1863. Sum of All Subset XOR Totals | Bit Manipulation | Generating Subsets | Xor | 3 Approaches

[@ARYANMITTAL]

Of course not an easy one, if this comes in an interview & Interviewer expects 3rd approach, then best of luck from my side if you have not seen the video 👍🌚 . Money Laundering & Drug Trafficking Scam with me - kzbin.info/www/bejne/mJa3qqeJaLybiZI

[@ravipatel-xu5qi]

thank you Aryan for putting effort everyday. A small request, it will be really helpful if you can put chapter into your video. So that we can jump between the sections easily and can watch the specific part. Here if you are explaining approches ,than better to have chapters based on that. Thank you.

[@abishekupadhyay6357]

amazing explanation sir.

[@dewanshpatle9056]

TYSM for the explaination.

[@rohita6532]

if someone struggling in last explanation they can see neetcode solution video he explained in short and easy way

[@mumukshapant]

Your backtracking crash course is not available

[@cenacr007]

❤️

[@RohanSharma-tl9wh]

pooja ghanti at 11:00

[@kevinesh]

Thnk you for the explanation. However, I'm not really convinced with the contribution of every bit to be 2 ^ (n-1) you didn't show the formal proof, just an intuition based on a example.

[@ARYANMITTAL]

Ohh that's one liner after this example. Imagine ith bit of those n numbers are like 11000.....11010 i.e x 1's & y 0's at the ith bit, now this ith bit will have 1 in the end after doing xor only when 1 is occuring odd number of times, which is same way saying find me the number of subsets having odd number of 1s. Its counter proof is available here (Number of subsets with even number of ones = Number of subsets with odd number of ones) - math.stackexchange.com/questions/1338787/number-of-subsets-with-even-number-of-elements

[@ITACHIUCHIHA-dr8sz]

//find contribution of each bit in sum // let k elements have the ith bit set // ways1 = ways to choose ith bit not set = 2^(n - k) // ways2 = ways to choose ith bit set = kC1 + kC3 + kC5... kCk => 2 ^ (k - 1) since xor of odd count of 1's would contribute to ans // total ways = ways1 * ways2 = 2 ^(n - 1) // total contribution to ans = totalways * each way contribution which is (1

[@asimshah1824]

The tf is the leetcode easy

[@KADOfficial23]

kya bol rha hai kuch samajh mein nhi aa rha hai