Spiral Traversal of a Matrix | Spiral Matrix

[@shubhamagarwal1434]

#Free Education For All.. # Bhishma Pitamah of DSA...You could have earned in lacs by putting it as paid couses on udamey or any other elaerning portals, but you decided to make it free...it requires a greate sacrifice and a feeling of giving back to community, there might be very few peope in world who does this...."विद्या का दान ही सर्वोत्तम दान होता है" Hats Off to you man, Salute from 10+ yrs exp guy from BLR, India.....

[@navkaransingh3948]

Hi striver, i think you already know but these series will be legacy for future coding students So, i hope you don't rush this series I'm saying this because i am seeing a lot of comments saying to complete it very fast.

[@takeUforward]

Yes I will be going at my pace, don't want to compromise on quality.

[@shivamprajapati8662]

thank you @@takeUforward

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​@@takeUforwardthank you so much bro❤

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True that in 2024 still can't find a better DSA course than this 🫡

[@habeeblaimusa4466]

Understood. The height of this series is you knowing how the code will be after the dry run.. Thank you Striver

[@shashamnk2525]

True

[@AbhinayKrishna4]

Man you literally are a lifesaver, providing such a one stop destination for all the coding questions required to crack any interview.Thanks, striver for your dedication and hard work towards providing quality content.

[@juliechoudhary9582]

This course is really worth more than any course but still it is free.Thank you so much.

[@yashkhatri3636]

you are providing such an amazing content at free of cost really india needs man like you !🤠

[@ruchirtd1]

Finally was able to solve this problem with 100% understanding! Thanks a lot.

[@maccode07]

UNDERSTOOD!!!! HAPPY FRIENDSHIP DAY STRIVER

[@mdfaizanmdfaizan6041]

You are from which college bro...

[@mdfaizanmdfaizan6041]

You are from which college bro...

[@maccode07]

I2IT

[@indroneelgoswami5654]

Unbelievable Explanation!!

[@mdfaizanmdfaizan6041]

You are from which college bro...

[@indroneelgoswami5654]

@@mdfaizanmdfaizan6041 TINT, Kolkata

[@yashgarg1961]

Saw several videos on this question. This was the perfect and the easiest one. Hats off to you bhaiya. Thanks for such quality content :)

[@shreeyarao8265]

So well explained! Understood every bit of it!

[@tonylee1868]

great content... i know it will be very difficult for you but we need more videos as soon as possible

[@JK-de2gh]

we don't need like this...we need quality...take your own time striver...

[@enzaki7603]

Thank You Striver Bhaiya ! Your playlist is really helping me a lot :)

[@akashverma5756]

Simulation approach is more intuitive and clean if you uderstand direction co-ordinate. Direction co-ordinate help you solve many other matrix problems easily.

[@navkaransingh3948]

Bro, wdym by simulation approach?

[@shadowslayer2248]

Understood. Thank you for striving for us and surely we'll also strive to make your channel and the A2Z course be rated as the best in the world.

[@leetcoder6569]

Why there's so much rush in CB, RAJ BHAIYA is a human being too, we have sheet na, we canbdo it on our own pace..

[@cinime]

Understood! Wonderful explanation as always, thank you very very much for your effort!!

[@KesavaReddyKuppireddygari]

watch in 1.25 is a myth

[@iAnuragJha]

true that

[@Try_me_]

wdym? im watching all his videos in 1.5x

[@lucario404]

edge cases are mat[m][1] and mat[1][n], because of which 2 checks are needed

[@prathameshprabhu-ed8dl]

thanks i was not getting sense of if(left

[@neoh1471]

your solution is good, but i have another idea, i'm using javascript shift(), pop() and reverse() to get the array element into my answer array, then every time i process a row, i'll transpose the matrix and i will reverse() it if needed, then repeat untill the matrix become empty, and print the answer array. All code is written in a while loop with lambda functions. let ans = [] while(matrix.length != 0){ matrix.shift().forEach(x => ans.push(x)); if(matrix.length == 0) break; matrix = matrix[0].map((_,colIndex) => matrix.map(row => row[colIndex])); matrix.pop().forEach(x => ans.push(x)); if(matrix.length == 0) break; matrix = matrix[0].map((_,colIndex) => matrix.map(row => row[colIndex])); matrix.pop().reverse().forEach(x => ans.push(x)); if(matrix.length == 0) break; matrix = matrix[0].map((_,colIndex) => matrix.map(row => row[colIndex])); matrix.shift().reverse().forEach(x => ans.push(x)); if(matrix.length == 0) break; matrix = matrix[0].map((_,colIndex) => matrix.map(row => row[colIndex])); } // print ans here

[@mukeshKumar-go5cz]

thank you for providing best content for us. we really love it.

