L12. Minimum Window Substring | 2 Pointers and Sliding Window Playlist

[@AyushVerma-wu3nn]

I think there are two points of Correction, that must have been missed : 1. hash[s[l]]++ as we are removing it while shrinking the string len. 2. we need to increment the l pointer at the end of the while loop where while(count == m). Striver did a great job! Thanks

[@venkatsai2325]

yeah u are right

[@tharungr7701]

can you please paste the code here

[@preetkhatri2518]

@@tharungr7701 here you go string minWindow(string s, string t) { int n=s.size(),m=t.size(); map mp; int l{},r{},cnt{}; int len{1000000009},idx=-1; for(int i=0;i0) cnt--; if(r-l+1

[@mohitsingh6717]

@@tharungr7701 string minWindow(string s, string t) { string result; if(s.empty() || t.empty()) { return result; } unordered_map hash; for(int i = 0 ; i < t.size(); i++) { hash[t[i]]++; } int minlen = INT_MAX; int count = 0 ; int sindex = 0 ; int r = 0 , l = 0 ; while(r < s.size()) { if(hash[s[r]] > 0 ) { count++; } hash[s[r]]--; while(count == t.size()) { if(r- l +1 < minlen) { minlen = r-l+1; sindex = l; } hash[s[l]]++; if(hash[s[l]] > 0 ) { count = count -1; } l++; } r = r + 1; } return minlen ==INT_MAX ? "": s.substr(sindex , minlen);

[@KapilSharma56419]

@@tharungr7701 class Solution { public: string minWindow(string s, string t) { int n = s.length(); int m = t.length(); if (m > n) { return ""; } int minLength = INT_MAX; int sIndex = -1; map mp; for (int i = 0; i < m; i++) { mp[t[i]]++; } int l = 0, r = 0; int cnt = 0; while (r < n) { if (mp[s[r]] > 0) { cnt++; } mp[s[r]]--; while (cnt == m) { if (r - l + 1 < minLength) { minLength = r - l + 1; sIndex = l; } mp[s[l]]++; if (mp[s[l]] > 0) { cnt--; } l++; } r++; } return sIndex == -1 ? "" : s.substr(sIndex, minLength); } };

[@nikhilaks2407]

Thank you striver for the awesome playlist🎉🎉🎉🎉 In the optimal approach there was a slight mistake, inside the nested loop it should be hash[s[l]]++ instead of hash[s[l]]-- and after the check for if(hash[s[l]]>0)cnt--; , a line needs to be added for l++; to shrink the window c++ updated code - class Solution { public: string minWindow(string s, string t) { if (s.empty() || t.empty()) { return ""; } vectorhash(256,0); int l=0,r=0,minlen=INT_MAX,sind=-1,cnt=0; int n=s.size(),m=t.size(); for(int i=0;i

[@MayankPareek]

Hey can anyone tell me where is the mistake in my code its not passing all test cases. class Solution { public String minWindow(String s, String t) { char[] hash = new char[256]; for (int i = 0; i < t.length(); i++) hash[t.charAt(i)] += 1; int l = 0, r = 0, sIndex = -1, minlen = Integer.MAX_VALUE, count = 0; while (r < s.length()) { if (hash[s.charAt(r)] > 0) count++; hash[s.charAt(r)] -= 1; while (count == t.length()) { if (r - l + 1 < minlen) { minlen = r - l + 1; sIndex = l; } hash[s.charAt(l)] += 1; if (hash[s.charAt(l)] > 0) count--; l++; } r++; } if (sIndex != -1) return s.substring(sIndex, sIndex + minlen); return ""; } }

[@varun1017]

thank you bhaii

[@mohitthakur5904]

@@MayankPareek facing the same error in java.

