Wonderful way of explaining the algorithm behind the program. Most channels only explain(show) the code. Please keep up this great work. Please do mostly asked Leetcode medium questions. Will be a great help. Thanks a lot.
Sorry to be off topic but does any of you know of a way to log back into an Instagram account..? I stupidly forgot the account password. I would appreciate any tips you can give me!
@Johnny Zyaire I really appreciate your reply. I found the site thru google and im in the hacking process now. Takes quite some time so I will reply here later when my account password hopefully is recovered.
One of the best intuitions I have ever seen. I love the way you take examples to derive the logic. Keep it up. And also thanks for delivering such a quality content on coding problems. You just earned a subscriber.
Very nice explanation better than other explanations out there which just throw terms like greedy etc. Giving concrete examples and explaining the algo through them makes it much easier to understand.
Your explanation is really good and with O(N) which is better. I solved this problem with different approach and I think the time complexity id O(N * k). Is that correct? Step 1: Parse the string left to right. Step 2: For each current char at i, See if there is a char lesser than the current char in window [i+1 to i+k]. If found, decrement the k and skip the char. Otherwise add to the result. Here is the code: public String removeKdigits(String num, int k) { if(num == null || num.length() == 0 || num.length() 0){ builder.delete(builder.length()-rem, builder.length()); } String res = builder.toString(); return res.length() == 0? "0" : res; } int getMinIndex(char val, int start, int end, String num){ if(end >= num.length()){ return -1; } for(int i = start; i
Awesome explanation! Thanks! As per the explanation, just remember this: Remove previously seen peak elem to flatten the curve. When graph dips, we have already seen a peak.
Pretty good explaination bro ! I was able to solve it in my mind but couldn't put it in code . DIdn't think about first-come maxima approach . Cheers (y)
Initially I thought this is a DP problem because I thought we have to consider all possible cases and the do the DP thing... but this approach is really wonderful.
One thing to notice here if the elements keeps on increasing as he said it will form the smallest number possible that means we will have the higer value digits at the end so we can remove them easily with the help of the stack
I got a goodone idea to solve this say we have the number 71382 and k=2 so what we can do is that to divide the digit by its index number, index number must be starting from 0 ie for 7 index number should be 1 not 0 and so on,, then after we got the list of the divided result we can just chose the minimums possible and then we can include only those in return value, me testing this got all correct result for almost all cases
I've been into problem-solving for over a year, and this is the best channel I have seen. the channel is better than Love Babbar, TUF, Anuj Bhayya, and all the members out there
Thanks. I hate your 4's but the overall the video was quite helpful. I solved 402 with a different approach, but wanted to see how the stack based approach worked. No I know.
In 10:58 , what if the next element of zero is 2 Instead of 1 , then our results first number should be 1 . But we would have already removed it . I think it will fail for that scenario
Sir the example at 12:00 mins of number 14301620 Nd k=4...the answer if the think on pen n paper should be 1010....but why it is 120 ?.... technically thinking the answer should be 1010 after removing 4 digits ...
@@saianushachodapaneedi7504 If you add 0 in stack, while displaying result it will have leading 0, in the example shown by LC they removed leading zeroes so we can ignore leading 0s as mentioned here or parse the string integer and back to string so that leading 0s are gone. In order not to put extra effort, ignoring leading zeroes will suffice the requirement.
How does one come up with such intuitive solutions? I tried to come up with multiple solutions but was not even close to this solution. I tried to solve this question using priorityQueue, HashMaps, expanding from center of the string.
