Which one line function can be used to get index of the least significant bit that is on (first 1 from right to left in binary representation of an integer) ?
A.log2( N ^ -N ) + 1
B.log2( N | -N ) + 1
C.log2( N & -N ) + 1
D.log2( N ~ -N ) + 1
how C is the answer?
@hrit04
The answer should be log2( N & -N ) + 1 , if n is a positive number , so -n will be a negative number and negative number are represented in 2’s compliment ,which is ~n+1 , ~n has all the bits of n inverted due to which we get 0 in place of 1 and vice versa so when we add 1 to ~n then that 1 occupies the first rightmost 0 in (~n)(which is same as rightmost 1 value in n) and other bits have inverted values or zero values . so when we perform & operation we get only rightmost bit set in the function ,and then we perform logn (base2) to get corresponding index.
eg :- n = 1110010100
~n= 0001101011
~n+1=-n= 0001101100
(n&(-n))= 0000000100 = 4
log(4) =2
2+1=3 which is the answer
you should try applying the formula on some examples on your own for better clarity
dont forget to hit like and mark resolved if cleared 
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