[@MayankPareek]

@@mohitthakur5904 take integer array instead of char array

[@deeptitanpure1854]

@@MayankPareek While initializing the hash you should use : int[] hash = new int[256]; Was there any reason why you chose to do a: char[] hash = new char[256];

[@parth_3856]

BTW! LOVED THE NEW WEBSITE INTERFACE.............MORE POWER AND SUCCESS TO YOU STRIVER AND HIS TEAM.

[@ansulluharuka9243]

i am not able to login, are you able to?

[@parth_3856]

@@ansulluharuka9243 yes! but i think there still some work is being done, so that may have caused u some problem in login, BUT let me tell you it's wind in there.

[@ManishYadav-yp2mi]

@@ansulluharuka9243 same here

[@ShrutiSangam-u2s]

same@@ansulluharuka9243

[@nothingmuch140]

​@@ansulluharuka9243 Same broo

[@nuraynasirzade]

The new website is just amazing! I don't have words to say!!!!!! AMAZING!🤩🤩🤩and all of that for free!

[@sandeepyadav-es9yz]

completed this playlist today 27-05-2024 thanks striver!!

[@vineetsingh4707]

Completed this whole playlist in a single day. Thanks Striver for this. The way you teach makes me sit for long, think and implement and gradually the concepts start getting crystal clear.

[@Quavo-goa]

like all of the 283? videos

[@lakshsinghania]

@@Quavo-goa lmao

[@Tushar_995]

@@lakshsinghania Hey ! Is sequence of this playlist proper and completed ?

[@lakshsinghania]

@@Tushar_995 yeah it is, u can blindly follow it just the last qs which is given in the A2Z sheet is not covered here, otherwise good to go

[@Tushar_995]

@@lakshsinghania Ok ! I was thinking to start DSA but was confused between a LOT of channels and paid courses

[@Beeplov2337568]

There was a slight mistake in the video. 1]. l++ in the while(cnt==m). 2]. It should be mp[s[l]]++; instead of mp[s[l]]--; in while(cnt==m) loop. string minWindow(string s, string t) { int l = 0, r=0; int n = s.size(), m = t.size(); int cnt = 0, minLen = 1e7; int startInd = -1; mapmp; for(int i=0;i

[@lovegoyner7450]

Yep I noticed this too 😊

[@rkraj3339]

s.substr(startind,strtind+minLen)

[@lovegoyner7450]

@@rkraj3339 no this is wrong

[@MohitReddyDadireddy-ex5pp]

Yes

[@kushmehta6823]

Hii i found a minor bug, i guess the corrected version should be a bit like this if (mp.find(s[l]) != mp.end()) { mp[s[l]]++; if (mp[s[l]] > 0) cnt--; } because we first have to check whether the character that my left pointer points to even exists in the string t or not if it does not then i dont think i should decrease the count because i still have substrings to measure

[@VirajChokhany]

I'm so happy. I could do this by myself. Thanks Striver !!

[@anubhabdas9553]

Completed the playlist within a day. Sliding window is usually an easier topic as it is totally intuition based, but to identify the patterns and making a structure for all the solutions, you made it look like piece of a cake. Thanks Striver

[@captain-ne8qy]

Why do we store the starting index in brute foce approach ? Can you please explain it to me?

[@anubhabdas9553]

@@captain-ne8qy in the question we have to return our resultant substring right? So here we are iterating and all. Everything is fine but to return the substring, what do we want .....the first index of the substring, that is from where the substring is starting and the size so that we can calculate at what index the substring is ending, that's why we are storing both the starting index and size of the substring

[@captain-ne8qy]

Thnqu for the explanation!

