Lecture 41: Sliding Window Protocol on Strings

[@CoderArmy9]

Coder Army, How's the Josh?

[@SabyasachiGorai-g9t]

High sir 🔥

[@goandget7945]

Bhaiya I think you need to take some rest. I have seen that you are struggling with throat pain

[@narutodihargo]

High bhaiya

[@prasanna9440]

High sir 🎉

[@im_golu199]

High Sir 😎

[@sushantbhardwaj556]

Today Is Chaat puuja that's why I wake early 😅 and I attended this Lectures. ❤ Jai Chaati maayia . ❤😅😅

[@shrivvlogs6045]

​@@sohamjain8599itne jitne apke population ni h

[@GoogleAccount-sm9hz]

​@sohamjain8599 consider this world to be a project then they are many people who contribute to this making of the project. Then consider each developer as a God. 😂

[@sohamjain8599]

@@GoogleAccount-sm9hz 👍🏼

[@techforyou7518]

Ques -- 1 vector count(256, -1); int ans = 0, start = 0, end = 0; while(end < s.size()) { if(count[s[end]] != -1) { if(count[s[end]] >= start) start = count[s[end]] + 1; } count[s[end]] = end; ans = max(ans , (end-start+1)); end++; } return ans;

[@anujrawat3058]

Thanks bhaiya, you are the best teacher I have ever watched on KZbin for coding 😊 Your way of teaching is very defined and easy to understand

[@universalmeditationmusic9299]

Har din kuch new shikhne mil rah hi our ab coding bhi ashan ho Gaya hi ❤ thank you @coder army

[@shahzaib3469]

amazing explanation !!! much love & respect from Pakistan.

[@Pramod_saraswat001]

Bhai mene to abhi abhi aap se pdna start Kiya h Bhaiya m to app ke teaching style ka bhau bdaa fan ho chuka hu Kya pdate to aap Jese ek badaa bhai apne chote bhai ko pdaa rhaa ho (pure siddat k saath ) (Bina kese loobh lalach k) Thank you so much bhaiya Apna attitude essi trh ka rkhna App k ye Chote bhai aap ke mahanat ko Jaaya nhee jaane denge. hmm bhi tavad tood mahanat krenge aap k saath Aap jese bade Bhai mil jaaye to fir tnsn kis baat ke Bdee bhagyasali h hum coder army team k bande Jenhe aap jese bade Bhai ka saath Mila Thank you bhai ❤❤❤❤❤❤ You soo much

[@gauravkheriwal9559]

love your dedication underateed tutor

[@divyanshsharma673]

This man is always excited and consistent, he brings me motivation to stay hard over this series always 🌟🌟

[@subhojit627]

Motivated to do the lectures with consistency at 6 am❤

[@joydeep-halder]

1:16:08 +1 nhi kiya q ki age wala while loop me second++ ho gya tha loop se bahar ate bakt.

[@colabbro7789]

6:13 BrutForce Approach for the problem. i spend my 1 Hour on this problem to only find brute-force approach. #include using namespace std; #include int main() { string str="ababc"; //cout

[@tiieej3950]

1: brute force class Solution{ public: int findSubString(string str) { // Your code goes here sets; for(auto i:str){ s.insert(i); } int ans=INT_MAX; for(int i=0;i

[@dontreadmybanner2002]

Bro it's not similar to sir how explain but I think it will give ans

[@Ajeetkumar-uo4hf]

Bhai baap level ka explain karte ho. By far the best DSA course on youtube

[@CHAUDHARY_PRAYAS]

finally this video is completed after giving it 6-7 days.

[@waniyafarooq8675]

it means I am not the only one which take 4+ days to complete a video

[@Billy-gd1rm]

me completing this video in 1:30 hrs

[@satyam6943]

Bhailog jisko jisko coding seekhte rhna hai aur Jo abhi Tak consistent hai suno kya hum log ek group bana le jisme hum coding problem bhejenge agar humse solve nhi hota hai to group me daal denge aur jisko aata hai wo bata dega. Aur sath hi sath GFG weekly contest me hissa lenge ek sath kya bolte ho. 5 6 log like Karke reply do to grp banate hai

[@vipulkumar1830]

Best Explaination in the world bhaiya

[@architgupta6109]

hw question 3: Smallest window containing 0, 1 and 2 code : class Solution { public: int smallestSubstring(string s) { // Code here int first = 0 , second =0 , len = s.size(); int diff = 0; vector count(3,0); for(int i=0; i

[@HETPLAYZ]

hw question :- 3 int smallestSubstring(string s) { // Code here vectorcount(3,0); int first=0,second=0,len=s.size(),diff=3; bool check=0; while(second