leetcode gave me this edge case for the following problem : "9964143637881536115347130215819342018286368478941148499497648482711459533461004254405644237053081606083361703097681313409879305615510320200854934444651935440830730707105506511273844202022847156939479617415032308016122590196433883734152425324321850950025461922368309179936567772058238956815658522566619712309337787110000248140248621983725541138216249932119341652497227527347196915584874245747655643373728114771002178121013476532176128561227651191732455258556988215663509467036265356759614472865379500702323093381756610448863769645016608790510082369942579876359844432606935705284237991853585529691576005445931743352187877817181108107659316666309094802979311362685246271238811648377471342618391148123088439359424933182816550379826963424443077369303388270800024963285014879985932202469314657763554311465766241899886051752598919297325070163176559846505309761680481734434389501672456194786083611750491579701118513267425527823659774604213876847372305982594830156571943761073290766254549904295349986681374115730100337100520099231426507753102943794893125561715341714882235592831859851753324171964100271220487416100160426921656692822076747471313551671799749136336020476415426498900467136354109743348482211848364210754765858145061682176996476703252113885157082272913476246001426543322720172472400433849455239728009056816478610972157143620619838281484903385698733878747333577266693321881082248284899461049170566515551638479945941859455913682794110638768950164185110174329858257546630386490667378849662828892086742219395018081013139661291385111259380764915297206827929993411346366971457561364592936565292180883672568239002607555932099570488014976458337969750530347455002905982811683646920337042844933044228156313556893574266924334421860399441795570348505986213235968877629037821039295531078587452820520378855971549385240599138029027426814355797039844185115797752068944043026514402978978851104279587917456738135851069474951293893468797930509914957546422062980494255056416478680885689780986382412165954803439553973540706927945767861390922237184889229475493329909116465687108626908432452951254474743412354718972999375833762203809869944881570164493465129271906768322772743880895596954354231919788356736973386736425035313669786510718228292965591836221183232782757135478753561150173194385615500385373233981959493952471916956111069857167656232936080328221546753405850472812773151559894114363782701095557909245140582135212670655043831517604969231621049089970261307870253571673590180617152285302103559731670339047857148567799820792277393882937146083861121444641752891357528477673783704643969540852343441491634297968882019783645863769499154099829169034519420545243923982738295303981036771224459015594039438755491178665247811195429718554410638417459245168087508373787473581006876786621492463488551382880888016193098727660257087286075211981304241455039635843389359277754175667320688287674673170776696626809610432006193791350589302883359213754039606437515551397976472818092708306048112752211824002614062564731378390107393841924024992900096272203427395268363591954016973222085497810130812644667188518603229549084506011656716594567797567298132136216194941840585237878858460280261239887687428829375605555945753827119720586750631367716075599073634731471504260724387869378014436808380008096784296619353982377042796709113277023048503614322336338787624495889957753806917512300465195258871128700879115968204258194381296288237529334846252325708114045756734861206974694332984226429182357067126837458065169631111462435830423526194589462766826719275660644126448562809748092006285700764039691021497055662341656594078963665734973515004383619424206199423404426260428435029625839720828715873547773951561589009316755538926217057660908236519924235235619770675436108507917722314466270142484807060731907806830319044273720218636481802179286069057173343243951397675980777851315120680118430072968591076578558637383169959517835261015038328382345688129364776302241168625264064812069025112090263137082552535421329754943044198941936240688824218041364039700546228900283717808668314344125472252807531518791099498692946306328235067764410531248477081885126875518308672990452448890110228731016986585572535897645362817103841400441546963512425504489024589005011590124360348938492006792308704507061642951011458749395538490335711130206859554892150422217109609854841320808883156074499689978384411831818569414262079698400452210643442851321588188354275888886257603641542109776241390729041700493644160923820461710058687648706149758610663198374013955557327262668118696927211331534855305270845371631301081119472690423140645543286568447703696095356440639011578632338858560471650438477891281241072990894958114372212031895484984653567691286852652650107819350239352406247153415410489981573464865003560861111332722204086414609128602020530497009851004558213098998166539307648066090774246910719321947561845561811551635349521128959781556450619336828717871420898920647009920722717177061958022742777205857695854934254785056637106031433088913246626097291550078584270096661510394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Sir ji aap great ho. Is there any way we can contribute even a little towards your hard work ?? Sir please let us know. And again, thank you so much for such a wonderful explanation 💙
I was also approching the same way, but stuck in one TC i.e "1234567890" where k=9 op = "0" this will give output 1 according to this approch. Correct me if i am wrong
@@techdose4u its okay Usually, I complete the task in the noon and compare our solutions in the night But this i couldn't do it, so I was waiting for your solution 😂😘😘
You are doing too much hard work dude.Wish you best of luck for your future endeavours.
Thanks :)
Did you just fire the guy lol
Great job though Tech Dose, well explained!
Hey there, I am from year 2022, I came back in time to give you good news that ,Surya sir(tech dose founder) has joined Google.🥳
@@nirajgusain1452 wow ! 🎉
Can't be explained better than this... Million times thanks for making these questions so simple :)
Wonderful way of explaining the algorithm behind the program. Most channels only explain(show) the code. Please keep up this great work. Please do mostly asked Leetcode medium questions. Will be a great help. Thanks a lot.
Welcome :)
Sorry to be off topic but does any of you know of a way to log back into an Instagram account..?
I stupidly forgot the account password. I would appreciate any tips you can give me!
@Dallas Gannon instablaster :)
@Johnny Zyaire I really appreciate your reply. I found the site thru google and im in the hacking process now.