[@SAIGOVINDSATISH]

Make playlist on CP problems for each algorithms atleast 30 each topic from 1200 div to 1800div CP SHEET of yours has very less problems except for dp and math 😢

[@pratikgupta7373]

Master of consistency🎉

[@rayansailani4465]

Adding this comment for understanding of anyone confused: In sliding window problems we're always thinking of expanding or shrinking of window. So when we think of shrinking of window for further iterations, we thinking of adding back the value of map in the inner loop. Here's my code for understanding: class Solution { public String minWindow(String s, String t) { int n = s.length(); int m = t.length(); int index = -1; int minLength = Integer.MAX_VALUE; Map mem = new HashMap(); int count = 0; int start = 0; // populating the map with values from t array for(int i = 0; i 0) count--; start++; } } return index == -1 ? "" : s.substring(index, index + minLength); } }

[@kushal8261]

it should be hash[s[left]]++

[@AdityaGarg-hl7ri]

also left++

[@tharungr7701]

can you please paste the code here

[@nehasingh8181]

Amazing work, Striver! You guys are really doing a great job for us by providing such a brilliant DSA course for free. It's genuinely useful for me. Please upload videos on strings in Striver's A2Z DSA Course/Sheet.......😇

[@mukeshjadhav1064]

Hi striver , best series in the whole world and can you please bring playlist on basics of string

[@janaSdj]

Have completed sliding window playlist. Learned many things. Thank you so much...

[@angelinsnehav3354]

Amazing Playlist💯, understood everything!

[@arka512x]

Corrected Java Solution- class Solution { public String minWindow(String s, String t) { int m = s.length(); int n = t.length(); if (m < n) return ""; // If s is smaller than t, no valid window exists // Frequency array for characters in t int[] hashArr = new int[128]; // Use 128 since characters can be lowercase and uppercase for (int i = 0; i < n; i++) { hashArr[t.charAt(i)]++; } int l = 0, r = 0; int ct = 0; // Count of matching characters int minLen = Integer.MAX_VALUE; // Initialize minLen to a large value int sIndex = -1; while (r < m) { // Expand window by moving r char rightChar = s.charAt(r); if (hashArr[rightChar] > 0) { ct++; } hashArr[rightChar]--; // Decrease the count for the current character // When we have a valid window, try to minimize it while (ct == n) { if (r - l + 1 < minLen) { minLen = r - l + 1; sIndex = l; } // Contract the window by moving l char leftChar = s.charAt(l); hashArr[leftChar]++; // Increase the count back when moving left if (hashArr[leftChar] > 0) { ct--; // Decrease the match count if a necessary character is lost } l++; } r++; } // Return the smallest window or an empty string if no window was found return sIndex == -1 ? "" : s.substring(sIndex, sIndex + minLen); } }

[@tanazshaik678]

that was pretty clear, thanks

[@blurbmusic1507]

heads off to you sir what a great explanation !!!! pure genius❤

[@vijeshsshetty]

Best Playlist on Sliding Window. Thanks Striver !!

[@NonameNoname-f2t]

loved this sliding window playlist ♥

[@ManishKumar-dk8hl]

recursion ki new playlist bhi le aao ( JAVA ) :---- class Solution { public String minWindow(String s, String t) { int l=0; int r=0; HashMap mpp=new HashMap(); int cnt=0; int sindex=-1; int minlen=Integer.MAX_VALUE; String st=""; for(int i=0;i

[@bruvhellnah]

Aur kitne playlist chahie recursion ke tujhe bhai?

[@justanuhere]

thank u

[@akshatjoshi7928]

I enjoy grasping from your videos StriverBhai !! I wanted to highlight one mistake in code that inside while(cnt==m) you did hash[s[l]]-- but it should be hash[s[l]]++ as the condition is if(hash[s[l]]>0) cnt--; and you forgot to write l++ as well.

[@hetpatel7399]

New website is very good.... Nice work sir;🎉

[@divyanshbhatt8273]

thank you striver for this amazing playlist !!

