Tearing bit into bits (C++)

A gist of vital questions/concepts related to Bit wise manipulation : 

1. Turn off the first set bit(1) from right of a number N.
1. Application: Fast Bit Count 
2. Application: Single Number III
2. Odd/Even test
3. Get Kth bit of N / Checking if the Kth bit is set or unset: { bool bit_K = num & (1 << K); }
4. Set Kth bit of N:  {num |= (1 << pos);}
5. Unset Kth bit of N: {num &= (~(1 << pos));}
6. Toggling a bit at Kth position: {num ^= (1 << pos);}
7. Convert an integer (Decimal) into a bitwise (string) 
8. Difference between '&' and '&&' or for that matter '|' and '||'
9. Range of (a &b) , range of(a|b), range of (a^b) 

REMEBER the tools at our disposal are : 

• SHIFT operators : >>(right shift, div by 2) and <<(left shift, mul by 2)
• BOOL operators : ^ (XOR), &(and) , ~(NOT)
• Arithmetic operators like +, - , /, * are not used. 

int int_var;

cout<<sizeof(int_var)<<endl; // returns number of bytes

output is  4 

this means an integer has 32 bits or 32 placeholders to hold either 1 or 0. 

1.Turn off the first set bit(1) of a number N.

Two ways : 

A. 

• get ~N 
• add +1
• multiply this with N... log of this result will show the position of rightmost set bit

B

• get (N-1)
• flip the bit ~(N-1)
• multiply this with N... log of this result will show the position of rightmost set bit

Fast Bit Count : 

```Java

count=0;

while(n>0) {

    n= n &(n-1);

    count++;

}

return count;

```

2. Check whether n is even or odd: The naive approach to check a number is even or odd is to take the modulo with 2. The better and efficient method is to take the (n&1). If the last bit is set n is odd, else even.

3. If you want the k-th bit of n, then do : 

(n & ( 1 << k )) >> k

(n >> k) & 1 is equally valid 

5. 

A&B =[0,min(A,B)] A^B =[0,A+B] A|B =[max(A,B),A+B]

TIPS : 

•  mask = 0xffffffff  then ~(a^mask)

References : 

• https://hackernoon.com/bit-manipulation-in-c-and-c-1cs2bux
• https://www.codingninjas.com/codestudio/library/bit-manipulation-for-competitive-programming
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