We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
For every un-set bit , there are 2 posiiblities of a number that when added to the original number would give the same answer , if xored
So , count the number of unset bits , and then raise it to the power of 2 , to get all combinations of numbers that when added to n will give the same result as to when xored by n
def sumXor(n):
# Write your code here
ct = 0
while n:
if n&1==0:
ct+=1
n>>=1
return 2**ct
`
Cookie support is required to access HackerRank
Seems like cookies are disabled on this browser, please enable them to open this website
Sum vs XOR
You are viewing a single comment's thread. Return to all comments →
So , count the number of unset bits , and then raise it to the power of 2 , to get all combinations of numbers that when added to n will give the same result as to when xored by n
def sumXor(n): # Write your code here ct = 0 while n: if n&1==0: ct+=1 n>>=1 return 2**ct
`