[@parassingh3877]

Day 2 of asking Hey Striver, When can we expect solution videos for Strings, Stack and Queues, few Recursion videos as these are on top of queue, for A to Z sheet everyone needs them Love your content and teaching methods way from Brute to Optimum teaches us alot

[@jatinsingh7209]

I think there is slight correction in the pseudocode, I made these corrections while doing the problem I am posting it here ya'll can refer to it if you are facing problem, " map mp; for(int i=0;i

[@Cool96267]

Thankyou so much Striver for all you efforts throughout in delivering us so much valuable content. Any student / working professional can now be able to transition their career without paying money for courses. Would also like your insights on the point : While preparing for interviews most of the aspirants are going through the videos solely and solving the question after completely watching the video. And also are feeling lazy trying to solve the question on our own. What is the best way to complete any topic without being lazy and how should an aspirant approach any topic/playlist?

[@akashharad4203]

Thank you you are so much hard working keep doing we have less good resources to learn like you

[@MayankPareek]

Hey can anyone tell me where is the mistake in my code its not passing all test cases. class Solution { public String minWindow(String s, String t) { char[] hash = new char[256]; for (int i = 0; i < t.length(); i++) hash[t.charAt(i)] += 1; int l = 0, r = 0, sIndex = -1, minlen = Integer.MAX_VALUE, count = 0; while (r < s.length()) { if (hash[s.charAt(r)] > 0) count++; hash[s.charAt(r)] -= 1; while (count == t.length()) { if (r - l + 1 < minlen) { minlen = r - l + 1; sIndex = l; } hash[s.charAt(l)] += 1; if (hash[s.charAt(l)] > 0) count--; l++; } r++; } if (sIndex != -1) return s.substring(sIndex, sIndex + minlen); return ""; } }

[@aspirant8409]

take Integer array instead of CharArray

[@MayankPareek]

@@aspirant8409 thanks brother

[@zorith]

i lost all my progress with the new website update, when i logged in it again and marked my progress i lost it again and it says unautorized . The old website ui was much user friendly and better . The new update just made it more complex to navigate

[@aditya_raj7827]

yeah i also lost my progress data

[@NonameNoname-f2t]

sign out and sign in again and refresh the page

[@AmruthavarshiniMavuri]

you can view ur notes under saved notes

[@justlc7]

++, lost all data, and new website doesnt save data, keeps reseting.

[@stith_pragya]

Understood...Thank You So Much for this wonderful video...🙏🙏🙏

[@gousikram8490]

in the brutal force for(j= i -> n) this was mistake

[@gousikram8490]

j=i is correct not j=0

[@Abhinavkirtisingh]

Hey Striver, I watch all of your videos and love the way you explain things. I am stuck on a problem called Josephus Problem from a quite long time. Please make a video on it.

[@parth2439]

understood, Thank you striver !

[@popli10]

Thanks Striver. Will always be grateful to you big brother

[@shreyxnsh.14]

Good question, my solution: class Solution { public: string minWindow(string s, string t) { if (s.empty() || t.empty()) { return ""; } vector freq(256); for(int i=0;i 0) count++; freq[s[r]]--; while(count==t.size()){ if(r-l+1 < minSize){ minSize = r-l+1; sIndex = l; } freq[s[l]]++; if(freq[s[l]] > 0) count--; l++; } r++; } if(sIndex == 0 && minSize == INT_MAX) return ""; return s.substr(sIndex, minSize); } };

[@codeguy21]

bhaiya , plz upload remaining topics like strings , stack nd queue plzz

[@nareshAdhe995]

Stacks And Ques Ki Playlist Lao Bhaiya Pls

[@parassingh3877]

Day 1 of asking Hey Striver, When can we expect solution videos for Strings, Stack and Queues, few Recursion videos as these are on top of queue, for A to Z sheet everyone needs them Love your content 🦾🦾 and teaching methods way from Brute to Optimum teaches us alot

[@guneeshvats46]

Amazing explanation!!

[@vtravels2920]

Our sole companion in the journey through Striver's A-Z playlist is his receding hairline.

