Bitwise operator

 &:  if a&1 then last digit of a will always same as in  previous a.

a^1= compliment (a), 

a^0=a, (diff bit =a)

a^a=0, (same bit =0)

coversion :

Decimal --> n base = keep divide by base 

exp : (17)D = ()2 =  .......

n base --> decimal = multiply each digit with the power of base and add them.

  exp: 1* 2^3 + 0*2^2 + 0 * 2*1+ 1*2^0=..

Left shift : a<<b = a* (2^b)

Right shift :  a>>b = a/(2^b)

unique no. in duplicate array : xor all elements

find ith bit of a no. n = n & (1<<n-1)

set ith bit  = n | (1<<n-1)

reset ith bit= !(1<<n-1)

Find position of rightmost set bit

negative no. in binary= 1st bit (MSB) = if 0 (+ve), id 1(-ve)

neg no. by 2's compliment : find compliment of no. ..then add +1

1 byte range = -128 to 127 (total 256 no. can be stored..2^7) because we ignore 1st bit

formula of range = -2^(n-1) to 2^(n-1) -1

find uniq no. in odd repeating no. array

find nth magic no.

find no. of digit in no. n havining base b= (int) (log(n)/log(b)) +1

check no. is a power of 2 : n &( n-1) ==0

find the xor of 0 to a  =   

• a%4==0, ans= a
• a%4 ==1, ans= 1
• a%4==2,ans=a+1
• a%4==3,ans=0

• total no of digit -1 of int : digit =(int)log10(a)
• 1st digit of number :  (int)(a/pow(10,digit));