[@GOATSOFFOOTBALL]

completed ,revise the second problem code

[@Samreen-sam12]

Bhaya love from uttarakhand ❤

[@nitinduke]

37:55 in java class Solution { public int lengthOfLongestSubstring(String s) { boolean[] count = new boolean[256]; int first = 0, second = 0, len = 0; while (second < s.length()) { // Repeating character while (count[s.charAt(second)]) { count[s.charAt(first)] = false;//ascii value count array me change hogi first++; } count[s.charAt(second)] = true; len = Math.max(len, second - first + 1); second++; } return len; } }

[@YogeshwarGupta-r3u]

Bhaiya ek baat bolu.....aap na next video se last me ye bhi bta diya kariye ki kal konsa topic padhayenge jisse hm log bi thoda dekh ke baithe rahenge aur padhte time usko relate karke aur achhse se samajh payenge.......isse hoga ye ki apne se padh kr jo hm galtiya kr sakte hai vo video dekhte time sudhar sake.....aur agar next video me sirf questions lagwane ho to ek do usse milta julta question hme description me de diya kariye jisse hm log bhi thoda hath pao mar le at least brute force ka logic bna paye.......hope aap ise consideration me lenge........Thank You for this beautiful Course 🙏❤

[@HETPLAYZ]

hw question :- 4 int longestKSubstr(string s, int k) { // your code here int first=0,second=0,len=0; vectorcount(26,0); bool check=0; while(second

[@saitanush9453]

Tq bhaiya, Aaj kuch naya seekha😄😄

[@slouchysenpai1970]

you explain everything in depth bhayya, u should have explained this also and then shouldve given homework

[@RockyBahi70336]

Motivated to do the lectures with consistency at 6 am🍵🍵🍵

[@abhinavparashar1934]

Solution of 2nd question Smallest Distinct Window Code start from here :- #include #include using namespace std; int findSubString(string str) { vector count(256, 0); int diff = 0; for (int i = 0; i < str.size(); i++) { // string str="AABBBCBBAC"; found distinct characters. if (count[str[i]] == 0) { diff++; count[str[i]]++; } } fill(count.begin(), count.end(), 0); // for again making them zero int len = str.size(); int first = 0, second = 0; while (second < str.size()) { if (diff != 0) { if (count[str[second]] == 0) { count[str[second]]++; diff--; } else // if (count[str[second]]>=1) { count[str[second]]++; } } if (diff == 0) { while (diff == 0) { count[str[first]]--; if (count[str[first]] == 0) { diff++; len = min(len, second - first + 1); first++; break; } first++; } } second++; } return len; } int main() { string str; cout

[@deepanshuagarwal2471]

class Solution: def lengthOfLongestSubstring(self, s: str) -> int: count=[0]*256 first=0 second=0 len1=0 while second

[@joydeep-halder]

We can optimize the solution of second problem by removing the first inner while loop 🚀🚀 while(second

[@pradeepraghu1see]

second-first kyu instead of second-first +1

[@MohammedHasmi577]

Still I don't understand why this? ​@@pradeepraghu1see

[@newglobal7271]

Maja agya , chamka ❤❤❤❤🙏🙏🙏🙏🙏🙏

[@DevanshGupta-io7rl]

bahut hard challenge hai ye to maza aa raha

[@saketjaiswalSJ]

Hacker or wott bhaii❤❤

[@sohamjain8599]

Bhaiya, please make a detailed video for first year students. We need a 4 year roadmap video so that an can get an idea.

[@4minix643]

class Solution { public: int lengthOfLongestSubstring(string s) { int maxLen = 0; int n = s.size(); if(n == 0)return 0; //iterate over all possible //substrings for(int i=0; i

[@ankushladani496]

Done Bhaiya... Day 57/180 ✅

[@rashmimohan1863]

Great Explanation 😊

[@Vaibhavvashishtha123]

Can anyone pls share the code and logic for homework question longest substring with k unique character

[@Babita-y7w3o]

Ques 1___ BRUTE FORCE... string s; cin>>s; int ans=INT_MIN,w=1; while(w

[@Ikkhhku]

Good Morning 🌞

[@shrivvlogs6045]

Love u bhiya ❤️

[@shubhamkumarjha9192]

Good morning bhaiya..#day57

[@joydeep-halder]

Good morning bhaiya ❤❤

[@technicalraftaar1487]

Love from Bihar

[@AIWALABRO]

Brother I am learning Data Science, and most of the DS have the fear of DSA. i mean they don't have indept knowledge of DSA. . 1) let suppose, i have to continue from this video then can i continued or i have to follow some previous videos of yours. 2) most of the time DS code in python , so can I do DSA in python? or you suggesting some thing different. 3) if I want a code of this problem in python then where should I get. I already registered of 99 Rs in your website, so are you tech supporters provide support over there? I am waiting for your response...from last 3 days happened. if you have mail id then pls provide if any kind of queries so i can direct reach out to you!