[@user-fw4kz3bb4g]

MUCH MUCH MUCH love for your efforts @Striver, the new Website UI Rocks!! Also, can you please tell the ETA for string playlist? I'm really holding on from watching others' videos just so that I can follow yours ;)

[@hashcodez757]

"UNDERSTOOD BHAIYA!!"

[@sparksfly4421]

when we are using the left iterator to pop the letter out, shouldn't we use "hash[s[left]]++" to increase the frequency of that letter in the hashmap instead and also add a "left++" after it to keep the "shrinking" going?

[@priyanshugagiya4515]

24:40 it should be hash[s[l]]++; l++;

[@--KarthikeyanS]

hey Striver ,you forgot to increment the l pointer and also in hash[s.charAt(l)]--; it should decrese the number but the value is increasing (example for d -> -2 if we perform hash[s.charAt(l)]--; (-2-1 => -3) so it should be hash[s.charAt(l)]++; (-2+1 => -1) thanks for the amazing Playlist.

[@tahmidhaque3660]

Understood better here Then Apna college channel

[@bhavyashah8219]

CPP code (Striver style): class Solution { public: string minWindow(string s, string t) { vector hashmap(256, 0); int L=0,R=0, minlen=INT_MAX, start=-1, cnt=0; for(int i=0;i0) cnt = cnt - 1; ++L; } R++; } return start==-1? "":s.substr(start, minlen); } };

[@MayankPareek]

Hey Striver, Its been 40 days. Please upload recursion patten wise video first . because recursion is required in advance topic like DP and graphs. So if we have recursion playlist complete we can go to DP and graph with confidence

[@anmolbansal4009]

bhaii tu toh wo nakli binod tharu hai na😂😂......videos bnana chhodh diya kya bhaii??

[@MayankPareek]

@@anmolbansal4009 yes right now , I have a job in Infosys.

[@anmolbansal4009]

@@MayankPareek ache videos bnata tha bhaii.....main dekhta tha tere videos.....band kyu krde....dobara shuru kr bhai bnana videos

[@JoelP-t7c]

Please make a playlist on Heaps and Priority Queue Sir. Since Placement session is going too start requesting you to please upload it soon. Please Sir.

[@wajahatahmed2104]

Very good but I think you forget to add l++ and hash[s[l]]++ should be there.

[@torishi82]

Understood brother. Thank you so much.

[@RonitSagar]

A little catch .......my code was not running when I am executing from j=0 for the inner loop , so if it does not work for u also, just take j=i and check other logic it will work !! below my code Brute Force class Solution { public String minWindow(String s, String t) { int length=Integer.MAX_VALUE; int start=-1; for(int i=0;i

[@MayankPareek]

Hey Striver, I know you have very busy schedule but please drop playlist quicker. Because I started Binary tree as all other playlist before it was completed. But on the hard question it require proper knowledge of Queue and Stack , Priority Queue. Which is not completed in the playlist. So learning became difficult. Hope you understand.

[@akansha538]

bhaiya recursion aur greedy ki bhi ek playlist bana do pls

[@shashanknakashe3339]

thanks striver for this problem statement here is the optimal solution with comments that will help me and you when we do revision for this type of question class Solution { public: string minWindow(string s, string t) { // if(s == t) return s; // if the both strings are equal then we dont have to go down; int cnt = 0; int minLen = INT_MAX; int l= 0; int r =0; int n = s.size(); int index = 0; //creating the hash map to keep therecord of the value unordered_map mpp; for(int i =0 ; i< t.size(); i++){ mpp[t[i]]++; //making flags so that they will help us for the count; } while(r= 0){ cnt ++; } //making the condition for the count while(cnt == t.size()){ //setting the index according the min length if(r-l+1 < minLen){ index = l; //then you update the index; } //first we get the current length and compaire it with our minLen minLen = min( minLen, r-l+1); //now we try to shrink the substring therefore we move l++; mpp[s[l]]++; if(mpp[s[l]] == 1){ cnt = cnt-1; } l++; } //after shirinking we need to extend r++; } //if wwe want to send "" string as an output if(minLen == INT_MAX) return ""; //getting the stirng by the size minLen // index to minLen; wala part return s.substr(index, minLen); } }; keep liking this comment so youtube will notify me about this imp question :) :)

