Prison Cells After N Days

class Solution {
    public int[] executePrisonDays(int[] cells, int days) {
    
        HashMap<Integer, Integer> seenStates = new HashMap<>();
        boolean accelerated = false;
    
        // Convert cells to single integer bitwise representation
        int bitCells = 0x0;
        for (int cell : cells) {
            bitCells <<= 1;
            bitCells |= cell;
        }
    
        while (days > 0) {
            if (!accelerated) {
                if (seenStates.containsKey(bitCells)) {
                    days %= seenStates.get(bitCells) - days;
                    accelerated = true;
                } else {
                    seenStates.put(bitCells, days);
                }
            }
    
            if (days > 0) {
                days--;
                bitCells = nextDayStates(bitCells);
            }
        }
    
        // Reconstruct the cell array from the bit representation
        int[] result = new int[cells.length];
        for (int i = cells.length - 1; i >= 0; i--) {
            result[i] = (bitCells & 0x1);
            bitCells >>= 1;
        }
        return result;
    }
    
    private int nextDayStates(int currentState) {
        currentState = ~((currentState << 1) ^ (currentState >> 1));
        // Ensure the first and last bits are 0
        currentState &= 0x7e;
        return currentState;
    }
}

In this Java solution for the "Prison Cells After N Days" problem, follow these steps to understand the mechanics of the implementation:

• Start by converting the initial state of cells into a single integer using bitwise representation for efficiency.
• Use a HashMap to track states that have already been seen. This is essential for detecting cycles and accelerating the computation process.
• Loop through the number of days, adjusting the cycle if a previous state is detected by using modulo operation. This reduces unnecessary iterations and hastens the result.
• Each day, update the state of the cells by computing the next state, ensuring that the first and last cells are always inactive (0), as they do not have two adjacent neighbours. This is achieved using bitwise operations.
• After completing the loop for the required days, or identifying a repeat in cycle which ends the days' loop early, reconstruct the state of prison cells from the computed integer state.

Here are details regarding crucial parts of the implementation:

• seenStates HashMap: This stores each unique cell configuration and the day it first appeared. It's used to detect cycles in cell states over days.
• Bitwise operations:
  • Cells are converted into an integer (bitCells) by shifting bits left and setting the least significant bit by the value of the cell.
  • For finding the next day's state, both left and right bitwise shifts are used to represent neighbour checks, then perform a bitwise XOR and NOT to effectively apply the rule that a cell becomes inactive unless both its neighbors are in the same state.
  • The first and last bit are cleared as they have no two adjacent neighbours.
• Reconstruction of the output array uses bitwise right shift and bit masking to extract the state of individual cells from the integer representation.

This approach efficiently handles large numbers of days by skipping over repeated cyclic states, therefore optimizing the simulation beyond the brute-force method.

class Solution:
    def processPrison(self, cells: List[int], days: int) -> List[int]:
    
        state_history = dict()
        optimized = False
    
        # Convert cells list to a bitmask
        state = 0x0
        for cell in cells:
            state <<= 1
            state |= cell
    
        # Simulation with caching states
        while days > 0:
            if not optimized:
                if state in state_history:
                    days %= state_history[state] - days
                    optimized = True
                else:
                    state_history[state] = days
            if days > 0:
                days -= 1
                state = self.transform(state)
    
        # Decode bitmask to cells list
        output_cells = []
        for _ in cells:
            output_cells.append(state & 0x1)
            state >>= 1
    
        return reversed(output_cells)
    
    def transform(self, state: int) -> int:
        state = ~ (state << 1) ^ (state >> 1)
        state &= 0x7e  # Clear first and last bit
        return state

The given Python solution simulates the behavior of prison cells over a number of days, where each cell has two possible states: occupied or vacant. The transformation rules for each cell's state depend on its neighboring cells.

Here's a concise description of the approach and what each part accomplishes:

• A dictionary state_history caches previously seen configurations (states) to avoid re-computation by detecting cycles.
• The cells configuration is initially converted into a bitmask (state) to efficiently compute future states and check cyclic occurrences.
• The main simulation loop iterates over the number of days or until a cycle is detected:
  • When a previously seen state is encountered, the number of remaining days is optimized using modular arithmetic to jump forward in the simulation, effectively reducing the processing time.
  • If not cycling through, the state is updated using the transform method, which computes the next day's state based on the current state.
• The transform function applies the next state's rule by using bitwise operations:
  • The next state is calculated by considering each bit's neighbors using bitwise shifts and negations.
  • Bits on the boundaries (the first and last bit) are always zero, handled by masking with 0x7e.
• Finally, the last computed state is decoded back into the standard list form and reversed before returning, to match the original left-to-right order of the input.

The efficient use of bitwise operations and cycle detection via modular arithmetic ensures the solution runs much faster than a naive day-by-day computation approach when the number of days is large.