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

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[@probabilitycodingisfunis1]

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

[@satvikarya569]

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

[@onkarsangale6222]

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;
}
};

[@JaJa-tp9ck]

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

[@shikha2121]

Yours is the simpleest explanation I found

[@piyush2750]

Thankyou for super easy explanation

[@saikrishna-cy2tr]

the way you explained the solution was awesome dear

[@manideep7148]

Excellent Explanation Alisha. Subscribed ;)

[@AMITKUMAR-xv2ny]

Awesome ...

[@kunalkheeva]

ty

[@sreeharia2356]

Nice explanation 🔥

[@babushaikh6582]

Nice explanation 🎇

[@commandermsv3329]

Good Work

[@Rajat_maurya]

Thanks

[@prashantshukla7477]

thanks

[@sahityaraj9417]

Op❤

[@dhananjoydey1337]

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

[@khushil9988]

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

[@n1111shant]

yrr aap muche kaat leti