Store Two Numbers in One Byte Using Bit Manipulation in Java

Java's bit manipulation allows us to store two numbers in a single byte while streamlining our code. The technique of changing individual bits in a binary data representation is called bit manipulation. In this situation, bit manipulation can be used to condense two numbers into a single byte.

A Java byte is an 8-bit primitive data type, allowing it to store values between -128 and 127. However, we can store two smaller values in this range using just a single bit from a byte.

Example:

Input: x=3 (00000011), y=6 (00000110)

Output: 54 Binary representation: 00110110

Explanation:

We will combine the two numbers x and y using the bitwise OR operator 00000111 | 00000110 = 00000111, which is 7 in decimal. Now we can cast this value to a byte to get output as 00110110, 54 in decimal. Finally, we stored the values of x and y are stored in a single byte.

Approach 1: Using Bitwise OR (|) operator

We can use the bitwise OR operator to combine two numbers into one byte. Each bit in the byte is set to 1 if each of the corresponding bits in the operand is one due to the logical OR operation of this operator on the two bits of this operand.

ALGORITHM:

Step 1: Construct binary representations of two integer variables, x and y.

Step 2: Add the two x and y positions using the bitwise OR operator.

Step 3: Cast to bytes and truncate the result to 8 bits.

Step 4: Print the binary representation of the byte value.

Implementation:

Here is the implementation of the above steps

FileName: ByteStream.java

Output:

54 Binary representation: 00110110 

Complexity Analysis:

Time complexity: Since this algorithm is independent of input size, its time complexity is constant or O(1).

Space complexity: Because this algorithm uses only a few variables, regardless of the input size, its space complexity is also constant, or O(1).

Approach 2: Using bit masking

ALGORITHM:

Step 1: Give the integer names x and y for the two numbers to be added.

Step 2: Create a mask by storing 1 for the bits used to store the first number and 0 for the bits used to store the second number.

Step 3: To move the first number to the left based on the number of bits needed to store the second number, you can add the two numbers using bitwise OR.

Step 4: Dump the result to bytes and then truncate to 8 bits.

Step 5: Bit masking separates two numbers from the composite result. Use the bitwise AND operator with the mask to remove the first number, and use the bitwise AND operator after the mask to remove the second number.

Step 6: Convert the selected number to bytes to get the result.

Step 7: Print the results.

Implementation:

Here is the implementation of the above steps

FileName: Bytestream.java

Output:

54 Binary representation: 00110110 3 Binary representation: 00000011 6 Binary representation: 00000110 

Complexity Analysis:

Time complexity: Since this algorithm is independent of input size, its time complexity is constant or O(1).

Space complexity: Because this algorithm uses only a few variables, regardless of the input size, its space complexity is also constant, or O(1).