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 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
@@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..
@@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.
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......
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
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"); }
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
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.
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 :)
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).
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
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..
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
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 ?
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.
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
Your video playlist is so super easy when I decide to learn the Dsa I searched a lot of sites or TH-cam but your way and teaching is amazing and the Practicing on Note is so brilliant ❤
#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...
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 ..
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
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
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
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.
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.
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
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 .
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?
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)
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
There are some problem in question, you tech reverse number in question sheet , question is based on Reverse the Bits of the number..... Pls provide the proper solution for the question
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
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.
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
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
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
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;
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; }
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
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
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
bool checkArmstrong(int n){ int dup = n; int sum = 0; int cnt = 0; while (n > 0) { int lastDigit = n%10; n = n/10; cnt = cnt + 1; } n = dup; while (n > 0) { int lastDigit = n%10; sum = sum + pow(lastDigit, cnt); n = n/10; } return dup == sum; }
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..
Bhaiya i have good foundation in java can I do these sheet in java lang
@@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
@@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..
@@arunn121 dekh bro mid Jan se toh mera end sem hai tu bana le merko add kar dena
@@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.
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......
Which college are you
Are you from ap ?
Yee college meru
Iam also from Andhra Pradesh
From Vijayawada, VR Siddhartha Engineering College ❤
iam also from Vr sid clg bro,Aiml you ?@@Md_sadiq_Md
We are so fortunate to live in an era where striver lives........
❤ from Andhra Anna ......
striver bro andhra na bro
Iam also from Andhra Pradesh, Vijayawada
@@Md_sadiq_Md avunu bro maa kulapodey
@@rgvcultman Ok
Enjoy
teri wajah se usse english mai padhana padhra hai anna ke chacha
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
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
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");
}
Actually power =number of digits in the number so try with power 4 it will work .
Thanks it helped
thanks bro!
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
Anybody in Aug-2024 or after
Me 😂
me:)
Me
Within first 5 minutes I realised the value these videos will be adding in the coding journey of the geeks !!
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.
tier*
He meant tear only 🥲@@a_maxed_out_handle_of_30_chars
which college ?
got placed?
Kaha se structured iski to playlist me hi ulti seedi vdeo lgri h muje
Have seen many DSA courses, none of them taught me math and DSA techniques in so much detail. Very Helpful Thank you!
Sir upload videos as much as possible, as our placement season starts from June .Hope I will learn maximum till June 💝💝💝💝💝💝💝💝💝💝😭😭😭😭
Bro did you get placed?
ya bro in toshiba software @@Tihorcreation
@@Tihorcreation💀
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 :)
Understood each and every concept thanks Striver
You are just amazing bro,Never seen such intellectual person with a clear explanation.
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).
Did you complete all topics?
Bhai please tell how to you learn DSA because l can not able to understand nothing
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
Congrats Striver for 300K Sub. Let's hope striver see's this comment.
Aniket er bari giye
Aniket er barite code chapo
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..
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
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 ?
Yes
Try to complete the strivers a2z DSA course sheet
I started preparing DSA and Take you forward is a life saviour! Lots of love Striver.
kitta prepare ho gya bhai 8 months mei
Thanks a lot for your effort Striver! Best explanation for GCD algo so far I've seen!
You seem to be under the weather. And yet you are teaching, making videos with such effort making it look effortless. Hats off!
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.
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
Yes
@@takeUforward thank you for reply bhaiya your reply really motivate me to be Consistence
Your amazing playlists specially graph and dp helped me to improve a lot. Can you please create a playlist for CP as well.
The amount of effort you put into each of your video🙌
Your video playlist is so super easy when I decide to learn the Dsa I searched a lot of sites or TH-cam but your way and teaching is amazing and the Practicing on Note is so brilliant ❤
HEY Striver ! You are just amazing .The simplicity your course has it's too easy for the beginner to grasp the knowledge Thanks man !
no one teaches maths in theire dsa course...this is indeed an AtoZ dsa course.. blessed to have it❤
im from chennai and love to watch ur lectures and learning through ur dsa sheet bro ..
