Day 14: Docking Data

After traveling on the bus, you realize your laptop can no longer synchronize its data bus with the school's central server.

The communication protocol uses a data bus equipped with a 24-bit binary mask. Your puzzle input contains the raw instruction stream transmitted by the system.

Two types of instructions appear in the stream:

• mask = ... : defines the active bitmask (a 24-character string composed of 0, 1, and X).
• mem[address] = value : writes an integer value into a specific memory address.

Part 1

In version 1 of the protocol, the mask is applied directly to the value before writing it to memory:

• If the mask bit is 1, the corresponding bit of the value is overwritten with 1.
• If the mask bit is 0, the corresponding bit of the value is overwritten with 0.
• If the mask bit is X, the corresponding bit of the value is left unchanged.

Example

Consider this short instruction sequence:

mask = XXXXXXXXXXXXXXXXX1XXXX0X
mem[8] = 11
mem[7] = 101
mem[8] = 0

Execution trace:

• For mem[8] = 11 : 11 in binary is 1011. With the mask, bit position 6 becomes 1 and bit position 1 becomes 0, resulting in 73 written to address 8.
• For mem[7] = 101 : 101 becomes 101 written to address 7.
• For mem[8] = 0 : 0 becomes 64, overwriting the previous value at address 8.

At the end of this example, the sum of all values left in memory is 101 + 64 = 165.

Execute your full input stream. What is the sum of all values left in memory after the initialization program completes?