[@Pramod_saraswat001]

Good morning bhaiya

[@motivationalmajakkhel]

Homework: Smallest window containing 0,1 and 2 int smallestSubstring(string s) { // Code here int length = -1, size = s.size(); vector a(3, 0); for(int i=0; i

[@sanketbiradar7978]

Best video

[@rhitam_biswas]

18:03 Rohit bhaiya's "kuch bhi ho sakta hain" accent bohut zyada similer hain Bhaibav's "Kuch Bhi Ho sakta hain " ( Kota Factory) 😂

[@Sharmila-j4e]

guys ,the 1st question was asked in accenture interview

[@ahmadhassan90590]

high sir 🔥

[@Aranyak.123]

unique character ke liye set use kr skte the kya??

[@Iamvijay23]

Jab Tak todenege nahi tab tak chorenge nahi Day 57/180

[@cosmickingthor6676]

good content bhai....

[@anuragyadav5065]

v.clear() run karne prr saari values 0 nhi hoke 1 ho jaati hai aisa kyu?

[@cybrarygamer486]

First comment bhaiya 🎉🎉🎉🎉 in this video

[@GopalKumar-xu3iw]

Day 57/180 done ✅✅✅✅✅✅✅✅

[@allinonemoviesyt]

good morning sir

[@firecrusher1384]

chamak gaya bhaiya.

[@Mr.FalseChecker]

Thank you very much

[@deepanshuagarwal2471]

str = "aaabcccccccaaaabbbb" count = [0] * 256 first = 0 second = 0 diff = 0 let = len(str) # Count the number of unique characters while first < len(str): if count[ord(str[first])] == 0: diff += 1 count[ord(str[first])] += 1 first += 1 # Reset variables for sliding window for i in range(0, 256): count[i] = 0 first = 0 # Initialize let to store the length of the smallest substring containing all distinct characters while second < len(str): # Slide the window until all distinct characters are present while diff and second < len(str): if count[ord(str[second])] == 0: diff -= 1 count[ord(str[second])] += 1 second += 1 # Update let let = min(let, second - first) # Slide the window until there's a character repetition while diff != 1: let = min(let, second - first) count[ord(str[first])] -= 1 if count[ord(str[first])] == 0: diff += 1 first+=1 print(let)

[@Pramod_saraswat001]

Bhaiya Mene DSA playlist ko last week se start Kiya h M abi 14 lecture pr hu Kya m app k saath ess ko continue kr skta hu Kya yhaa se Mere samjh m aa skta h Yaa fir m lec 14 se hee continue rakhu

[@lyricalboy9281]

day 57 completed #180dayscodingchallenge #coderArmy

[@yashdabral2773]

Homework question - 3 int smallestSubstring(string S) { // Code here vectorcount(3,0); int first = 0; int second = 0; int length = S.size(); int diff = 0; for(int i = 0; i< S.size();i++){ if(count[S[i]-'0'] == 0){ diff++; } count[S[i]-'0']++; } first = 0; if(diff < 3){ return -1; } for(int i = 0; i

[@a.gcrazy555]

why can't we use map instead of vector it will save memory in first question;

[@heetpatel3037]

Chamka par Aaj thoda hard Laga😊

[@drlovephukn]

Day 57/180 , full power full power

[@asad_iliyas]

Homework question 4: int longestKSubstr(string s, int k) { // your code here int n=s.size(); vectorcount(26,0); int first=0,second=0; int uniquechar=0; int length=0; //check if answer possible or not for(int i=0;i

[@yashubisen7131]

Gm bhaiya ❤

[@ayeshasolanki5386]

Are you not going to upload homesheet on strings?

[@AakashParvani]

Day 57 👍✅

[@air1topped738]

Bhaiya kai recession hoga if yes than when it will end

[@dipakmali9817]

Day 57/180 Done ✅

[@farhansadik7236]

why not second-first+1? can anyone explain this

[@ankushladani496]

Thanks Bhaiya...❤🎉

[@motivationalmajakkhel]

Homework 2: Longest K sub string int longestKSubstr(string s, int k) { int length = 0; int diff = 0; vector count(256, 0); for(int i=0; i

[@vasupratapsingh3652]

👌🏻👌🏻

[@akashkewat1219]

🎉thank bhaiya😊.