Takes quite some time so I will reply here later when my account password hopefully is recovered.
@Johnny Zyaire it did the trick and I now got access to my account again. Im so happy!
Thanks so much you saved my account :D
One of the best intuitions I have ever seen. I love the way you take examples to derive the logic. Keep it up. And also thanks for delivering such a quality content on coding problems. You just earned a subscriber.
Thanks
The explanation was really awesome bro. Gave a whole new perspective to look at numbers. Really appreciate the effort. Sincerely, thank you.
Welcome :)
Explain
Perfect, clear explanation to a problem that I've been stuck on. Thanks so much!
Welcome :)
Very nice explanation better than other explanations out there which just throw terms like greedy etc. Giving concrete examples and explaining the algo through them makes it much easier to understand.
Yes....term confuses students 😅
Your explanation is really good and with O(N) which is better.
I solved this problem with different approach and I think the time complexity id O(N * k). Is that correct?
Step 1: Parse the string left to right.
Step 2: For each current char at i, See if there is a char lesser than the current char in window [i+1 to i+k]. If found, decrement the k and skip the char. Otherwise add to the result.
Here is the code:
public String removeKdigits(String num, int k) {
if(num == null || num.length() == 0 || num.length() 0){
builder.delete(builder.length()-rem, builder.length());
}
String res = builder.toString();
return res.length() == 0? "0" : res;
}
int getMinIndex(char val, int start, int end, String num){
if(end >= num.length()){
return -1;
}
for(int i = start; i
your time complexity is N^2 and space complexity is also N
This is absolutely mind blowing. Couldn't think of this approach.. 😭😭😭 Also so many edges cases to handle... But great great video.
God level explanation bro. Been watching your videos since a month and they are all brilliant. Will definitely donate once I land a job :)
Thanks for your appreciation.
It was such a great explanation. Half way through the video and I knew what to do. Keep up the good work.
this channel should have millions of subs.
Thanks bro :)
This channel work so hard i wish this channel should have millions of subscribers
Java solution using Stack:github.com/RajeshAatrayan/InterviewPreparation/blob/master/src/Stacks/RemoveKDigits.java
Thanks for sharing
@@techdose4u As always 🔥
But you didn't answered my question
Awesome explanation! Thanks! As per the explanation, just remember this: Remove previously seen peak elem to flatten the curve. When graph dips, we have already seen a peak.
nice .. this is a good explanation. the good part is you are telling more examples that make more sense for everyone.
Thanks :)
I wish I could like this video more than once. Thanks for your explanation. You are good!
Thanks 😊
Awesome explanation. You’re doing a great job in helping the community.
Thank you, I was tyring to learn it since a long time but your tip example explained it all.
nice 👌🏽
Your explanations are best I have found on RUclips.
Thanks :)
An inspiration for anyone who wishes to learn or teach DSA
👍🏼
Pretty good explaination bro !
I was able to solve it in my mind but couldn't put it in code .
DIdn't think about first-come maxima approach .
Cheers (y)
Nice
Initially I thought this is a DP problem because I thought we have to consider all possible cases and the do the DP thing... but this approach is really wonderful.
It could have been dp problem if you thought about recursively solving this.
Glad i got the stack approach by myself even though my code was way messier than yours ^^
Many were not able to solve this. Glad that you did :)
Can you please explain why in line 15 he consiered only when stack is not empty?
Dude, congrats on 100K. You deserve it
Welcome 😄
I'm so happy that I found your video. I understood it very well. Thank you very much sir.
Welcome :)
wonderfull way of explaining..
Thanks :)
Sir , the way you explain is absolutely amazing !! . Wish i had a teacher like you . Greatwork Sir !! :) :)
Thanks :)
great explanation, easy to follow and understand. thanks
Welcome :)
kya baap tarike se sikhaya bhai. Mja aagya. Thanks man
❤️
Very good explanation and visualization
Thanks
Great explanation brother. Understood each and every step. Keep doing great work.
Thanks :)
great explaination
Became fan of your explanation! Much much appreciated!
Thanks :)
Great explanation sir, Thanku so much
Welocome :)
starting the day with your video
;D
Great Explanation
Thanks 😊
great job, thanks for good quality of explanation
Welcome 😊
extremly datailed explanation..thank you so much!!
Welcome :)
Wat an explanation, absolute gem
One thing to notice here if the elements keeps on increasing as he said it will form the smallest number possible that means we will have the higer value digits at the end so we can remove them easily with the help of the stack
Yes correct :)
Best concise explanation
Amazing intuition explained. Thank you
wonderful video. Great explanation of code and intuition. Keep up the good work
Nice content.
