For the second problem I'm actually surprised the first guy didn't have a clue of the most efficient way. I'm a CS Junior and one of my first class last semester was about assembly. It helped me so much with the understanding of problems like these.
# One of the better ways to find if the number is a power of 2:- #include using namespace std; bool f(int n) { if(n == 0) { return false; } int count = 0; while((n & 1) == 0) // when the lsb is set the loop stops. { count++; n >>= 1; } if(n == 1) { return true; } return false; } int main() { int n = 16; cout
why bother doing that? just use bitwise to solve it public static void main(String[] args){ int x = [INPUT]; int ct = 0; while(x > 0){ if(x & 1 == 1){ ct ++; x >>= 1; } if(ct == 1) System.out.println("is a power of 2"); else System.otu.println("not a power of 2"); } very easy
Hey yours will be wrong as for ex of we take 12 it will return 6 and 12//2 also returns 6 even though they are same we know that 12 is not a power of 2
@@Vvarsha007 you are absolutely correct ma'am and i agree as now I can see I am purely wrong there btw I started to learn programming 3 months ago so no hardships 😁 Let's see what this world has to offer for me 😌
The O(1) solution is limited by the address space of the registers in the CPU and ALU. So for very large powers of two, like above 2^64 it may not be as efficient. Depends on how many bits the registers can store. Disclaimer: This is just thought aloud. Might need more verification.
Honestly i enjoy the series It's great to know the foreign students how they study in college and what apporch they think 💬. And sir one request is there , can you make a video for Bca students . How they can start to do coding and after doing bca they should do masters for multinational companies . And for multinational companies only coding is important or any more thing we have to study. Because we are learning basic things of coding . And one more thing i am noob in coding so how can i start coding can you explain in video or reply here . And for coding you can prefer any book .
the solution you just gave for was very creative for checking the power of the two , but after that the more simple answer would've been just checking the LSB of that binary number
8:45 here's my solution, Please tell me is this a good solution (I am just a beginer) counter = 0 a = int(input("enter any no.")) b=a while True: if b%2==0: counter += 1 b = b/2 else: break if (2**counter)==a: print("True") else: print("False")
BTW all these solutions are actually correct... The full adder circuit is itself O(number of bits) so subtracting 1 takes O(number of bits which is same as number of digits)
Aryender is doing such an amazing job as Harnoor's new partner! Also giving away 100 or 50 USD to these students who can afford leetcode premium doesn't makes much sense! Why don't you give your Indian subscribers a question who can't afford leetcode premium?
For the power of 2 one you can just do log(number)/log(2) and if its an int then you know the number has a base of 2. However, I may have also misunderstood the question lol.
i am from pakistan i don`t no what an diffecult level(mean at logically level not craming ) iit take off the test but an incridble change in india education really appricated ....
For 2 to the power n, can’t we just take base 2 log of the number. If we get an integer back. Then it’s true else it’s false. That would also be constant time. Assuming log2 function has constant time.
I have a question for all of you. Q1. Given the root of a binary tree where every node contains a natural number, return the maximum path sum from node to leaf.
x="101" y="111" list=[] def binary(num): for index in range(len(num)): list.append(index) list.reverse() return list def bin_cal(daxil): result=0 for index, eded in enumerate(daxil): eded=int(eded) result+=int(eded)*(2**binary(daxil)[index]) return result print(bin_cal(x))
6:54 if a number can be written as a power of 2 then it means that it is divisible by 2 ...so I think so we can do it by simply testing as we test an even number 🥇 or not ??
Question: im bad at math but i want walk in technology path, should i learn math first or codding(Frontend) but i havent decide what i want to focus with is it website or the other.
I suggest sticking to your passion, if you love technology then pursue that, allow everything else to follow suit. I shall give you an example, a strategy game like Yu-Gi-Oh requires a lot of reading, however reading is not a necessity, if one has a passion in Yu-Gi-Oh, they simply need to play the game, memorising the cards they need to know, eventually memorising what words relate to what meaning, after decades of playing Yu-Gi-Oh, do you not think he would have grasped basic reading, enough to play Yu-Gi-Oh effectively? Analogous to you grasping maths, enough to be effective in the path of tech
Mai bba ka student hu pr Mai ek code developer bna cahta hu to kya Mai bn skta hu agr bnta hu to bhi kya mujhe achi job milegi jaise ki btech walo ko milte hai
This would work right ? base_number = 2 power_of = int(input("Enter the power of number: ")) user_calculated_output = int(input("Enter the number to check if the required result is True/ False: ")) multiply_power = base_number ** power_of if multiply_power == user_calculated_output: print("True")
harnoor? please tell me,, are you on your OPT right now or what? in what status are you there right now? you had got F1 again in dropbox from delhi,, so you are doing masters?
honestly, I really enjoy such content harnoor, please continue doing such type of vids. I really appreciate your effort man
The way SHE told that why she was able to solve the 'leetcode' questions, shows us the harsh reality of being poor. Kudos woman.👍🏿👍🏿
Good to see people appreciating difficulty at IITs ....
