Count the number of one's in binary representation of given no. || Competitive Programming ||

In this video we are going to discuss very interesting algorithm in bitwise operator to count the number of one's in the binary representation of given number.competitive programming. codecehef solution also in hello world.
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#competitiveprogramming #helloworld #codechef

[@devanshgoel3433]

Bro, you have explained amazingly. But, one point to note is that if number is zero then it will give output as 1.

[@Ji-yoon]

Your channel is awesome......continue making videos of this kind.....

[@AnuragSingh-iy7fz]

sir log question kare ya na kare, ye unki problem hai, aap videos banate rehna competitive coding par, ham sab aapke saath hain.
Appreciate your efforts

[@manojkumar-qm4mx]

What a nice job
Sir AAP bahut achha padate hain 🙌🙌

[@HelloWorldbyprince]

Thanks Manoj

[@aniketjaiswal3874]

sir
c=0;
while(n)
{
if(n%2==1)
c++;
n=n/2;
}
this will do in O(log n) which in worst case better than O(k) (=O(n))....
#completed

[@samarth9467]

public class Main {
public static void main(String[] args) {
int number = 23; // Change this to the number you want to count 1's in binary for
int count = countOnes(number);
System.out.println("Number of 1's in binary representation of " + number + ": " + count);
}
public static int countOnes(int n) {
int count = 0;
while (n > 0) {
count += n & 1; // Check the least significant bit
n = n >> 1; // Right shift to check the next bit
}
return count;
}
}

[@HelloWorldbyprince]

Amazing 🤩

[@nimssinha5373]

Good job Brother✌

[@7949RAJ]

sir,
can we run a loop and check if(n&(1

[@pankajnavale744]

#Completed

[@ketandoshi5641]

#completed

[@pankajnavale744]

Completed

[@rockytapader4643]

nice

[@HelloWorldbyprince]

thank you ... Happy coding :)

[@pradeexsu]

we can also use __builtin_popcount() no header file require
#completed

[@RANJEETSINGH-dd8ru]

sir please give some good question in home work. It is too easy .

[@Dragoon102]

Thank you sir for ur help

[@HelloWorldbyprince]

Your welcome buddy ... Please share this channel to different college and with your friends

[@teores8312]

It works, thx Dude!

[@HelloWorldbyprince]

You're welcome!

[@AnuragSingh-iy7fz]

#completed
#include
using namespace std;
int main()
{
int t,n;
cin>>t;
while(t--)
{
cin>>n;
if(n

[@surajkumarsaw7398]

#complete

[@rishabhpandey8892]

#include
using namespace std;
int main() {
int t;
cin>>t;
while(t--){
int n;
cin>>n;
if(n

[@abhishekshetty9138]

question is repeated

[@ShubhamChhimpa]

#completed
#include
using namespace std;
int main()
{
int T;
cin >> T;
while(T--){
int x;
cin >> x;
if(x < 10){
cout

[@parikshitsharma5651]

#done

[@itz_me_imraan02]

count=0;
while(n!=0)
{ n=n&(n-1);
count++;
}
cout

[@HelloWorldbyprince]

Amazing

[@chetankumar1487]

why count is taken as 1...it should be initialised with 0.

[@chetankumar1487]

understood

[@azrallex]

homework question is repeated, sir.

[@amansohani4620]

sir par agar sare 1 ho binary representation me toh worst case me toh is algorithm ki bhi log(n) complexity hogi na.

[@HelloWorldbyprince]

But Aman, aap ye v to dekhiye at max aap sareee one's wale number kya lijiyega .. for example : 111111111111111111111111111111 --> mann lo aaapne ye liya ...and first of all ye itna badda number hai ki aap calculate nahi Kar sakte but fir v ..iska complexity O ( 30 ) jayega jo ki mathematically bahut kam aata hai .. your point is right but in actuality itna big number se hmlog deal nahi karte ... Thank you ..

[@amansohani4620]

@@HelloWorldbyprince ha samagh gaya sir thanks

[@chidambarjoshi3470]

#servantcompleted

[@vaibhavsingh964]

how and why this logic?

[@HelloWorldbyprince]

because its simple and basic

[@md.azharulkarimanik1273]

#completed

[@sourjyapal945]

#completed

[@rajumolla4212]

#completed

[@pawansinghal585]

#completed

[@rohananand9905]

#completed

[@tarunprajapat2240]

#completed

[@amitjaiswal6594]

#completed