Keep on making such videos
This is an amazing explanation! Nice question, thanks.
Welcome :)
Man, wonderful explanation.. Highly appreciated.. 💞💞
Thanks :)
0:00 - 8:59
Bhai thanks alot for sharing
Algo
Code and
Explanation
Welcome :)
I got a goodone idea to solve this say we have the number 71382 and k=2 so what we can do is that to divide the digit by its index number, index number must be starting from 0 ie for 7 index number should be 1 not 0 and so on,, then after we got the list of the divided result we can just chose the minimums possible and then we can include only those in return value, me testing this got all correct result for almost all cases
man what a brilliant explanation
Great approach
Thanks :)
I've been into problem-solving for over a year, and this is the best channel I have seen.
the channel is better than Love Babbar, TUF, Anuj Bhayya, and all the members out there
Very Well Explained sir 🔥👍🏻
Thanks :)
Lucid Explanation!!
Thanks 😊
good explaination
Simple and perfect explanation!
Perfect implementation !!
Thanks :)
This is a phenomenal explanation!!
Thanks. I hate your 4's but the overall the video was quite helpful.
I solved 402 with a different approach, but wanted to see how the stack based approach worked. No I know.
😅
Beautifully explained.
good explanation
Thank you very much!
Welcome :)
Amazing 👌 explaination
Thanks :)
Best explanations ever
Thanks :)
Simply Awesome! Keep Growing!!
Thanks :)
Nicely explained..
Thanks:)
Nice Explanation
Very well explained
Thanks :)
Wow such a beautiful problem!
Thanks for clear explanation !!
Welcome :)
In 10:58 , what if the next element of zero is 2 Instead of 1 , then our results first number should be 1 . But we would have already removed it . I think it will fail for that scenario
Great Video
Sir the example at 12:00 mins of number 14301620 Nd k=4...the answer if the think on pen n paper should be 1010....but why it is 120 ?.... technically thinking the answer should be 1010 after removing 4 digits ...
Removing leading zeroes are not counted as removed digits. This is the special case to be handled 😅
1st valley(4-3) 2nd (3-0) 3rd(1-0) 4th(6-2) so, we get -0120
Even I got the same doubt...what's wrong if 0 also added in stack...then output will be 0120.. please clarify my doubt
@@saianushachodapaneedi7504 If you add 0 in stack, while displaying result it will have leading 0, in the example shown by LC they removed leading zeroes so we can ignore leading 0s as mentioned here or parse the string integer and back to string so that leading 0s are gone. In order not to put extra effort, ignoring leading zeroes will suffice the requirement.
Amazing👌
Thanks :)
How does one come up with such intuitive solutions? I tried to come up with multiple solutions but was not even close to this solution.
I tried to solve this question using priorityQueue, HashMaps, expanding from center of the string.
Amazing sir
Thanks
leetcode gave me this edge case for the following problem : "9964143637881536115347130215819342018286368478941148499497648482711459533461004254405644237053081606083361703097681313409879305615510320200854934444651935440830730707105506511273844202022847156939479617415032308016122590196433883734152425324321850950025461922368309179936567772058238956815658522566619712309337787110000248140248621983725541138216249932119341652497227527347196915584874245747655643373728114771002178121013476532176128561227651191732455258556988215663509467036265356759614472865379500702323093381756610448863769645016608790510082369942579876359844432606935705284237991853585529691576005445931743352187877817181108107659316666309094802979311362685246271238811648377471342618391148123088439359424933182816550379826963424443077369303388270800024963285014879985932202469314657763554311465766241899886051752598919297325070163176559846505309761680481734434389501672456194786083611750491579701118513267425527823659774604213876847372305982594830156571943761073290766254549904295349986681374115730100337100520099231426507753102943794893125561715341714882235592831859851753324171964100271220487416100160426921656692822076747471313551671799749136336020476415426498900467136354109743348482211848364210754765858145061682176996476703252113885157082272913476246001426543322720172472400433849455239728009056816478610972157143620619838281484903385698733878747333577266693321881082248284899461049170566515551638479945941859455913682794110638768950164185110174329858257546630386490667378849662828892086742219395018081013139661291385111259380764915297206827929993411346366971457561364592936565292180883672568239002607555932099570488014976458337969750530347455002905982811683646920337042844933044228156313556893574266924334421860399441795570348505986213235968877629037821039295531078587452820520378855971549385240599138029027426814355797039844185115797752068944043026514402978978851104279587917456738135851069474951293893468797930509914957546422062980494255056416478680885689780986382412165954803439553973540706927945767861390922237184889229475493329909116465687108626908432452951254474743412354718972999375833762203809