 # NCR codewars question doubt

Hello Community!
So problem was like this —
find x in such way that after performing Xor operation with each element of given array the sum of array is maximized and if the there are more than one x which is giving the maximum array sum (after operation) then we have to return the minimum x.

x can be upto 2^63.

input
1
5
10 12 5 7 19

output
92233720368…(some 18 digit number, i don’t remember)

How can i solve this?

thanks for replying! but can you explain what you are doing in your code?

first of all i have checked that how many numbers out of n have their ith bit set .

if more than or equal to half of the numbers have their ith bit set then the ith bit of our ‘X’ will be 0 else the ith bit of our ‘X’ will be 1.

I hope u got it know ,still not dry run the code or ask me again …!!!

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You actually have to use equal to here too because if half the numbers have ith bit set then we have to take minimun ‘X’.

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ohh yes , my bad …!!!

Got it. thanks!!

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Very nice explanation . Thanks.

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can we also do this like, first find the sum of array element then set the 0 bit to one and one bit to 0 that will be our X…

No we can’t do like this .

eg let’s take 3 numbers 1,2 ,4

1 -> 0 0 1
2 -> 0 1 0
4 -> 1 0 0

s -> 1 1 1 (sum of 1,2,4)

according to you the value of all these bits will be ‘0’ but all these will be ‘1’ in the ‘X’.

hope it helps …!!!

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Thanks for replying, what is the value of x in this case…

and what is the value of maximum sum in this case

As all the bits will be set to 1 in this case , so value of ‘X’ will be 2^63-1.

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