UNDERSTOOD.......Thank You So Much for this wonderful video........🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻
your teaching skills are top notch🔥
understood, this guy is an absolute champion!
#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...
Thank you striver. Basic maths completed. Coded along with you. Made the notes for the lecture .
Can you send me notes pdf??
Raj, Thanks a lot for This Amazing Video about C++ Basic Maths
Lecture - 7 Completed ✅
I can't believe he is doing this for free, amazing sir. Love from Bangladesh.
Thankyou striver...you made all the things understand in the simplest way,......
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
..
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
Great learning from the video ❤
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
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
Exactly
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.
Euclidean algo is so easy this way...
@@mizzzile yes 🥹
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.
Hi @Raj sir. Thanks for the video also Congratulations for your 300k subscribers. Looking forward for your more wonderful videos. Truly enjoying. ❤
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.
Congrats striver bhai for 300K subscribers.
Hope it cross million soon.
Great content for any professional or an engineering student❤💯. Thank you sir for taking this much of efforts 💯
understood!!!!!!!!!!!!!!!!!!
may god bless you striver bro.
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
Finally, here we go ! Thank you very much striver !!! 🤩🥳🔥❤️
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 .
The best I repeat the best explanation ever💫
Thank you for creating such valuable and informative content! It's greatly appreciated.
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
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?
me too :(
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)
so far, this is better than the paid courses 🔥🔥🔥.
I really like the explanation, specially the pattern in which you have explained starting from count digits till gcd.
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
thnx
Thanks a lot Striver!! you make very stuff so easy to understand in simple terms..
Perfect and simplest explanation found ever
There are some problem in question, you tech reverse number in question sheet , question is based on Reverse the Bits of the number.....
Pls provide the proper solution for the question
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
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.
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
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
Bhaiya maza aa gya
Fully understood 😃😃
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
What a brilliant way of teaching hats off to you and thank you so much for the content ;)
Understood! Super fantastic explanation as always, thank you very much!!
🙏 🙏 🙏 🙏 🙏 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.
Unstoppable striver for a reason 🙌❤
God bless you brother, i am in better place all thanks to you!!!
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;
started working on this a-z dsa course.. Have a job but want to start over these topics .. To switch to a better one!!!
Make videos frequently bro,it will help for placements
your teaching skills are awesome bro
leetcode is asking for premium for solving the armstrong problem
started dsa with your this course,thank you!
Amazing. I loved the GCD part, understood very well. Thank you so much
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;
}
I learnt many things in this video
Congratulations for 300k bro 💕🥳.
And I've a doubt, is OOPS concept included in A2Z dsa series or not. A small doubt.
no bro
@@sachinnegi2924u knw the reason why it is excluded?
@@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
freecodecamp ka video achha hai
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 )
Ahh yes my bad :(
#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");
}
}
hey bro,your code is correct but at the end it will show TLE
@@VIKASKUMAR-pc4jz
👍@@anantagrahari2547
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)?
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
I wish i would have known
abt this channel earlier rather than spending money on paid course thank u striver.
I was waiting from last 5 days. Finally come 😌😌
UNDERSTOOD THANK YOU STRIVER
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
Excellent Sir,My honorable thanks for making these videos with excellent explanation❤.
It's great to learn from this channel ❤️Thank you:-)
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
Very nice explanation...God Bless You
This Is Just Awesome video🔥🔥🔥🔥🔥🔥🔥
understood the assignment :) thanks striver
bool checkArmstrong(int n){
int dup = n;
int sum = 0;
int cnt = 0;
while (n > 0) {
int lastDigit = n%10;
n = n/10;
cnt = cnt + 1;
}
n = dup;
while (n > 0) {
int lastDigit = n%10;
sum = sum + pow(lastDigit, cnt);
n = n/10;
}
return dup == sum;
}
I understood everything and i am also a student of your your college JGEC😊
best one till now!!! 🙏🙏🙏
very clear and crisp explanation.