Binary Left Shift (Multiplies):
a &lt;&lt; b = a * (2^b)
Original : 0 0 0 0 1 0 1 : Int (5)
Shift Left: 0 1 0 1 0 0 : Int (20)
if a = 5, b = 2, hence 5 &lt;&lt; 2 and 5 * (2 ^ 2) Therefore 20.
Binary Right Shift (Divides):
a &lt;&lt; b = a / (2^b)
Original : 0 1 0 1 0 0 : Int (20)
Shift Left: 0 0 0 0 1 0 1 : Int (5)
if a = 20, b = 2, hence 20 &lt;&lt; 2 and 20 / (2 ^ 2) Therefore 5.
Sample Interview Questions:
1) Odd Even
<pre><code class="lang-java"> if(x&amp;1){
 System.out.println("Odd");
 }
 else{
 System.out.println("Even");
 }
</code></pre>
2) Getting the iTH bit
Take a empty bit, and move the 1 to ith location, such that we can perform AND (&amp;) to the <code>n</code> number.
Step 1:
Given: n=5, i=2
n in 0 0 0 1 0 1
Step 2:
Build a bit, such that it only has <code>1</code> in the ith Location using LeftShift.
mask=0 0 0 0 0 0 0
x&lt;&lt;i i.e. 0 0 0 1 0 0
Step 3: Perform AND with n and mask,
res = n &amp; mask (this will multiply)

**Binary Left Shift (Multiplies):**

a << b = a * (2^b)

Original : 0 0 0 0 1 0 1 : Int (5)
Shift Left: 0 1 0 1 0 0 : Int (20)

if a = 5, b = 2, hence 5 << 2 and 5 * (2 ^ 2) Therefore 20.


**Binary Right Shift (Divides):**

a << b = a / (2^b)

Original : 0 1 0 1 0 0 : Int (20)
Shift Left: 0 0 0 0 1 0 1 : Int (5)

if a = 20, b = 2, hence 20 << 2 and 20 / (2 ^ 2) Therefore 5.

Sample Interview Questions:

1) Odd Even


```java
	if(x&1){
		System.out.println("Odd");
	}
	else{
		System.out.println("Even");
	}
``` 

2) Getting the iTH bit

Take a empty bit, and move the 1 to ith location, such that we can perform AND (&) to the `n` number.

Step 1:
Given: n=5, i=2
n in 0 0 0 1 0 1

Step 2:
Build a bit, such that it only has `1` in the ith Location using LeftShift.

mask=0 0 0 0 0 0 0
x<<i i.e. 0 0 0 1 0 0

Step 3: Perform AND with n and mask,
res = n & mask (this will multiply)

Java Interviews: DSA - Bit Manipulation