Basic Maths for DSA | Euclidean Algorithm | Strivers A2Z DSA Course

[@takeUforward]

Let's march ahead, and create an unmatchable DSA course! ❤ There is a slight mistake on the logic of Armstrong, it will be 1634 = (1^4 + 6^4 + 3^4 + 4^4), basically digits raised to the power count of digits, so sum = sum + pow(ld, cntDigits) will be the change, where the cntDigits is the number of digits. Do consider to give us a like, and a subscribe, means a world to us..

[@himanshusharma3382]

Bhaiya i have good foundation in java can I do these sheet in java lang

[@brajeshmohanty2558]

@@himanshusharma3382 Han bhai mil ke try karte hai main bhi java Bala hi hun par bhaiya ne bola concept toh same hai par kahi kahi toda muskil hota hai cause c++ ke kuch function java mai nahi hai par toda manage karna padega

[@arunn121]

@@brajeshmohanty2558 yes bro... functions and Collection frameworks implementation syntax vgera b thora different rehta hai. .. ek group discord telegram ya kuch b bnalo Java k liye best rhega..discussion help vgera and sath m A2Z sheet solve krenge toh consistency b bni rhegi..

[@brajeshmohanty2558]

@@arunn121 dekh bro mid Jan se toh mera end sem hai tu bana le merko add kar dena

[@takeUforward]

@@techyouknow8026 lol frequency, bro the entire sheet is at your hand, you can do it by self also, simple google search will give you answers if you stuck, but the thing is, you just want to be a keyboard warrior. The though topics are fully covered, so stop crying, I will upload at my speed, because I have a full time job, its not easy to work 9 hours and then create content.

[@NGANTALALITHA]

I'm from Andhra Pradesh, when I decided to learn DSA I searched a lot of streams, but I'm still stuck somewhere. After Watching your videos, it's just awesome. It very much helpful to beginners like me.... Thank you so much sir. After this I'm become a fan of striver......

[@uday2159]

Which college are you

[@asish4554]

Are you from ap ?

[@Ramesh-ramesh-03]

Yee college meru

[@Md_sadiq_Md]

Iam also from Andhra Pradesh From Vijayawada, VR Siddhartha Engineering College ❤

[@hydershaikh18]

iam also from Vr sid clg bro,Aiml you ?​@@Md_sadiq_Md

[@dinimeegada9167]

We are so fortunate to live in an era where striver lives........ ❤ from Andhra Anna ......

[@unanimousu]

striver bro andhra na bro

[@Md_sadiq_Md]

Iam also from Andhra Pradesh, Vijayawada

[@rgvcultman]

@@Md_sadiq_Md avunu bro maa kulapodey

[@Md_sadiq_Md]

@@rgvcultman Ok Enjoy

[@rgvcultman]

@@Md_sadiq_Md vurikae anna bro nak telidhu

[@akris_adi]

Within first 5 minutes I realised the value these videos will be adding in the coding journey of the geeks !!

[@Tbm4545]

Whats ur status bro r u placed or still in college where u have reached ins triver dsa sheet

[@akris_adi]

@Tbm4545 I've placed through off campus in an mnc

[@akris_adi]

I've been*

[@gopalakrishnan618]

@@akris_adi how helpful would you say these dsa sheets were overall

[@akris_adi]

@@gopalakrishnan618 I would say 4.5/5

[@Vardhan.coding]

I started preparing DSA and Take you forward is a life saviour! Lots of love Striver.

