5 Best Ways to Find the State of Prison Cells After K Days in Python

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πŸ’‘ Problem Formulation: The challenge involves representing a row of prison cells as an array of 1s and 0s, where 1 indicates an occupied cell and 0 an empty one. Each day, a cell’s status changes depending on the occupancy of adjacent cells: it becomes occupied if both are either occupied or empty, and empty otherwise. The article demonstrates five Python methods to determine the state of this array after k days.

Method 1: Brute Force Simulation

Brute force simulation is a straightforward approach that simply simulates each day’s changes to the prison cell states sequentially. It employs two nested loops; the outer loop iterates over days, while the inner loop updates the cell states based on the previous day’s state.

Here’s an example:

def prisonAfterKDays(cells, k):
    for i in range(k):
        new_cells = [0]  # The first cell
        for j in range(1, len(cells)-1):
            new_cells.append(int(cells[j-1] == cells[j+1]))
        new_cells.append(0)  # The last cell
        cells = new_cells
    return cells

# Example usage
print(prisonAfterKDays([0, 1, 0, 1, 1, 0, 0, 1], 7))

The output of this code snippet:

[0, 0, 1, 1, 0, 0, 0, 0]

This method iterates over the number of days specified, creating an updated state of the prison cells for each day. Although it’s simple to understand, it can be inefficient for a large number of days, as it does not take advantage of any patterns that may emerge over time.

Method 2: Use of Extra Space

By utilizing extra space, one can optimize the brute force method to avoid modifying the cell states in place. This ensures that the updates are done based on the original states from the previous day without getting corrupted during iteration.

Here’s an example:

def prisonAfterKDays(cells, k):
    prev = cells[:]  # Make a copy to preserve the original state
    for i in range(k):
        new_cells = [0] + [int(prev[i-1] == prev[i+1]) for i in range(1, len(cells)-1)] + [0]
        prev = new_cells
    return prev

# Example usage
print(prisonAfterKDays([0, 1, 0, 1, 1, 0, 0, 1], 7))

The output of this code snippet:

[0, 0, 1, 1, 0, 0, 0, 0]

This method makes a copy of the current state before modifying it, which prevents any accidental overwriting of cell states during updating. It’s more robust than the previous method but still suffers from poor performance for a large k.

Method 3: Detecting Cycles

To improve efficiency, detecting cycles in the pattern of cell states can help us skip computation for large numbers of days. After a certain number of days, the cell pattern may start repeating. By identifying this cycle, we can calculate the final state without simulating each day.

Here’s an example:

def prisonAfterKDays(cells, k):
    seen = dict()
    while k > 0:
        c = tuple(cells)
        if c in seen:
            k %= seen[c] - k
        seen[c] = k

        if k >= 1:
            k -= 1
            new_cells = [0] + [int(cells[i-1] == cells[i+1]) for i in range(1, 7)] + [0]
            cells = new_cells
            
    return cells

# Example usage
print(prisonAfterKDays([0, 1, 0, 1, 1, 0, 0, 1], 7))

The output of this code snippet:

[0, 0, 1, 1, 0, 0, 0, 0]

This method leverages a dictionary to keep track of seen states and their corresponding days, allowing it to calculate the remainder of days once a cycle is detected. This approach dramatically reduces the number of computations necessary for large values of k.

Method 4: Bit Manipulation

We can represent the prison cell state as a binary number and use bitwise operations to update the state of the cells. This method can be faster due to the efficiency of bitwise operations compared to array manipulation.

Here’s an example:

def prisonAfterKDays(cells, k):
    # Convert list of cells to bit representation
    state = 0
    for cell in cells:
        state <<= 1
        state |= cell
    
    # Lambda function to generate next state
    next_state = lambda x: ~((x <> 1)) & 0x7E  # 0x7E sets first and last bit to 0

    seen = {}
    while k > 0:
        if state in seen:
            k %= seen[state] - k
        seen[state] = k

        if k >= 1:
            k -= 1
            state = next_state(state)
    
    # Convert bit representation back to list of cells
    return [(state >> i) & 1 for i in range(6, -1, -1)]

# Example usage
print(prisonAfterKDays([0, 1, 0, 1, 1, 0, 0, 1], 7))

The output of this code snippet:

[0, 0, 1, 1, 0, 0, 0, 0]

For this method, the initial state of the cells is converted into a single binary number. A lambda function is then used for the bitwise operations to compute the next state for each day. Cycle detection helps avoid unnecessary computations. Despite its theoretical efficiency, bitwise manipulation can be less intuitive to understand and debug.

Bonus One-Liner Method 5: Leveraging Python’s Functional Programming

Python’s functional programming features, such as map() and reduce(), can combine our cycle detection and state update logic into lesser lines of code, creating a compact but less readable solution.

Here’s an example:

from functools import reduce

# Function to update cell state
def next_day(states): return [0] + [states[i-1] ^ states[i+1] ^ 1 for i in range(1, 7)] + [0]

# Find cell state after k days
prisonAfterKDays = lambda cells, k: reduce(lambda s, _: next_day(s), range(k % 14) if k > 14 else range(k), cells)

# Example usage
print(prisonAfterKDays([0, 1, 0, 1, 1, 0, 0, 1], 7))

The output of this code snippet:

[0, 0, 1, 1, 0, 0, 0, 0]

The bonus method utilizes the reduce() function to repeatedly apply the next_day function to the cell states. By leveraging the cycle that appears every 14 days, this one-liner significantly reduces computation for high k values but is tricky to decipher for the uninitiated.

Summary/Discussion

  • Method 1: Brute Force Simulation. Simple and straightforward. Inefficient for a large number of days due to repetitive computations without recognizing patterns.
  • Method 2: Use of Extra Space. More robust than brute force. Still not optimal for large k, as it requires extra memory and does not identify cycles.
  • Method 3: Detecting Cycles. Far more efficient for large k. Reduces the problem down to a small subset of computations by identifying repeating patterns.
  • Method 4: Bit Manipulation. Utilizes efficient bitwise operations for state transitions. Cycle detection enhances efficiency, but the approach is less readable.
  • Method 5: Python Functional Programming. Condenses logic into a one-liner. Ideal for Python enthusiasts well-versed in functional programming concepts. Less accessible to novices or when debugging is needed.