869944881570164493465129271906768322772743880895596954354231919788356736973386736425035313669786510718228292965591836221183232782757135478753561150173194385615500385373233981959493952471916956111069857167656232936080328221546753405850472812773151559894114363782701095557909245140582135212670655043831517604969231621049089970261307870253571673590180617152285302103559731670339047857148567799820792277393882937146083861121444641752891357528477673783704643969540852343441491634297968882019783645863769499154099829169034519420545243923982738295303981036771224459015594039438755491178665247811195429718554410638417459245168087508373787473581006876786621492463488551382880888016193098727660257087286075211981304241455039635843389359277754175667320688287674673170776696626809610432006193791350589302883359213754039606437515551397976472818092708306048112752211824002614062564731378390107393841924024992900096272203427395268363591954016973222085497810130812644667188518603229549084506011656716594567797567298132136216194941840585237878858460280261239887687428829375605555945753827119720586750631367716075599073634731471504260724387869378014436808380008096784296619353982377042796709113277023048503614322336338787624495889957753806917512300465195258871128700879115968204258194381296288237529334846252325708114045756734861206974694332984226429182357067126837458065169631111462435830423526194589462766826719275660644126448562809748092006285700764039691021497055662341656594078963665734973515004383619424206199423404426260428435029625839720828715873547773951561589009316755538926217057660908236519924235235619770675436108507917722314466270142484807060731907806830319044273720218636481802179286069057173343243951397675980777851315120680118430072968591076578558637383169959517835261015038328382345688129364776302241168625264064812069025112090263137082552535421329754943044198941936240688824218041364039700546228900283717808668314344125472252807531518791099498692946306328235067764410531248477081885126875518308672990452448890110228731016986585572535897645362817103841400441546963512425504489024589005011590124360348938492006792308704507061642951011458749395538490335711130206859554892150422217109609854841320808883156074499689978384411831818569414262079698400452210643442851321588188354275888886257603641542109776241390729041700493644160923820461710058687648706149758610663198374013955557327262668118696927211331534855305270845371631301081119472690423140645543286568447703696095356440639011578632338858560471650438477891281241072990894958114372212031895484984653567691286852652650107819350239352406247153415410489981573464865003560861111332722204086414609128602020530497009851004558213098998166539307648066090774246910719321947561845561811551635349521128959781556450619336828717871420898920647009920722717177061958022742777205857695854934254785056637106031433088913246626097291550078584270096661510394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Nice Explanation! Thank you 😄
nice work sir, please keep it up. Kudos
Thanks :)
pure genius!
That was amazing,
Thanks :)
hey, can you please tell me about how you are wrinting are you using digital pen and which software you are using for writing
thanks so much, great explanation!
Thanks
please make a video on heavy light decomposition and centroid decomposition
Sure....will add it to my to-do list
This helps alot ThankYou sir
Welcome :)
amazing explanations man...
Thanks :)
nice explanation ❤❤
Thanks :)
Sir ji aap great ho. Is there any way we can contribute even a little towards your hard work ?? Sir please let us know. And again, thank you so much for such a wonderful explanation 💙
You love and motivation is everything I need :)
@@techdose4u We love you 3000 Sir 🥺 keep up the good work and we'll always make you proud.
Thanks :)
Amazing stuff! Thank you.
Welcome :)
good explanation man
Thanks man :)
I used deque instead of stack so as to maintain the order in which we fetch the result.
Nice
plzz make a video on create maximum number of leetcode.It is one of the hardest problem of leetcode and no solution is present
Will do it later. Not in MAY atleast. Hope leetcode doesn't extend the challenge to June 😅
I was also approching the same way, but stuck in one TC i.e "1234567890" where k=9
op = "0" this will give output 1 according to this approch. Correct me if i am wrong
Good
👌
👍
Hi sir amazing explanation the method looked very nice, but i dont think it will work for numbers with trailing zeros in the end
Thank You Sir!
Welcome :)
Man you are doing very good work keep it up
Also I did the same in C but C is very messy TBH should I go for c++ or python?
C++ has lower runtime in general. So go with C++
at 7:08 is 2 also a peak element
also tc is o(n^2) I hope, How can one think of these many corner cases in a interview
I was waiting for your explanation,
:) This video was little longer and I was very tired after office 😅
@@techdose4u its okay
Usually, I complete the task in the noon and compare our solutions in the night
But this i couldn't do it, so I was waiting for your solution
😂😘😘
As in, so i was
😅
me tooo....