Tho tu kyu khush ho raha
@@rohanIVY INKO IIT ME NAUKRI LAGA HAI. JANITOR KA
@@rohanIVY dikhai nahi deta woh professor hai !!🙂🥲
@@rohanIVY Because I was at one point involved in studying for these institutes. I understand how hard it is to get in.
@@akashpaul4143 Bhai phir toh muje kal paaka asli Willi smith ne comments meh gaali di
Please make more videos like this 🥹 really enjoy watching them. Very insightful
um with u
I’m loving this series of asking questions to top College students❤
bhai indians hi indians dikh rahe sab jagh..hahah..its like u r in dilli sarijini nagar..so happy seeing indian roots at all places !!
All of these videos are so good, especially at Stanford and with the Meta guy!!
Keep posting such videos. They are very inspiring.
We want this type of content More and More and Most
For the second problem I'm actually surprised the first guy didn't have a clue of the most efficient way. I'm a CS Junior and one of my first class last semester was about assembly. It helped me so much with the understanding of problems like these.
Yeah but after 3 months you’ll forget it 😂
14:03 best part 😂
# One of the better ways to find if the number is a power of 2:-
#include
using namespace std;
bool f(int n)
{
if(n == 0)
{
return false;
}
int count = 0;
while((n & 1) == 0) // when the lsb is set the loop stops.
{
count++;
n >>= 1;
}
if(n == 1)
{
return true;
}
return false;
}
int main()
{
int n = 16;
cout
Everytime I watch his video, I start doing leetcode for a week and then stop. 😂
Relatable😂
@@rohanIVYhow do you decide on the list of problems?
@@boy0607boy no. Of likes on the problem
Me tooo
Let me take your mock interview if you are still alive!😊😂
x=int(input(enter the no;))
y=x/2
if y==x//2 :
print("is a power of 2")
maybe the second question cqn be solved like this
why bother doing that? just use bitwise to solve it
public static void main(String[] args){
int x = [INPUT];
int ct = 0;
while(x > 0){
if(x & 1 == 1){
ct ++;
x >>= 1;
}
if(ct == 1) System.out.println("is a power of 2");
else System.otu.println("not a power of 2");
}
very easy
@@Sanyu-Tumusiime that seems more hard though 😅😅🤣🤣
Hey yours will be wrong as for ex of we take 12 it will return 6 and 12//2 also returns 6 even though they are same we know that 12 is not a power of 2
@@Vvarsha007 you're right. it will be wrong. use my solution
@@Vvarsha007 you are absolutely correct ma'am and i agree as now I can see I am purely wrong there btw I started to learn programming 3 months ago so no hardships 😁
Let's see what this world has to offer for me 😌
The O(1) solution is limited by the address space of the registers in the CPU and ALU. So for very large powers of two, like above 2^64 it may not be as efficient. Depends on how many bits the registers can store. Disclaimer: This is just thought aloud. Might need more verification.
yeah but the input itself is stored in int/long format so this solution would work
int range 2^31-1
This is just a pure understanding of how binary operations work…. Nothing too technical on algorithm.
This students are quite intellegent .
Bhai won Stanford hai
Wahan to honge hi
Stanford Universities not all students r as briliant as iitians
As 40-50% goes to iit due to their social activities in international level
He never stop to motivate people 🥺
Honestly i enjoy the series It's great to know the foreign students how they study in college and what apporch they think 💬. And sir one request is there , can you make a video for Bca students . How they can start to do coding and after doing bca they should do masters for multinational companies . And for multinational companies only coding is important or any more thing we have to study. Because we are learning basic things of coding . And one more thing i am noob in coding so how can i start coding can you explain in video or reply here . And for coding you can prefer any book .
Same
Super video Bro Nice Intro
the solution you just gave for was very creative for checking the power of the two , but after that the more simple answer would've been just checking the LSB of that binary number
bro that string question is cake walk level 😂
My son is doing CS from UTS in sydney and Its my dream to see him get his masters from MIT.