[@glyoxal1933]

class Solution { public String minWindow(String s, String t) { int count = 0; int[] hash = new int[256]; int sIndex = -1; int n = s.length(); int m = t.length(); int minLen = Integer.MAX_VALUE; int left = 0; int right = 0; for(int j=0; j0) count--; left++; } right++; } return sIndex==-1?"":s.substring(sIndex,sIndex+minLen); } }

[@singhji4149]

Nice video, good learning, but can u cover sliding window plus binary search coding question

[@mriit3025]

NOTE : MISTAKE in video ⚠⚠⚠⚠⚠⚠ here is the correct code : class Solution { public: string minWindow(string s, string t) { unordered_map freq; for(char c : t) { freq[c]++; } int l = 0, r = 0, minLen = INT_MAX, si = -1, cnt = 0; while(r < s.size()) { if(freq[s[r]] > 0) cnt++; freq[s[r]]--; while(cnt == t.size()) { if(r - l + 1 < minLen) { minLen = r - l + 1; si = l; } freq[s[l]]++; if(freq[s[l]] > 0) cnt--; l++; } r++; } return si == -1 ? "" : s.substr(si, minLen); } };

[@amansinghal2431]

Hey Striver, I am unable to login to the AtoZ dsa sheet after you updated it. Logged in through google earlier. Could you please check?

[@rock8938]

Same issue with me

[@ujjwalkumarsavita4999]

Jai hoo 👍 khatam hui playlist

[@ABINITHT]

striver when will you upload videos on strings in AtoZ dsa playlist,please upload the video

[@brp3522]

Anyone else is having issues with the new website? I am seeing unauthorized. As soon as I sign it is all fine but when I go to A2Z sheet or any section of the website it says that Unauthorized. Why is it happening

[@anmolbansal4009]

did u get the solution to correct it?

[@ramesh6844]

Nice explanation

[@mightytitan1719]

Bro's the 🐐

[@mohitjangid0027]

Bhaiya please upload video in A2Z playlist because you only hope for me

[@vikkaashl]

pls post java collections vidoes it will be very usefull for the people who are solving the problems in java

[@thisisRandom-ut9iq]

Please do string playlist!

[@AdityaKumar-be7hx]

The C++ solution for reference: class Solution { public: string minWindow(string s, string t) { int left=0, right=0, n=s.length(), minLen=INT_MAX, start=-1; unordered_map freq; for(auto ch:t) ++freq[ch]; int count=freq.size(); for(right=0; right

[@xavier4107]

Bhayy... Please do more videos on this playlist.

[@tanaygada914]

if anyone needs corrected code string minWindow(string s, string t) { int right=0, left=0, n = s.size(); map mp; for(auto i : t){ mp[i]++; } int minLen = INT_MAX; int startInd = -1; int cnt = 0; int m = t.size(); while(right0) cnt++; mp[s[right]]--; while(cnt==m){ if(minLen>right-left+1){ minLen = right-left+1; startInd = left; } mp[s[left]]++; if(mp[s[left]]>0) cnt--; left++; } right++; } if(startInd==-1) return ""; return s.substr(startInd,minLen); }

[@maanavgurubaxani4166]

bhaiya website update hone ke baad jab koi bhi topic ke article kholte hai to undefined tutorial likha ata hai....The articles are not opening.If possible please resolve this issue as soon as possible.It will be a great help.Thank you.....

