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.

[@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

[@SaGaR-is1jg]

teri wajah se usse english mai padhana padhra hai anna ke chacha

[@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 ?

[@Educator1723]

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

[@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!

[@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

[@user-dj4sm9gd8k]

which college ?

[@cs_harshitha__088]

got placed?

[@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

[@akris_adi]

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

[@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💀

[@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 :)

[@AlwaysAStranger]

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

[@nizarahamed.m1004]

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

[@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

[@user-zn4bg9bp8a]

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.

[@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..

[@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

[@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

[@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

[@rajkumarvb5197]

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

[@shivanshsingh176]

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

[@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

[@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

[@komal6816]

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

[@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

[@niranjanniru4115]

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

[@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 .

[@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

[@shwetachoudhary9003]

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

[@user-ff2uu2wz4w]

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 ❤

[@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)

[@paulbarsha1740]

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

[@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 !

[@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

[@stith_pragya]

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

[@Santhoshmani1]

Thank you striver. Basic maths completed. Coded along with you. Made the notes for the lecture .

[@kinjalshah6308]

Can you send me notes pdf??

[@NHSanjay]

Why in the "Print all divisors" problem, we are solving it in a way where the time complexity is O(sqrt(N))+O(n log(N))+O(N) and why not O(N)?

[@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

[@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 :(

[@oye_pritpal]

understood, this guy is an absolute champion!

[@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...

[@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 ..

[@rohanprabhakar1991]

In your DSA sheet, the Reverse Number Question is different. It is changing the decimal into binary and then writing binary for reversed binary again

[@abudeepa0474]

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

[@PrithaMajumder]

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

[@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

[@amitkumarghosh]

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

[@adrianfernandes2004]

The best I repeat the best explanation ever💫

[@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 🥹

[@thodupunurisathwik5589]

Hey Striver I wanna highlight one point about Armstrong number logic you have explained might not correct... You said any number if we make cube of its digits and if it's equal to original number then it's an Armstrong number but this is not always the case..For example you said 1634 is an Armstrong number (which is true) if you calculate cubes of each digit like 1**3+6**3+3**3+4**3 is not equal to 1634 Armstrong number is a number that equal to the sum of its own digits each raised to the power of the number of digits in original number So for 1634 Total number digits is 4 so we should calculate each digit raise to 4 so 1**4+6**4+3**4+4**4==1634 I see the gfg link you given has a test cases where 100

[@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

[@rakhibhagwat670]

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

[@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 ❤

[@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]

@@amogu_07 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.

[@amogu_07]

@@amansinghal4663 got it tnx!!!

[@DivineVision201]

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

[@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.

[@AnkushGhosh-x5t]

there is a small edge case left in case of reverse number we should check the condition for the overflow if (reverse > Integer.MAX_VALUE / 10 || (reverse == Integer.MAX_VALUE / 10 && lastDigit > 7)) { return 0; } if (reverse < Integer.MIN_VALUE / 10 || (reverse == Integer.MIN_VALUE / 10 && lastDigit < -8)) { return 0;

[@karunasagarks5228]

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

[@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.

[@sumitstark7119]

I think this way is simple for reversing number without the occurrence of '0'; void count(int n){ while(n>0){ int last_digit; last_digit= n%10; if(last_digit!=0){ cout

[@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

[@virajdeshpande5043]

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

[@somnathrooj]

Prime no : natural number starting from 2 , having 1 and itself as factors is a prime number.. Note : Natural Number is mandatory, otherwise it will go for -ve integers as well

[@jayanth1844]

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

[@kashafkhan9726]

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

[@nurulain5534]

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

[@udit5490]

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

[@AmartyaPardeshi]

Great teaching, understood all the concepts effortlessly

[@ShwetaSingh-iw6dk]

Unstoppable striver for a reason 🙌❤

[@adityakumardixit8779]

Bhaiya maza aa gya Fully understood 😃😃

[@user-mc8sc1gf8q]

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

[@cinime]

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

[@ayushtandon1719]

question on reverse number of coding studio is wrong as it is question of reverse bits which requires bit manipulation. no gfg links in this a2z sheet :(

[@AbhayBagoriyahacker]

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

[@AbhijeetKumar-sb1sr]

armstrong code bool checkArmstrong(int n){ int sum =0; int dum=n; int count=0; int temp=n; while(temp>0){ count=count+1; temp=temp/10; } while(n>0){ int ld=n%10; n=n/10; sum=sum+ pow(ld,count); } return dum==sum; }

[@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.

[@vYadav16380]

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

[@tanisha2504]

started working on this a-z dsa course.. Have a job but want to start over these topics .. To switch to a better one!!!

[@madhavamelkote4554]

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

[@012srishtisaini7]

The palindrome problem can also run without including using namespace std; as --- bool palindrome(int n) { int reversenum = 0; int dup = n; while (n > 0) { int ld = n % 10; reversenum = (reversenum * 10) + ld; n = n / 10; } return dup == reversenum; if (dup == reversenum) cout

[@Shwetasingh-yz4fx]

Perfect and simplest explanation found ever

[@ItsAbhiDestiny]

Bhaiya aapne website pe reverse a number question m coding ninja ka Reverse bit question dal diya h

[@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.

[@VijayKumar-l2v4x]

leetcode is asking for premium for solving the armstrong problem

[@syedhamzashah3803]

Print all divisors in Python def foo(n): x = 1 while x

[@rahuljain9748]

gcd was too good!!!!!!!!!!!❤❤❤❤

[@saravanakumar7342]

Make videos frequently bro,it will help for placements

[@shashanks6906]

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

[@uday2159]

010 is palindrome or not How can we prove it's palindrome. Code gives its not palindrome gcd(0,5) code Will give 0 but actual answer is 5 if we are using brute force method, should we need to specify in code to check any number is 0. Please anybody help me....!

[@msdarshankirthic9990]

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

[@technicaldoubts5227]

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

[@YoyoChintu-wt8gw]

best one till now!!! 🙏🙏🙏

[@anishakumari4077]

Sir , your A2Z DSA course website is not visible. FILE NOT FOUND. It was really good and i was dependent on the same for my dsa preparation. Kindly do something about it.

[@supriya_codes]

started dsa with your this course,thank you!

[@arnd12940]

much more needed these topics

[@dr.stephen.strange]

understood ❤

[@abudeepa0474]

I learnt many things in this video

[@sahilsingh3542]

For the countDigits algorithm I can't understand how the function runs in log base 10 of N. For every iteration it decreases a digit, from the whole number. So for a number, let's say 576, it will run three times, after running 1st, you will have 57, then 7 and then it will exit. So it is running in O(n) where n is the number of digits in a number. For n = 53627834653278 (14 digits) the loop will run 14 times and you will have answer as 14. Let me know if I am missing something.

[@anantagrahari2547]

Yes bro, I am also thinking the same thing

[@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

[@yamini436]

understood the assignment :) thanks striver

[@watch2-grow]

UNDERSTOOD THANK YOU STRIVER

[@shivanitayde9177]

your teaching skills are awesome bro