Bitwise.md

Bitwise Operations

Bitwise Operations are always faster that the arithmetic operations

Bitwise Operators

• AND - &
• OR - |
• NOT - ~
• XOR - ^
• Left Shift - <<
• Right Shift - >>

public class bitwiseOperator {

    public static void main(String[] args) {
        int a = 3;
        int b = 4;
        System.out.println("a & b = " + (a & b));
        System.out.println("a | b = " + (a | b));
        System.out.println("a ^ b = " + (a ^ b));
        System.out.println("~a = " + ~a);
        System.out.println("a << 1 = " + (a << 1));
        System.out.println("a >> 1 = " + (a >> 1));
    }
}

Even or Odd

If the last bit is 1 then the number is odd, otherwise even.

____1&1=1 odd
____0&1=0 even

Using Arithmetic - (NOOB)

if(num%2==0){
    System.out.println("Number is Even")
}
else{
    System.out.println("Number is Odd")
}

Using Bitwise (PRO)

if(num&1==0){
    System.out.println("Even")
}
else{
    System.out.println("Odd")

}

Questions

1. Count the number of 1 bits in an integer.

2. Counting Bits - Leetcode

Given an integer n, return an array ans of length n + 1 such that for each i (0 <= i <= n), ans[i] is the number of 1's in the binary representation of i.

Input: n = 5
Output: [0,1,1,2,1,2]
Explanation:
0 --> 0
1 --> 1
2 --> 10
3 --> 11
4 --> 100
5 --> 101

class Solution {
    public int count(int n){
        int c = 0;
        while(n>0){
            int lastbit = n&1;
            if(lastbit ==1){
                c++;
            }
            n= n >> 1;
        }
        return c;
    }
    public int[] countBits(int n) {
        int[] arr = new int[n+1];

        for(int i=0;i<=n;i++){
            arr[i]= count(i);
        }
        return arr;
    }
}