[@ImRich-xo8cm]

Whenever I see you I got inspired ❤

[@suraj_patwa]

Really this is a better to paid course Thank you striver bhiya

[@bhagyashreekhairnar683]

great explanation as always!! really clear!! Please continue making these videos.

[@Manishgupta200]

Understood ❤ I do it by two different technique. Approach is same but condition is different

[@mdfaizanmdfaizan6041]

You are from which college bro...

[@chinnadhodlekavitha6540]

Please add videos related to strings as soon as possible.

[@rishabh1S]

Very good explaination!! Thanks Striver.

[@indroneelgoswami5654]

unbelievable explanation!!

[@amankumarverma8426]

amazing approach!

[@SibiRanganathL]

Understood.

[@hrushi_borhade]

Thank you striver for great content!!

[@harshilpatel3205]

Understood 🙏🏻

[@who__are___you____26__]

Thank you very much for this course learnt a lot and your dedication is on another level you were recording this video on 3 am thanks a lot 🙌🙌

[@rohandhalpe9084]

Sir please , upload videos as fast as possible. 📤📤📤🕔🕔🕔 No more time remaining for our placement

[@notcliche4864]

Are bro, wo bhi toh insan ha ek saath kitna kaam krega

[@alokkumarmishra5172]

bhai sath me dp series ya graph ya tree series padh sakta hai it also help for your upcoming placements

[@aman_singh__]

bro tum wait karoge to nhi hoga khud se karne ka try karo. ek banda akele kitna hi kar lega

[@rohandhalpe9084]

@@aman_singh__ 🙌

[@Ri-hg6hs]

providing these well structured series by you feeling like that giving water to tree thnx by core of heart.

[@milindjarsaniya]

UNDERSTOOD as always!

[@AshishSingh-he2qo]

Is there any way so that i can like this video more than 1 time ❤❤

[@AniketKumar-hf2bo]

understood ,thnx for excellent explanation ❤❤❤❤❤❤

[@NitinKumar-wm2dg]

Thank you bhaiya, understood

[@AmareshwarBhat800]

Hi @takeUforward I did not get the last contraint from your explaination as you are too fast at the end. But I got it. Thanks for the video.

[@AjeyaKumaraK]

understanding

[@a_maxed_out_handle_of_30_chars]

this was good, thank you bhrata :)

[@raynwani137]

Bhai U are an Inspiration kasam se Career wise!

[@hareshnayak7302]

Understood,Thanks striver for this amazing video.

[@infernogamer52]

Understood Bhaiya!

[@NazeerBashaShaik]

Great explanation as usual. Thank you.

[@diyaghosh3721]

understood! amazing explanation!!

[@ishangujarathi10]

lovedd the intuition

[@MMirabel]

Understood..Very clean explanation!

[@arnavkumar4631]

Understood

[@user-cq6en6rd4q]

This is one of the hardest questions I have come across in a while

[@ce038_divanshsingh3]

And u r also looking hard.😊

[@Yup728]

@@ce038_divanshsingh3 chill bro wth u saying

[@khalasianiket816]

understood❤

[@konankikeerthi]

Thank you bro. understood clearly.

[@Sachin-x4m7c]

I was able to solve this by myself but used one pointer only, this is very easy than my code lmao

[@priyachauhan813]

Understood at one go Thank You Striver

[@AbhishekKumar-cd4gg]

At times i feel by watching your videos that i am just seeing the solution and at the end it just feel like spoon feeding , there is no feeling if growth in myself so i request @takeUforward aka striver to please make a video the correct way to approach a problem and how should be approach to your series and problem . if felt same do give like to ensure the comment reaches out to striver

[@sayakghosh5104]

Understood! Awesome explanation.