8:00 return n & (n-1) == 0
Hey bro I need more videos like mit students solving iit question paper and iit students as same as mit question paper. Need from you
Their names were Sanjay,Sahil & saurav and they were not from India🙃.even foreigners give Indians name to their children.
They are from Indian origin not Indian their parents are probably 2nd Gen Immigrants
@@sxmeersharma Sahil from Bangladesh
@@mrkiba1781 it’s a common name in both countries he could be even Pakistani
I guess they are OCI's
@@sxmeersharma he is my relative, don't claim anyone Indian origin
8:45 here's my solution, Please tell me is this a good solution (I am just a beginer)
counter = 0
a = int(input("enter any no."))
b=a
while True:
if b%2==0:
counter += 1
b = b/2
else:
break
if (2**counter)==a:
print("True")
else:
print("False")
Idk
@@quandledingleiv6082 why comment "idk"
its TC is still o(log n) its not the most optimal solution the & operation solution is O(1) so that is the most optimal solution
@@chickenstrangler3826 my choice
@@quandledingleiv6082 Stupid choice.
BTW all these solutions are actually correct... The full adder circuit is itself O(number of bits) so subtracting 1 takes O(number of bits which is same as number of digits)
mm not really , it's more efficient because it's a native thing for the processor.
@@ziedbrahmi4812 but it is still under the assumption that number fits in the circuits
I need this kind of videos more ...
good job , keep it up.
Aryender is doing such an amazing job as Harnoor's new partner! Also giving away 100 or 50 USD to these students who can afford leetcode premium doesn't makes much sense! Why don't you give your Indian subscribers a question who can't afford leetcode premium?
Admission in IIT is much tougher than MIT, People like you who not able to clear JEE moved to US/UK for further degree course.
First go see selection process of these college then talk
# Adding 2 Binary Strings:-
#include
using namespace std;
string bin(string &s1,string &s2)
{
int i = s1.size() - 1;
int j = s2.size() - 1;
int carry = 0;
string res = "";
while(i >= 0 || j >= 0 || carry)
{
int n1 = (i >= 0) ? s1[i] - '0' : 0;
int n2 = (j >= 0) ? s2[j] - '0' : 0;
int sum = n1 + n2 + carry;
carry = sum / 2;
res.insert(res.begin(),(sum % 2) + '0');
i--;
j--;
}
return res;
}
int main()
{
string s1 = "101";
string s2 = "0010";
cout
This is good content bro but please post the selected question or any question to solve for the viewers ✌
thanks for everything!
For the power of 2 one you can just do log(number)/log(2) and if its an int then you know the number has a base of 2. However, I may have also misunderstood the question lol.
Wow awesome 👌
Appreciating for Last girl 👧
Maja e aa gaya vlog dekh ke toh
i learnt whole of this binary system in grade 10th and after joining coaching institute it all went to vein due to extreme pressure of examsss
I thought MIT < Stanford. Anyways bro you got to get a dedicated mic since the device is picking up sounds of the background as well
lmao why , MIT is #1
For powers of 2 Idk
bool is_power_of_two(int number) {
return number > 0 && (number & (number - 1)) == 0;
}
Could someone check please 10:31 10:32 10:35
And me thinking the optimal solution for power of 2 => (ceil(log2(n)) == floor(log2(n)))
I really enjoy it 😀😀🙃.
For checking if the number of 2 to the power of some number:
def check(x):
m = x/2
while m > 1:
m = m/2
if m == 1:
return True
else:
return False
14:00 "IIT" be like Raula hai hamra , kabhi kabhi to lgta hai apun hicch bhagwan hai.
Still not in top 50 in global list where as MIT and Staford are in top 3
i am from pakistan i don`t no what an diffecult level(mean at logically level not craming ) iit take off the test but an incridble change in india education really appricated ....
For 2 to the power n, can’t we just take base 2 log of the number. If we get an integer back. Then it’s true else it’s false. That would also be constant time. Assuming log2 function has constant time.
Where can you get that log function from ? Not every lang is python.
@@spiderop2125 exactly...That is what I was thinking
Yeah it can be possible in Java and CPP both they have log functions
They’re basically asking them to code the log function from scratch. In the most efficient way possible
12:00 *That's why I'm at Stanford not MIT*
As if Stanford is the LPU of USA.
I have a question for all of you.