[@MuskanSingh-vw2rt]

kitta prepare ho gya bhai 8 months mei

[@bharathslip9]

I think this course can be directly included in any bachelor/maters courses by collages. Nobody can teach better than this. Please keep up this good work going. KUDOS

[@DR-mq1le]

For all those that couldnt get the 1634 test case right in the armstrong questoin, the power of each digit is equal to the no. of digits in the number for example : if it is a 4 digit number (1634) , then (1^4)+(6^4)+(3^4)+(4^4)=1634 if it is 3 digit then instead of 4 power will be 3 heres the code: #include bool checkArmstrong( int n) { string digits=to_string(n); int digitcount=digits.size(); int num=n; int r=0; int pal=0; while(n>0) { r=n%10; pal=pow(r,digitcount)+pal; n=n/10; } if(pal==num) { return true; } else return false; } might not be optimal but it is how i tried and it works

[@Iammuslim947]

yeah u r right even i got this mistake in most of the google search results but chatgpt and gfg gave me the right answer

[@MemoryFlick]

is this correct as well ?? what about time complexity??? { int i=0; int N = 371; int duplicateOfN1 = N; int d2 = N; while(N>0) { N=N/10; i++; } int sum = 0; int digitcount=i; while(duplicateOfN1>0) { int lastdigit = duplicateOfN1%10; duplicateOfN1 = duplicateOfN1/10; sum = sum+((int)Math.pow(lastdigit, digitcount)); } if(sum==d2) System.out.println("it is an armstrong number"); else System.out.println("it is not an armstrong number"); }

[@manishagrawal7104]

Actually power =number of digits in the number so try with power 4 it will work .

[@AshikPoojary]

Thanks it helped

[@shikharpandey9265]

thanks bro!

[@anshujaiswal5622]

Thanks a lot Striver for this amazing content. Honestly, the simplicity and the way of enplaning algorithm by breaking them down does help a lot of Beginner Folks to grasp DSA concepts. Understood Striver :)

[@sarveshneekhra5611]

I'm from a tear-3 college, current i am 6th sem. Student and i followed your videos and your A2Z DSA playlist is amazing bcz everything is well structured and easy to understand, your way of explaining is truly amazing, Thanku striver ❤, huge respect brother and keep make it more DSA video and guide us.

[@a_maxed_out_handle_of_30_chars]

tier*

[@bunnytheweebster]

He meant tear only 🥲​@@a_maxed_out_handle_of_30_chars

[@ojasvivashisht-f9m]

which college ?

[@cs_harshitha__088]

got placed?

[@sukuxsharma]

Kaha se structured iski to playlist me hi ulti seedi vdeo lgri h muje

[@AlwaysAStranger]

Have seen many DSA courses, none of them taught me math and DSA techniques in so much detail. Very Helpful Thank you!

[@surajbaranwal56.]

I already have been read advanced dsa, because of striver's teaching pattern I watched this complete lecture ,even I knew all these topics instead learn something. That's how this bootcamp is game changer for all beginner ninja's who want to live conding

[@komal6816]

You seem to be under the weather. And yet you are teaching, making videos with such effort making it look effortless. Hats off!

[@nizarahamed.m1004]

You are just amazing bro,Never seen such intellectual person with a clear explanation.

[@vinaykumarmaryala9041]

Hi @Raj, Thanks a lot for all your efforts, I am one of the people who are benefitting from your Sde sheet and your youtube videos, also I feel you deserve the name striver. you are the real savior for many people like me. May god bless you... and you are a pious soul for sure. Thanks striver..

[@niranjanniru4115]

im from chennai and love to watch ur lectures and learning through ur dsa sheet bro ..

[@rajkumarvb5197]

Thanks a lot for your effort Striver! Best explanation for GCD algo so far I've seen!

[@preethamreddy9610]

Understood. Today is November 19th, 2024 and i have just completed watching this video. I will come here periodically to update my status and my offers! Wish me luck!

[@amanshah1995]

HEY Striver ! You are just amazing .The simplicity your course has it's too easy for the beginner to grasp the knowledge Thanks man !

[@Akash-yr2if]

Congrats Striver for 300K Sub. Let's hope striver see's this comment.

