Maximum XOR for Each Query | Leetcode 1829

[@bhavanbharadwaj7901]

thanks a lot!!

[@techdose4u]

welcome :)

[@ashishsharma2341]

Nice explanation

[@techdose4u]

thanks :)

[@AfrinJahanBME]

Nice explanation 👏👏

[@techdose4u]

Thank you 🙂

[@fightermad]

Thanks so much

[@techdose4u]

Welcome :)

[@sailendrachettri8521]

Thank you :) :)

[@techdose4u]

No problem 😊

[@ssquare1]

👏👏

[@techdose4u]

:)

[@devmahad]

thanks :)

[@techdose4u]

welcome :)

[@21flame72]

3:19 why 2^7 could you explain that part was not clear

[@techdose4u]

We will try all possible values from 0 to 2^maxBits-1 for bruteforce.
Later I have explained why 7 works in optimal solution :)
Using Bitmasking.

[@ayushjaiswal997]

how did you pic value of k = 101 = 5 that is less that 8 instead of 010 = 2 which is xor_result of all array element or why not 3,4,5,6,7. Is it Hit and Trial method or any logic behind it. if yes then please explain.

[@techdose4u]

timestamp ?

[@ayushjaiswal997]

@@techdose4u 2:25

[@techdose4u]

It is explained ahead in operation strategy part.
This is just an example and not solution.
The Xor maximizing strategy will tell uou the answer explained ahead.
It is based on Xor property.
It is not trial method.

[@anjali3063]

but what if after masking i got number which is greater than range of k

[@techdose4u]

It is impossible as the contraint mentions all array items are less than equals to max value of K.

[@anyapatel50]

"Thank you for the solution, sir. I have a question: should I solve the daily challenge, or are there specific questions I should focus on because in my college there is no campus placement no guidence 😞?"

[@techdose4u]

You should defnitely solve POTD (daily challenge).
Moreover, you must practice a techstack.
Popularly followed is fullstack.
You can ping me on LinkedIn for more questions.

[@scientificabhishek4332]

please make videos in java also so java people can also understand

[@techdose4u]

Java code is present in code.
C++
Java
Python

[@ZahiidKhan262]

I am not able to understand 🥲🥲

[@techdose4u]

Please watch.
you will; definitely :)