{x}
blog image

Prison Cells After N Days

There are 8 prison cells in a row and each cell is either occupied or vacant.

Each day, whether the cell is occupied or vacant changes according to the following rules:

  • If a cell has two adjacent neighbors that are both occupied or both vacant, then the cell becomes occupied.
  • Otherwise, it becomes vacant.

Note that because the prison is a row, the first and the last cells in the row can't have two adjacent neighbors.

You are given an integer array cells where cells[i] == 1 if the ith cell is occupied and cells[i] == 0 if the ith cell is vacant, and you are given an integer n.

Return the state of the prison after n days (i.e., n such changes described above).

 

Example 1:

Input: cells = [0,1,0,1,1,0,0,1], n = 7
Output: [0,0,1,1,0,0,0,0]
Explanation: The following table summarizes the state of the prison on each day:
Day 0: [0, 1, 0, 1, 1, 0, 0, 1]
Day 1: [0, 1, 1, 0, 0, 0, 0, 0]
Day 2: [0, 0, 0, 0, 1, 1, 1, 0]
Day 3: [0, 1, 1, 0, 0, 1, 0, 0]
Day 4: [0, 0, 0, 0, 0, 1, 0, 0]
Day 5: [0, 1, 1, 1, 0, 1, 0, 0]
Day 6: [0, 0, 1, 0, 1, 1, 0, 0]
Day 7: [0, 0, 1, 1, 0, 0, 0, 0]

Example 2:

Input: cells = [1,0,0,1,0,0,1,0], n = 1000000000
Output: [0,0,1,1,1,1,1,0]

 

Constraints:

  • cells.length == 8
  • cells[i] is either 0 or 1.
  • 1 <= n <= 109

Solution Explanation and Code for Prison Cells After N Days

This problem involves simulating the daily changes in the state of prison cells based on the given rules. A naive approach would be to simulate each day for n days, but this would be extremely inefficient for large values of n. The key observation is that the cell states eventually enter a cycle. We can leverage this to optimize the solution.

Approach:

  1. State Representation: We can represent the state of the 8 cells using an integer or a bitmask. This allows for efficient state transitions and comparison.

  2. Cycle Detection: We maintain a hash map (dictionary in Python) to store previously seen cell states and their corresponding day numbers. When we encounter a previously seen state, we know we've entered a cycle.

  3. Cycle Calculation: Once a cycle is detected, we can calculate the remaining days by taking the modulo of the remaining days with the cycle length. This avoids unnecessary simulations.

  4. State Transition: The core logic involves calculating the next day's state based on the current state using the given rules. This can be implemented using bit manipulation or array operations.

Time Complexity Analysis:

  • The time complexity is dominated by the number of days before a cycle is detected. In the worst case, before a cycle is detected, we might iterate through at most all possible states, which is 28 = 256. After cycle detection, the modulo operation takes constant time. Therefore, the overall time complexity is O(min(n, 256)), where 256 represents the upper bound on the number of states before the cycle repeats. This is practically constant time for all reasonable inputs.

  • The space complexity is O(min(n, 256)) because we need to store the seen states in the hash map, which will be at most 256 if no cycle is detected before then.

Code Implementation (Python):

def prisonAfterNDays(cells, n):
    """
    Calculates the state of prison cells after n days.
 
    Args:
        cells: A list of integers representing the initial state of cells (0 or 1).
        n: The number of days.
 
    Returns:
        A list of integers representing the state of cells after n days.
    """
    seen = {}  # Hash map to store seen states
    n = min(n, 100000) #We limit n to a large number which will surely contain the pattern
 
    for day in range(n):
        state = tuple(cells)
        if state in seen:
            first_seen = seen[state]
            remaining_days = (day - first_seen)
            cycle_len = day - first_seen
            days_to_simulate = (n - day) % cycle_len
            for _ in range(days_to_simulate):
                cells = next_day(cells)
            return cells
 
        seen[state] = day
        cells = next_day(cells)
    return cells
 
def next_day(cells):
    """Calculates the next day's cell state."""
    new_cells = [0] * 8
    for i in range(1, 7):
        new_cells[i] = 1 if cells[i - 1] == cells[i + 1] else 0
    return new_cells
 
#Example Usage:
cells = [0,1,0,1,1,0,0,1]
n = 7
result = prisonAfterNDays(cells,n)
print(f"Prison cells after {n} days: {result}")
 
cells = [1,0,0,1,0,0,1,0]
n = 1000000000
result = prisonAfterNDays(cells,n)
print(f"Prison cells after {n} days: {result}")

Code Implementation (Java):

import java.util.HashMap;
import java.util.Map;
 
class Solution {
    public int[] prisonAfterNDays(int[] cells, int n) {
        Map<Integer, Integer> seen = new HashMap<>();
        n = Math.min(n, 100000); // Limit n to a large number to handle cycles
 
        int state = convertToInt(cells);
        for (int day = 0; day < n; day++) {
            if (seen.containsKey(state)) {
                int firstSeen = seen.get(state);
                int cycleLength = day - firstSeen;
                int remainingDays = (n - day) % cycleLength;
                for (int i = 0; i < remainingDays; i++) {
                    cells = nextDay(cells);
                    state = convertToInt(cells);
                }
                return cells;
            }
            seen.put(state, day);
            cells = nextDay(cells);
            state = convertToInt(cells);
        }
        return cells;
    }
 
    private int[] nextDay(int[] cells) {
        int[] nextCells = new int[8];
        for (int i = 1; i < 7; i++) {
            nextCells[i] = (cells[i - 1] == cells[i + 1]) ? 1 : 0;
        }
        return nextCells;
    }
 
    private int convertToInt(int[] cells) {
        int num = 0;
        for (int cell : cells) {
            num = (num << 1) | cell;
        }
        return num;
    }
}

Similar implementations can be done in C++, Go, or any other language, using the same core logic and data structures. The choice of using bit manipulation or array operations for the state transition is mainly a matter of style and potential minor performance differences. The cycle detection optimization is crucial for handling large n values efficiently.