[@Lucifer0872]

Aniket er bari giye

[@Lucifer0872]

Aniket er barite code chapo

[@sheikmuizzahmed2308]

Thanks alot! Please complete this DSA series for beginners asap. It helps alot for beginners in dsa like us 🙏. The only complain I have from your other DSA playlist is it's not begginer friendly and needs abit dsa knowledge

[@Z71990]

This year I got Civil branch in my state NIT. Can I understand this course as a begineer and also as a non-cs student? Also is this playlist is enough for placement or from any other playlist I have to study ?

[@hemanth9928]

Yes

[@hemanth9928]

Try to complete the strivers a2z DSA course sheet

[@khushiupadhyay686]

hey, can you please guide me. should i watch this playlist or should i follow the series from their website ,are they same or what? please reply

[@rohandhalpe9084]

Sir upload videos as much as possible, as our placement season starts from June .Hope I will learn maximum till June 💝💝💝💝💝💝💝💝💝💝😭😭😭😭

[@Tihorcreation]

Bro did you get placed?

[@rohandhalpe9084]

ya bro in toshiba software @@Tihorcreation

[@TarunPurohit02]

​@@Tihorcreation💀

[@rohanprabhakar1991]

Sir, you are helping me in doing hard work and making me experience the relief after doing the hard work we get Thanks for all your support Only have one request. I know you have been trying to make videos as fast as you can, but actually the placement is in the next 6-7 months, so if you can try to upload the beginner part as soon as possible, it would be beneficial to move for the other playlist of you which you have been uploaded in the past about different topics. Thank you for all the resources bhaiya(sir).

[@uday2159]

Did you complete all topics?

[@shayamkumar5929]

Bhai please tell how to you learn DSA because l can not able to understand nothing

[@RozinaKakar-p6m]

Your video playlist is so super easy when I decide to learn the Dsa I searched a lot of sites or KZbin but your way and teaching is amazing and the Practicing on Note is so brilliant ❤

[@paulbarsha1740]

The amount of effort you put into each of your video🙌

[@Dhanasri_Parisam]

amazing bro,Never seen such intellectual person with a clear explanation.

[@Joy-x9m]

1.3 - C++ STL and Java Collection should be a single topic like C++ STL / Java Collection. Not separate topic. So one can move forward with everything checked.

[@sonalshinde4960]

Hey Striver, its been 1 month since i started following u...i must say u r the best....i don't think anyone would have explained the dp problems or any problem for that matter the way u have explained them!!!Thanks a million ....words are not enough to praise u...the kind of passion u have to teach students for free is just awesome...may god bless u ..

[@karanbhoi5895]

23:55 One correction !! Armstrong number is the number in any given number base, which forms the total of the same number, when each of its digits is raised to the power of the number of digits in the number. 1634 is not equal to 1^3 + 6^3 + 3^3 + 4^3 it is equal to 1^4 + 6^4 + 3^4 + 4^4

[@chicha8482]

Can u provide the code for it am doing cmath and using pow function to use the number of powers equal to total no. Of digits but in 1634 its giving false

[@tt-sb8ez]

Exactly

[@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...

[@riteshsharma4229]

bhaiya in question Print All divisors insted of vector we can also use set but time complexity remins same for both ds (nlog n for sort) and (nlogn for set) btw Thank you for all The Lectures and sheet

[@takeUforward]

Yes

[@riteshsharma4229]

@@takeUforward thank you for reply bhaiya your reply really motivate me to be Consistence

[@andrea-kg9yx]

Thank you for such amazing videos. It helps me a lot to grasp new concepts and revise old ones.

[@DivineVision201]

Hi @Raj sir. Thanks for the video also Congratulations for your 300k subscribers. Looking forward for your more wonderful videos. Truly enjoying. ❤

[@amitkumarghosh]

I can't believe he is doing this for free, amazing sir. Love from Bangladesh.

[@shivanshsingh176]

Your amazing playlists specially graph and dp helped me to improve a lot. Can you please create a playlist for CP as well.