[@pratyushtripathi1728]

Understood 😊

[@VineetKumar-fk2rl]

Understood amazing explanation as always❤❤

[@moharramansari2594]

The way you explain

[@shristipurusarthi6675]

thank you soo much brother for the wonderful explanation

[@kingbadshah452]

thanks striver understood everything

[@culeforever5408]

understood partially 🤔

[@dinimeegada9167]

Understood I have been waiting for this video from Striver anna ❤️ U Anna

[@sarangkumarsingh7901]

Another Awesome Lecture Sir............

[@tarushidubey2973]

you are amazing sir huge respect ...

[@VarunGodiya]

please complete it as soon as possible .

[@inspiringzone12]

Quality is rare with speed

[@NithinvKumar-uk2he]

Yes please do complete fast it will be so much helpful for our upcoming placements please

[@lakshsinghania]

I saw at the bottom "last saved on" 3:57 AM 3:58 AM 4:00 AM" orz striver bhaiya, staying awake and recording the lecture at late night

[@ru_mum77]

Sir Thanks a lot for the excellent explanation 😁

[@karanveersinghdeora5085]

understood

[@p4avinash]

Understood, Great explanation.

[@gautamsaxena4647]

understood bhaiya

[@oyeesharme]

thanks bhaiya

[@architmishra0057]

I Think the last IF condition is irrelevant because by the time the control reaches the last loop, the TOP will be greater than BOTTOM

[@abhishek__anand__]

Great Explanation

[@samuelfrank1369]

Understood. Thanks a lot

[@sauravchandra10]

Instead of checking several conditions multiple times, can we maintain a counter which tells us what to print? This the the implementation, and this code runs on both LC and CN int n=matrix.size(),m=matrix[0].size(); int top=0,left=0,right=m-1,bottom=n-1; int counter=0; vector temp; while(top

[@23cash86]

Intersting

[@placement123]

@@23cash86 good catch buddy..

[@ProgrammingKT]

Nice explanation!

[@priyankavasam8769]

wonderful explanation

[@falaknaaz789]

Thank you sir❤

[@Heartless-he8km]

Ericsson ???

[@SHAKTI2.O]

It is very confusing to understand where the right left top and bottom corner lies...I think it is very basic but at the time very confusing also

[@alheraahmad5481]

understood!

[@DeadPoolx1712]

UNDERSTOOD;

[@ysabhishekmishra]

class Solution { public: vectorans; int n,m; void dfs(int i,int j,vectormatrix,vector&visited){ if(i=n || j=m || visited[i][j]){ return; } visited[i][j]=1; ans.push_back(matrix[i][j]); if(i

[@reddygopichand2002]

Understood thanks 👍🙏

[@hariomatkar4689]

Itne aasan se bataya ki aadha dekhke hi pura pta chal gya

[@Sumeet_100]

just amazing !!

[@GhostVaibhav]

Understood🔥

[@student_03]

understood sir thanks a lot

[@djpro6413]

bhie bhut kadak smja bas tumhi bas last ka spiral ki diagram smaj lo lotebook mai or dry karo smj jauoge viase usne batya ki firat 2 condiotion cheak ho jati hai bas ek line agar print karni hai first two condition se ho jata hai bas topp++ hotnepar us mai vo cheak lagate hai or bottom ke lieye ek cheak taki bottom to tpo value replaced ho jayegi uske liye matlab ki value print nahi honi chiye. thanks ..like me if you understood

[@star-yz3xv]

If there is already a while loop that says top

[@charan123rams3]

thankyou sir love you so much 🥰

[@SYCOA12CHAITANYAASOLE]

Understood !! 😍😍

[@quickknowledge1401]

Best approach

[@computerscience68]

Thank You Bhaiya 😊

[@torishi82]

Samaj aa gaya!!

[@PSMADHURIHSIIPCMCREG]

understood👍

[@prabhagaikwad4849]

Understood!! This problem link is not updated in A2ZDSA course sheet. Thanks

[@shubhammeena2316]

thank you so much bhaiya

[@shreyasinha4016]

just a doubt while loop me condition h fir wapas q check kr rhe,? top bottom ki values tpo mana change ho rhi pr ek bar condition dalne ke bad dubara q dalte wo update nhi hoti values ke sath

[@sushma6314]

kyunki after first two for loops, top will be increased and right will be decreased. toh check karne ke liye ki while loop ka condition abhi bhi true hold hain ki nahi,