[@lyricalboy9281]

Bhaiya homework questions nhi hote aur leetcode pr jao toh naye questions bante hi nhi consistent bhi hai code ko dry run bhi karte hai aur homework solve karne ki bhi koshish karte hai uske baad bhi😢😢

[@GopalKumar-xu3iw]

Same bhai wahi problem h

[@kartik5134]

34:44 chamak gya bhiaya

[@Aniruddha_Mukherjee]

Josh is always high Bhaiya.

[@HARSHRAJ-gp6ve]

Bhaiya smallest distinct widow ke ques ka solution de do plese c++ ka 0(n^2) wala

[@shrivvlogs6045]

Chamka++

[@omsingh00017]

2nd problem to bilkul samjh nahi aayi...aaj laga recorded or live me difference...jitne baar bahiya kah rahe ..samjh a gaya hoga utni baar gussa lag rahi...

[@carcompares9022]

har har mahadev

[@indian_gaurav_8648]

33:11 bhiya ise hum while loop se pehle likhna chahiye na or while loop ke andr hum first ko 0 kar rhe h wo to pehle se hi zero h

[@divyanshsharma673]

But also count[S[second]] is zero so that loop to make first 0, wouldn't execute.

[@Billy-gd1rm]

❤❤

[@adityapawar6174]

Lecture 41 check ✅

[@procrastinator0811]

2. PYTHON Solution (Smallest Distinct Window) #User function Template for python3 class Solution: def findSubString(self, s): # Your code goes here #find num of distinct char in the string unique_char = len(set(s)) if unique_char == 1: return 1 min_len = len(s) # print(unique_char) freq = {} first = 0 second = 0 while second < len(s): if s[second] not in freq: freq[s[second]] = 1 unique_char -= 1 if unique_char == 0: #we got one probable ans min_len = min(min_len,second - first + 1) #let's try to reduce the min length further while True: freq[s[first]] -= 1 if freq[s[first]] == 0: unique_char += 1 freq.pop(s[first]) first += 1 break else: first += 1 min_len = min(min_len,second - first + 1) else: freq[s[second]] += 1 second += 1 return min_len

[@deep_singh01]

Day 57/180 👍👍

[@nitinduke]

01:14:24 in java class Solution { public int findSubString( String str) { // Your code goes here int[] count = new int[256]; int first=0,second=0,len=str.length(),diff=0; //Calculate all the unique char while(first

[@ShaikSharma]

chamak gaya bayya

[@AbhishekKumar-lu3lp]

Bhai india har gyi

[@colabbro7789]

Bhaiya first problem ka brutforce approach hi nahi bani . Nahi o(n^3) time aur na hi o(n^2).

[@awadhigeet9692]

Bhai longest sub string ka to O(n^2) hoga n kaise aya samaj nahi a eha

[@joydeep-halder]

✅✅ HW Solution for problem 3: int smallestSubstring(string s) { // Code here if (s.size() < 3) return -1; vector count(3, 0); int first = 0, second = 0, len = INT_MAX, diff = 3; while (second < s.size()) { if (count[s[second] - '0'] == 0) diff--; count[s[second] - '0']++; second++; while (diff == 0) { len = min(len, second - first); count[s[first] - '0']--; if (count[s[first] - '0'] == 0) diff++; first++; } } return (len == INT_MAX) ? -1 : len; }

[@dayashankarlakhotia4943]

public int findSubstring (String str){ int n=str.length (); Setset=new HashSet (); for(int i=0;i

[@procrastinator0811]

1. PYTHON CODE def lengthOfLongestSubstring(self, s: str) -> int: l = len(s) if l == 0 or l == 1: return l first = 0 second = 0 char_freq = {} max_len = 0 while second < l: if s[second] in char_freq: while True: if s[first] == s[second]: char_freq[s[first]] = second first += 1 second += 1 break else: char_freq.pop(s[first]) first += 1 else: max_len = max(max_len,second-first+1) char_freq[s[second]] = second second += 1 return max_len

[@nownow1025]

//smallest distinct window logic same just code part different #include #include using namespace std; int findSubString(string str) { vectorcount(256,0); int diff=0; for (int i = 0; i < str.size(); i++) { // string str="AABBBCBBAC"; found distinct characters. if (count[str[i]]==0) { diff++; count[str[i]]++; cout

[@Rishi_s.corner]

#revision done of string lect 41 on date 3/3/2024

[@averagecoder7]

Day 57/180 ✅