[@shwetachoudhary9003]

no one teaches maths in theire dsa course...this is indeed an AtoZ dsa course.. blessed to have it❤

[@dhanraaj]

24:14 minor error bhaya armstrong number = sum of everydigit^(total no. of digits in number) for 1634(1^4 + 6^4 + 3^4 + 4^4 )

[@takeUforward]

Ahh yes my bad :(

[@VIKASKUMAR-pc4jz]

#include #include void main() { int n,c=0,ans=0,re; scanf("%d",&n); int copy=n; int a=n; while(n>0) { n=n/10; c++; } while(copy>0) { re=copy%10; ans=ans+pow(re,c); copy=copy/10; } if(ans==a) { printf("true"); } else { printf("false"); } }

[@anantagrahari2547]

hey bro,your code is correct but at the end it will show TLE @@VIKASKUMAR-pc4jz

[@VIKASKUMAR-pc4jz]

👍@@anantagrahari2547

[@oye_pritpal]

understood, this guy is an absolute champion!

[@saravanakumar7342]

Make videos frequently bro,it will help for placements

[@rushboba5883]

You are the best Striver! I didn't see anyone explaining this well!!❤❤😭😭

[@16yogeshyadav31]

hey i am also learning DSA from him would youu like to connect i still get stuck on some of topics..

[@pavan305]

Congratulations for 300k bro 💕🥳. And I've a doubt, is OOPS concept included in A2Z dsa series or not. A small doubt.

[@sachinnegi2924]

no bro

[@pavan305]

@@sachinnegi2924u knw the reason why it is excluded?

[@sachinnegi2924]

@@pavan305 cause oops ka khali basic pta hona chaiye bilkul low level baki dsa m khi kaam nhi ati development m use hoti h wo

[@arnd12940]

freecodecamp ka video achha hai

[@stith_pragya]

UNDERSTOOD.......Thank You So Much for this wonderful video........🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻

[@farhansadidzihan]

Finally come!

[@Rakibulislam-bp3kv]

Your video is very informative. I can learn new things from every video. I knew all of this topic. but your video taught me new things, and how to think clearly. Thank you, sir.

[@amansinghal4663]

Guyss... For finding LCM, you can simply use the below formula after getting the GCD: LCM(a,b) * GCD(a,b) = a*b

[@amansinghal4663]

@@techkunbyamogh It is returning 1 only. Dry run and check. Gcd (13,11) = gcd(13%11, 11) = gcd(2,11) Now, gcd(11,2) = gcd (11%2, 2) = gcd(1,2) Now, gcd(2,1) = gcd (2%1, 1) = gcd (0,1) Hence one of the element has become 0, so the other element is the answer. Similarly you can dry run the gcd(a,b) = gcd (a-b, b) formula also , you will get the same answer. I think you are making mistake while doing "a-b", instead of minus sign you may have put the comma.

[@techkunbyamogh]

@@amansinghal4663 got it tnx!!!

[@Manishgupta200]

Striver your video is superb.. and in each video you taught in a very energetic way which makes your video very interesting. All the part in the video is amazing but the last.. i.e. Euclidean’s theorem and reducing loop by half by sqrt.. method is wonderfull.. Now, I try to build-up this type of logic in every Q. to reduce time & space complexity. Thankyou Striver ❤

[@HarshitMalang]

Anybody in Aug-2024 or after

[@Dhruvvvvvv56]

Me 😂

[@susonly2004]

me:)

[@vkgamer9305]

Me

[@MamathaN-yr5hl]

Me in September

[@Codegirl_01]

Yeah bro 🎉

[@myyoutube5237]

20:10 for reverse number, you can use Number! =0 condition in while loop, this condition also reverse negative numbers

[@Hachiman-nf6zc]

in the site when i click prime concept it opens to a different question called "minimum number of jumps" and i find the solution to be difficult. Could you explain that?

