First non-repeating character in a Stream | LinkedList | GFG | Love Babbar DSA Sheet | Amazon🔥🔥

#Linkedlist #competitiveprogramming #coding #dsa
Hey, Guys in this video I have explained with code how we can solve the problem 'First non-repeating character in a stream '.
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[@RajatSingh-vs6wu]

If you are getting TLE on gfg there are two things you can do you optimize your code : 1) Currently the time complexity of the code in the video is O(n^2) since the size of the both the "vis" vector and "v" vector is n. This can be optimized down to a time complexity of O(26*n) by only adding unique characters in the "v" array instead of all the characters in the string. This will make sure that the size of "v" vector does not exceed 26. 2) Instead of iterating over the whole "v" array again and again from the 0th index to the last index, you can store the index of the previous "first unique" character found, say "X", and in the next iteration start the loop from "X" rather than the 0th index since if you won't be find your unique character in between 0th index and "X"th index anyways. Below is the updated code with these optimizations( I have used a map instead of "vis" vector) : string FirstNonRepeating(string A){ mapmp; vectorv; int n = A.size(); int cnt=0; string ans=""; for(int i =0;i

[@MritunjayKumar-rd2es]

lekin main to linked list approach dekhne aya tha🥴🥴

[@rishabsharma5307]

Simple C++ Solution string FirstNonRepeating(string A){ // Code here string output; vector vect(26, 0); queue mq; for(auto x : A) { int idx = x - 'a'; if(vect[idx] == 1) vect[idx] = 2; else if(vect[idx] == 0) { vect[idx] = 1; mq.push(x); } while(!mq.empty() and vect[mq.front() - 'a'] == 2) mq.pop(); if(mq.empty()) output.push_back('#'); else output.push_back(mq.front()); } return output; }

[@navendraagrawal]

thanks bro your solution is much better than this video

[@maharishicoding440]

Approach: 1. we have to find the 1st non repeating character not non repeating character only keep in mind. 2. So for that we have to initialize 3 thing 1. Array to keep the count of non repeating character, 2. Queue to Keep the record of 1st non repeating character,(you can use arraylist as well) 3. We also need a variable to store a result. 3. Now iterate increase the frequency of the array for repeating character and add the 1st non repeating char to the queue then add it to the String variable at each iteration. here is my Java code: class Solution { public String FirstNonRepeating(String A) { StringBuilder result = new StringBuilder(); // initialize a string to store the output int[] freq = new int[26]; // initialize an array to store the frequency of each character Queue queue = new LinkedList(); // initialize a queue to store the non-repeating characters // process each character in the input stream for (char c : A.toCharArray()) { freq[c - 'a']++; // increment the frequency of the current character // if the current character is non-repeating, add it to the queue if (freq[c - 'a'] == 1) { queue.add(c); } // remove all characters from the queue that are no longer non-repeating while (!queue.isEmpty() && freq[queue.peek() - 'a'] > 1) { queue.remove(); } // if there are no non-repeating characters in the queue, append "#" to the output string if (queue.isEmpty()) { result.append('#'); } else { // otherwise, append the first non-repeating character in the queue to the output string result.append(queue.peek()); } } return result.toString(); // return the output string } }

[@faizannasimhyder9011]

Thanks for the excel sheet more than the video bro. I will practice all the questions in sequence. In order to remain organized.

[@hemantsurariya3239]

IDK , its showing TLE. I tried several time.

[@DurgaShiva7574]

hi, thankyou for this amazing video... one quick question, i think the space complexity in your code is 52 ...and not 26..right?... because we have an visited array of 26 + another vector which might also range upto 26 characters at same time.. please let me know

[@dhirendrasingh6071]

It's considered constant space

[@randomtalks1234]

great explanation brother.......

[@bhaveshsoni6684]

if the same logic applies in python then where we store characters bcoz if we can store characters and their counter in the array, then its output is TLE....

[@shahidagarwal2868]

Hello Yogesh. I have been seeing that you are doing some good to the community.. But instead of just running and explaining through your code which u already coded.. You can code in front of them so that they can understand it much more clearly.

[@tusharsolanki8564]

yes definitely or ye pta chl jayega logic kese click hua

[@CodeLibrary]

If I code in front of them then video will become too long that's why I explain in the already written codes😀

[@Yash_Parashar]

@@CodeLibrary please explain for last c in test case how your code give b??

[@debdhritiroy6868]

12:28 very true, same algo java me TLE derha h, aur cpp me accept ofc, time limits java me zyada hone k bawajud

[@hannanathar3627]

use stringBuilder for ans instead of string , it wont give TLE then

[@rutujahande5850]

Here, output is coming as wrong - aabc ####

[@yash_verma]

string ans will take O(n) space complexity

[@malharsurangalikar6224]

Same approach gives TLE for java on GFG

[@nimratkaur6153]

try using StringBuilder

[@vishalsingh2408]

Good job brother...

[@payalsagar1808]

Thanks a lot😁

[@shivamdalania]

Can someone please explain me how this is related to linked list?

[@AnkitSingh-wq2rk]

Doubly Linked list ka question hai yeh Ek queue maintain karenge Doubly LL type jisme possible candidates for First non repeating char rakhenge in case agar element repeat hota hai toh woh ab possible candidate nhi hoga first non rep ke liye hence hame use queue se nikalna hoga so for O(1) deletion we are using LL

[@Piyush-yd1ez]

Ye question queue section me bhi hai.

[@mdsaifhussainiqbal2236]

queue approach code string FirstNonRepeating(string str) { int n = str.size(); queue q; unordered_map umap; string ans = ""; for(int i=0;i

[@MrCoder-xj9yi]

good job, keep it up!

[@yuvrajsharma2134]

noice

[@deepanshukumar7290]

bhai quick sort

[@prajwalingole1584]

opppp