Q1. Given the root of a binary tree where every node contains a natural number, return the maximum path sum from node to leaf.
recursion
i like your content Harnoor. Keep visiting the college
Please make video on MIT VS STANFORD
just stupid idea bro
x="101"
y="111"
list=[]
def binary(num):
for index in range(len(num)):
list.append(index)
list.reverse()
return list
def bin_cal(daxil):
result=0
for index, eded in enumerate(daxil):
eded=int(eded)
result+=int(eded)*(2**binary(daxil)[index])
return result
print(bin_cal(x))
6:54 if a number can be written as a power of 2 then it means that it is divisible by 2 ...so I think so we can do it by simply testing as we test an even number 🥇 or not ??
Non Indians wouldn't know about IIT -- some are just being polite and picking IIT as the host is Indian.
no , they're picking IIT coz they wouldn't wanna deal with MIT's question. I would do the same if i was in their place.
Lots of Indian names like Sanjay , sahil etc love to see Indians doing great
Theyre from Indian parents so their descent is indian but their nationality is ig american so i don’t really think we should address them as indians
@@xPhilosophyy i know that they are American citizen
just because someone looks Indian with indian names, does not mean that person is actually an Indian
please share the result such as valid answers of all questions at the end
Ivy league kids are literally out of this world
Facts
Ayender is genius!!!!
I would have gone for;
int a ;
if (2 modulus a == 2) {cout
No...
BCA student here ❤️
GREAT HARNOOR BRO
Amazing content❣️❣️I also wanna be like you
In my college out of 100 almost 99 can't solve these problems. Here almost every one have knowledge
Question: im bad at math but i want walk in technology path, should i learn math first or codding(Frontend) but i havent decide what i want to focus with is it website or the other.
I suggest sticking to your passion, if you love technology then pursue that, allow everything else to follow suit. I shall give you an example, a strategy game like Yu-Gi-Oh requires a lot of reading, however reading is not a necessity, if one has a passion in Yu-Gi-Oh, they simply need to play the game, memorising the cards they need to know, eventually memorising what words relate to what meaning, after decades of playing Yu-Gi-Oh, do you not think he would have grasped basic reading, enough to play Yu-Gi-Oh effectively? Analogous to you grasping maths, enough to be effective in the path of tech
You can visit mission college to contest code in 5 minutes
sir it's good to go
imagine how selectvie stanford would be if it was in washington
MIT admissions process pa ek video banaow bahi
Mit: if(log2(x) - round(llog2(x))==0): return True
wrong
kuch samjah nehi aya but sunke achaa lagaa :)
Please visit LAC too
Video seems to be lagging. Anyways harnoor great content.
What is Lewis Hamilton doing in Stanford??🤔
Both were super easy 😉
I Agree
They literally said it’s easy questions
Mai bba ka student hu pr Mai ek code developer bna cahta hu to kya Mai bn skta hu agr bnta hu to bhi kya mujhe achi job milegi jaise ki btech walo ko milte hai
Bro my brain 🧠 is hanged after watching match . Any mathematic solution for this?
I have seen that purple t-shirt guy with glasses in one of your other video
10:55 he has a kalawa on his wrist! Indian origin
Can't you tell by the face?
Multiple of 2 question soo easy...
This would work right ?
base_number = 2
power_of = int(input("Enter the power of number: "))
user_calculated_output = int(input("Enter the number to check if the required result is True/ False: "))
multiply_power = base_number ** power_of
if multiply_power == user_calculated_output:
print("True")
else:
print("False")
Man I just saw this video to make me realize how dumb I am.
why not popcount(x)==1 to find pow of two.
harnoor? please tell me,, are you on your OPT right now or what? in what status are you there right now? you had got F1 again in dropbox from delhi,, so you are doing masters?
bruh these questions are so easy
brother i want to know more about cs engineering so please can you give idea of a good plate-form for coding.
the power of 2 soln is brian kernighan algorithm , just fyi
Wedding begins 😃😃 next year
IIT--jalba h hamara 🤣
IIT Bombay CS students be like bruh that is most basic question u gonna get...
Jalwa hai .
Whole Budget Of This Video:- $300 =₹24,501.14 😄
and revenue just 5 times of the investment
bhai jan. Full stack web developer kha be scope ha canada ma
plz reply
why do Indians in US are too tall than an average Indian
Maybe it's all about food
Indians in US are not tall by any means.Average Indian is short.My cousins in US are shorter than me and my brother.It's more about food and sport.
Lagta hai bhai ko bhi startup karne ka hai.
Can I know which programming language are they using?
shouldnt N & (N-1) be O(log2N) and not O(1)
where can I get those IIT exams pdf am studying computer science.
from students at iit
No longer Meta Engineers😣😣
What is not clear in the question is if the result is to be in binary format or decimal.
1100 or 12?