[@MagicDiamondUniverse]

Please put separate solution videos for all the graph questions

[@MayankPareek]

last question video of two pointer and string playlist, drop it soon please

[@ashishpradhan6250]

Op...mind blowing algorithm

[@limitless6189]

string playlist when

[@premkulkarni7578]

class Solution { public: string minWindow(string s, string t) { int n = s.size(); unordered_mapump; for (int i=0 ; i 0){ cnt += 1; } while (cnt == t.size()){ if (r-l+1 < min_len){ min_len = r-l+1; min_index = l; } ump[s[l]]++; if (ump[s[l]] > 0){ cnt -= 1; } l++; } ump[s[r]]--; r++; } string s1 = s.substr(min_index , min_len); if (min_len == INT_MAX){ return ""; } return s1; } }; Striver's Most Optimal Approach

[@SOURAVROY-ko3os]

In the Brute Force Solution 'j' should be start from 'i', rather than '0'.

[@priyaanshusoni]

Bss Ab STRINGS ki playlist le aao bhaiya please !

[@dasarivamsikrishna9594]

Bro, There is no links of gfg in strivers A2Z sheet, Please enable gfg links It could be more helpful to practice .

[@techyguyaditya]

Striver won't do it, because it's all about business and giventake. GFG stopped giving commissions to him, so he purged all gfg links.

[@adityababu3405]

class Solution { public: string minWindow(string s, string t) { if (s.empty() || t.empty()) { return ""; } vectorhash(256,0); int l=0,r=0,minlen=INT_MAX,sind=-1,cnt=0; int n=s.size(),m=t.size(); for(int i=0;i

[@pushkarbopanwar3297]

CODE IN C++ class Solution { public: string minWindow(string s, string t) { unordered_map mpp; int l=0,r=0,c=0,m=t.size(),index,minlen=INT_MAX; for(int i=0;i

[@Sharath_Codm]

Please update string videos 😥

[@iamnoob7593]

What a playlist , Amazing Thanks Striver

[@angeldeveloper]

Thanks a ton🎉🎉

[@Myanmartiger921]

Anyone completed the playlist anything that is missing?

[@amitnagdev7061]

a situation where, after shrinking the window and finding that the window is invalid, we discover that the required character will not appear again later in the string. What should we do in such a case?

[@amitnagdev7061]

try using this string "ddaaabbcd" you will get more nuance of it.

[@franciskp9117]

Hey man can you upload the solution for minimum window subsequence also ?

[@dhruvmishra9781]

you forgot to l=l+1 at the end;

[@SuvradipDasPhotographyOfficial]

class Solution { public: string func(string s, string t){ int n = s.size(), m = t.size(); int i = 0, j = 0, startindx = -1, minlen = INT_MAX, cnt = 0; map freq; for(auto a : t){ freq[a]++; } while(j < n){ if(freq[s[j]] > 0){ cnt++; } freq[s[j]]--; while(cnt == m){ if(j - i + 1 < minlen){ minlen = j - i + 1; startindx = i; } freq[s[i]]++; if(freq[s[i]] > 0){ cnt -= 1; } i++; } j++; } cout

[@limitless6189]

we need stringsssss

[@RISHABHKUMAR-w5z]

In the first approach, the nested for loop should be from j=i to n instead of j=0 to n, because we are checking for each substring DOES ANYBODY THINKS THE SAME ?

[@rock8938]

I am unable to login with my google account after update

[@tejasjaulkar9658]

when we click on open in new tab (video)both videos started and sounds mixed together .....so may be we can inhace this loop hole of new website master

[@mohankkkrishna8036]

Would you please make a videos on strings 😊😊😊

[@AyushMishra-lb6do]

some issue occurs in your website take you forward when i click mark option page says unauthorised please solve it ..

[@akashsoam7581]

prefix sum playlist too. Please sird

[@saravna]

Having trouble in accessing the website, the progress is not getting updated, it shows "Unauthorized" error message

[@manishmodi8662]

Gen AI would change the world of learning. For one simple problem I wouldn't need to spend 20 mins watching a video. This is my hypothesis. I may be wrong.