Жыл бұрын

Thank you so much. And regarding approaches, I always believe in making story first. And trust me, with time, it will come to you too. Keep practicing

Java code using for-loop-recursion & regex : ``` class Solution { public List restoreIpAddresses(String s) { List result = new ArrayList(); int N= s.length(); if (N>12) return result; //empty List solve(s, 0, N, new StringBuilder(), 0, result); return result; } public void solve(String s, int idx, int N, StringBuilder ipSoFar, int octetCount, List result) { // defining : base or termination condition if ( octetCount == 4 && idx == N ) { String validIP = ipSoFar.substring(0, ipSoFar.length()-1);//removes trailing delimeter '.' result.add(validIP); return; } // exploring different considerations : max 3 possible at each stage for(int r=idx+1; r

❤ Ye kal ke question jaisa hua na bhaiya? Ki ham backtracking karre har step pe, ki lenge ya nahi lenge string ko, agar valid hai to append karenge warna nahi

@codestorywithMIK

Жыл бұрын

Yep, you can call it backtracking too.

Itni sari conditions sochna ....fir usko Code me convert krna kaise kr lete ho aap ....ho hi ni pata mujhse ye jitna bhi khud se likhne ka try kr lu .... 😥😭

@codestorywithMIK

Жыл бұрын

Trust me, keep practicing and it will come soon. Never stop 💪

@shashwats9273

Жыл бұрын

@@codestorywithMIK Ok bhaiya 🥰❤

Bro why on passing string curr by reference in solve() and isValid() gives an error? I am new to c++ so I didn't get it.

@codestorywithMIK

Жыл бұрын

Actually this is this will give error because, you are trying to pass s.substr() which will be of type std::__cxx11::basic_string to &str which is of type std::__cxx11::string& Both are of different type. So, whenever you pass anything by reference, make sure you evaluate the substr (or any operation) and save it in a variable and then pass it. For example : #include using namespace std; void check(string &str) { cout

@piyushacharya7696

Жыл бұрын

@@codestorywithMIK Bro thanks a lot for describing it so properly even in the comment section. May you get millions of subscribers 💪. 🐐 of DSA explanation.

@codestorywithMIK

Жыл бұрын

Thanks a lot Shrijal. Glad it helped ❤️

Please don't pause the graph concept playlist and complete it soon, so that we can start with other concept like dp,tree ,etc

@codestorywithMIK

Жыл бұрын

Yep Coming tomorrow

Need Help Can anyone tell me where am i wrong class Solution { public: bool isValid(string st){ if(st[0] == '0' || stoi(st) 255) return false; // leading 0 in 2 or 3 size parts AND range check return true; } void solve(string &s, int index, int &n, vector &answer, string ans, int parts){ if(index == n && parts == 4){ ans.pop_back(); // removing last dot . answer.push_back(ans); } if(index+1

We started index from 0 na So why we are using indx+1

@codestorywithMIK

Жыл бұрын

In C++ substr(idx, length) It includes the current element at idx too. For example : s = “abc” idx = 1 I want to take length of 2 substring s.substr(idx, 2) It will take “bc” i.e. start at idx = 1 And from here take 2 length i.e. “bc” so , idx+2 = 1+2

@twi4458

Жыл бұрын

@@codestorywithMIK Thanks for clearification bhaiya, I had the same doubt.

please help it is showing illeagal character class Solution { public List restoreIpAddresses(String s) { List ans = new ArrayList(); if(s.length() > 12){ return ans; } int n = s.length(); String p = ""; helper(p,s,ans,0,0,n); return ans; } static void helper(String p,String s,List ans,int parts,int index,int n){ if(index == n && parts == 4){ ans.remove(ans.size()-1); ans.add(p); return; } if(index+1

code fat jane ka matlab?

@codestorywithMIK

Жыл бұрын

It means “error ajaega” Code fatna is like a slang used in corporate 😅

@oqant0424

Жыл бұрын

@@codestorywithMIK ok😅

My approach: Only 3 options are possible when we are at an index i of str 1. Only option is to take the char at index i. 2. Only option is to put a dot. 3. Exploring both above options class Solution: def f(self,i,n,temp,result,s,last,dotcount): if dotcount >3: return if i == n: if dotcount == 3: result.append(temp) return if last == "":#only_option_is_to_take_the_current_character self.f(i+1,n,temp+s[i],result,s,last+s[i],dotcount) elif last == '0' or int(last+s[i])>255: #only_option_is_to_put_a_dot self.f(i,n,temp+'.',result,s,"",dotcount+1) else:#exploring_both_options_(taking_current_char_and_putting_dot) self.f(i+1,n,temp+s[i],result,s,last+s[i],dotcount) self.f(i,n,temp+'.',result,s,"",dotcount+1) def restoreIpAddresses(self, s: str) -> List[str]: slen = len(s) result = [] if slen>12: return result self.f(0,slen,"",result,s,"",0) return result