[@awalkingnosebleed]

me too :(

[@netrahindocha8423]

Understood ☺ Best channel for DSA

[@HimanshuYadav-e9i]

there is a small mistake you take a number 1634 is equal to the 1^3+6^3+3^3+4^3 but armstrong number is a number which is equal to the sum of digits to the power of how many digits in a number ... 1634 = 1^4+6^4+3^4+4^4 thanks 🙏🙏😇

[@anirudhcodes]

I never though i will be able to solve these so easily, thanks a lot striver for make it very very simple and easy to understand.

[@mizzzile]

Euclidean algo is so easy this way...

[@anirudhcodes]

@@mizzzile yes 🥹

[@uttam.py2]

who is starting in 2025

[@SamiKhan-y6m]

Understood each and every concept thanks Striver

[@msdarshankirthic]

bro do some videos on maps, there are only very very few videos on youtube, please do one playlist

[@proneetkumarseth6880]

I am very much grateful for the work you are doing for programming community. ❤ Below is the "Print Divisors" code with Sorting in JAVA, which is not available in A2Z DSA course platform under optimal approach section. Thank You 😊 import java.util.*; public class PrintDivOptimizedSort { public static void main(String[] args) { Scanner scn = new Scanner(System.in); System.out.println("Enter the number for which you want the divisors: "); int n = scn.nextInt(); System.out.println(); printDivisorsOptimalSort(n); scn.close(); } public static void printDivisorsOptimalSort(int n) { List list=new ArrayList(); System.out.println("The divisors of "+ n + " are: "); for(int i = 1; i

[@sk36]

thnx

[@vaibhavkadam476]

Thank you striver for explaining all the basic math stuff mostly in other courses the instructor assume that we already know they simply explain the code.

[@hemanthpothala356]

hi bro, doing masters(in CS) in the US costing 40-50 lakhs is really worth it? Are we able to repay the loan?

[@Rational_Human1]

That's not the question , 40 - 50 lakhs u can repay with 7-8 months salary , if u get a job , loan is not a problem. Real question is h1b visa , with that lottery system , visa seems impossible to get.

[@Bobby-vs3zc]

took me more than 2 days to finish this lecture. But the quality was top notch. Understood.

[@karunasagarks5228]

Finally, here we go ! Thank you very much striver !!! 🤩🥳🔥❤️

[@kashafkhan9726]

Congrats striver bhai for 300K subscribers. Hope it cross million soon.

[@pravargupta6285]

Solution for print all divisors at 26:24 without sorting #include using namespace std; void printDivisors(int num, int n) { if (n * n > num) { return; } if (num % n == 0) { cout

[@avipatel7106]

Understood! Love your content. Thank you for explaining it in simple terms.

[@sujayshanbhag2055]

Hey Striver, I am a 5th semester student of a tier-3 college, first of all thankyou for all your videos, I have completed your graph series, almost completed your SDE sheet. I needed your help so I joined your channel. I see that your last members only post was a year ago, I understand that u have a full time job and therefore you are busy. I just wanted to know if you will continue making meeting sessions, because I need your guidance. Thanks again for your hard work.

[@PrithaMajumder]

Raj, Thanks a lot for This Amazing Video about C++ Basic Maths Lecture - 7 Completed ✅

[@CinematicClips-uz3mk]

25:25 this Armstrong number works only for 3 digits If you want to find less or more than 3 the approach should be 1. Find the number of digits using int i = (int) log10(n)+1 2. While loop n> 0 3. Extract digits using n%10 4. In the while loop use pow(digits,i) function to get power of that digit 5. Check with the original n outside the loop

[@shazam8095]

nice explanation much needed

[@janarddanmandal2984]

Time Complexity Section: 1. 14:40 (log10(n), log2(n), log5(n) when iterating condition value is divided by 10 or 2 or 5 respectively. And similarly for dividing by other numbers)

[@sanatwalia1432]

Here is the soltion I came p with a little bit of trial and error. This solution also checks the integer overflow and also finds reverse for the negative number class Solution { public: int reverse(int x) { long long dup=x; long long n=0; if(dup0) { long long temp=dup%10; dup=dup/10; n=n*10+temp; if(n>INT_MAX) { return 0; } } if(x

[@AmartyaPardeshi]

Great teaching, understood all the concepts effortlessly

[@paramjjethwa8734]

all divisor explained beautifully

[@nurulain5534]

so far, this is better than the paid courses 🔥🔥🔥.

