Leetcode 338. Counting Bits | Bit Manipulation | Brute Force to Optimal | 2 Solutions with full code 

vectorans(n+1,0); for(int i=0;i

This makes much more sense instead of % 2 we can use &1 which return 1 if bit is 1 and 0 when bit is 0; and we can right shift num by 1 class Solution { public: vector countBits(int n) { vector ans; for(int i=0;i>1; } ans.push_back(sum); } return ans; } };

Can you please tell me how the binary value of a number is getting assigned to the variable 'num'?

Thank you so much for effort . it is simple logic and so easy to get understand. 💌

the way you explained the solution was awesome dear

Yours is the simpleest explanation I found

Awesome job Alisha. I just found out about this channel and I'm definitely here to stay.

thanks its very usefull

ty

Excellent Explanation Alisha. Subscribed ;)

how diving the num by 2 removing 1 value from the unit place i think it's done by dividing it by 10 1011/2 = 505 1011/10 = 101 and that's what we want here

N is given in decimal but when we are iterating over N how we can find 1 and 0s?

Awesome ...

Thankyou for super easy explanation

thanks

Good Work

Nice explanation 🎇

Op❤

yrr aap muche kaat leti

Thanks

Nice explanation 🔥