[@MayankVerma-e9x]

Understood.. for armstrong number here's my java solution : public static boolean isArmstrongNumber(int n){ int x = n; int sum = 0, count = 0; for(int i = 1; i < n; i *= 10){ count++; } // while(x > 0){ int ld = x % 10; sum += Math.pow(ld, count); x = x / 10; } return sum == n; }

[@virajdeshpande5043]

Thank you for creating such valuable and informative content! It's greatly appreciated.

[@paramjjethwa8734]

dude amazing i am doing with python the count digit logic is really well explained

[@udit5490]

I really like the explanation, specially the pattern in which you have explained starting from count digits till gcd.

[@abudeepa0474]

Thankyou striver...you made all the things understand in the simplest way,......

[@hasnainmemon_01]

your teaching skills are top notch🔥

[@ShwetaSingh-iw6dk]

Unstoppable striver for a reason 🙌❤

[@supriya_codes]

started dsa with your this course,thank you!

[@adityakumardixit8779]

Bhaiya maza aa gya Fully understood 😃😃

[@payalthakur6065]

Great learning from the video ❤

[@webtestingui7187]

🙏 🙏 🙏 🙏 🙏 hats off to you, preparing this videos require a lot of time and from your busy schedule you are preparing for us Thank you, Dada.

[@AbhinavTiwari-u7b]

very well explained sir i have just started the series and I am getting each and every thing . Your teaching methodology is just insanely superb. Please asap bring a linked list series as well sir

[@Shwetasingh-yz4fx]

Perfect and simplest explanation found ever

[@cinime]

Understood! Super fantastic explanation as always, thank you very much!!

[@jayanth1844]

Thanks a lot Striver!! you make very stuff so easy to understand in simple terms..

[@rakhibhagwat670]

Great content for any professional or an engineering student❤💯. Thank you sir for taking this much of efforts 💯

[@YoyoChintu-wt8gw]

best one till now!!! 🙏🙏🙏

[@madhavamelkote4554]

God bless you brother, i am in better place all thanks to you!!!

[@technicaldoubts5227]

I was waiting from last 5 days. Finally come 😌😌

[@vYadav16380]

I wish i would have known abt this channel earlier rather than spending money on paid course thank u striver.

[@PriyadarshiniAddakula]

understood!!!!!!!!!!!!!!!!!! may god bless you striver bro.

[@letcodez]

24:15 bro did a mistake which was noticed already, An Armstrong number is a number that is equal to the sum of the nth powers of its digits, where n is the total number of digits in the number. For a four-digit number, the power used for each digit would indeed be 4, not 3. Refer to the link for better understanding, kzbin.infozrrLbXFEjtY?feature=share

[@ByteBuilder-b6u]

In the eucledian algo why we are taking min(a,b) in time complexity like the time complexity should be logphi(max(a,b)) Because we take max of a, b and divide that. Like in count digit we write log10(N) where N is the number divided by 10, so here max(a,b) is the dividend in eucledian algo

[@PrabalSingh2908]

understand striver thanks for making it much clearer

[@shashanks6906]

Amazing. I loved the GCD part, understood very well. Thank you so much

[@adrianfernandes2004]

The best I repeat the best explanation ever💫

[@asiftech4u859]

your gcd explanation is awesome

[@KoushikDas2005]

Understood , but there are a few things that can be changed according to the logic given in the question as per the solution discussed , like in armstrong number . Btw, got to know about time complexities , the approach for optimizing solutions and all other things .

[@AbhayBagoriyahacker]

What a brilliant way of teaching hats off to you and thank you so much for the content ;)

[@yamini436]

understood the assignment :) thanks striver