LeetCode 957 - Prison Cells After N Days

The problem presents a simulation scenario with a row of exactly 8 prison cells, where each cell is either occupied (1) or vacant (0). Each day, the state of a cell changes based on the states of its immediate neighbors.

leetcodemediumarrayhash-tablemathbit-manipulation

LeetCode Problem 957

Difficulty: 🟡 Medium
Topics: Array, Hash Table, Math, Bit Manipulation

Solution

Problem Understanding

The problem presents a simulation scenario with a row of exactly 8 prison cells, where each cell is either occupied (1) or vacant (0). Each day, the state of a cell changes based on the states of its immediate neighbors. Specifically, a cell becomes occupied if both of its neighbors are in the same state (both occupied or both vacant) and becomes vacant otherwise. The first and last cells in the row only have one neighbor and thus cannot satisfy the condition for two neighbors, so their updates are always based on the rule for neighbors (in practice, they always become vacant after the first day if the standard rules are followed). The input is a list of 8 integers representing the initial state of the cells and an integer n representing the number of days to simulate. The expected output is the state of the cells after n days.

The constraints are small in terms of the number of cells (always 8), but n can be very large, up to $10^9$. This implies that a brute-force simulation for each day would be too slow for large n. The key insight is that with a finite number of states (2^8 = 256 possible configurations), the system must eventually enter a cycle, which allows us to compute the final state efficiently without simulating all n days.

Important edge cases include the smallest n = 1, the largest n = 10^9, configurations where all cells are initially the same (all 0s or all 1s), and alternating patterns (e.g., [0,1,0,1,0,1,0,1]) that may reveal cycles immediately.

Approaches

The brute-force approach simulates each day in sequence. For each day, a new list of cell states is created by applying the neighbor rules for every cell. This method is simple and guarantees correctness because it directly implements the problem statement. However, with n up to $10^9$, iterating day by day is infeasible, as it would require far too many operations.

The optimal approach leverages the fact that there are only 256 possible states for the 8 cells. By storing previously seen states in a hash map along with the day they occurred, we can detect cycles. Once a cycle is detected, we can compute the remainder of n modulo the cycle length and simulate only the remaining steps, drastically reducing the number of iterations.

Approach Time Complexity Space Complexity Notes
Brute Force O(n * 8) = O(n) O(8) = O(1) Simulates each day sequentially; feasible for small n but not for large n.
Optimal O(cycle_length) ≤ O(256) O(256) Detects repeating states to avoid full simulation; uses a hash map to store previous states.

Algorithm Walkthrough

  1. Initialize a hash map to store previously seen states, mapping each state to the day it occurred. Also, initialize a variable to keep track of the current day.
  2. Convert the initial list of cells to a tuple so it can be used as a hashable key in the map.
  3. Iterate while n > 0. On each iteration, check if the current state has been seen before. If it has, calculate the cycle length as the difference between the current day and the day it was first seen, and reduce n modulo the cycle length.
  4. For each day, compute the next state by applying the rule: each cell (except the first and last) becomes occupied if its neighbors are both the same; otherwise, it becomes vacant. Set the first and last cells to 0 because they only have one neighbor.
  5. Update the current state and decrement n by 1. Store the current state in the hash map with the current day.
  6. Repeat until n reaches 0. Return the current state as a list.

Why it works: Since there are only 256 possible states, a cycle is guaranteed. By detecting the cycle, the algorithm reduces redundant simulation. Applying the neighbor rules ensures that the transition between states is consistent with the problem's rules, and handling the first and last cells separately prevents boundary errors.

Python Solution

from typing import List

class Solution:
    def prisonAfterNDays(self, cells: List[int], n: int) -> List[int]:
        seen = {}
        current = tuple(cells)
        
        while n > 0:
            if current in seen:
                cycle_length = seen[current] - n
                n %= cycle_length
            seen[current] = n
            
            if n >= 1:
                next_state = [0] * 8
                for i in range(1, 7):
                    next_state[i] = 1 if current[i-1] == current[i+1] else 0
                current = tuple(next_state)
                n -= 1
        
        return list(current)

This implementation first converts the list to a tuple for hashing. The seen dictionary stores states and the corresponding remaining days. If a cycle is detected, n is reduced modulo the cycle length, which avoids unnecessary computation. The next state is calculated using the neighbor rule, while the first and last cells are set to 0, as they have only one neighbor.

Go Solution

func prisonAfterNDays(cells []int, n int) []int {
    seen := make(map[string]int)
    current := make([]int, 8)
    copy(current, cells)
    
    for n > 0 {
        key := fmt.Sprint(current)
        if val, ok := seen[key]; ok {
            cycle := val - n
            n %= cycle
        }
        seen[key] = n
        
        if n >= 1 {
            nextState := make([]int, 8)
            for i := 1; i < 7; i++ {
                if current[i-1] == current[i+1] {
                    nextState[i] = 1
                } else {
                    nextState[i] = 0
                }
            }
            current = nextState
            n--
        }
    }
    
    return current
}

In Go, slices are used for the current state, and fmt.Sprint converts the slice to a string for hash map keys. Copying the slice avoids modifying the input. The handling of cycles and next state computation is analogous to Python, with explicit slice creation for the next state.

Worked Examples

Example 1: cells = [0,1,0,1,1,0,0,1], n = 7

Day State
0 [0,1,0,1,1,0,0,1]
1 [0,1,1,0,0,0,0,0]
2 [0,0,0,0,1,1,1,0]
3 [0,1,1,0,0,1,0,0]
4 [0,0,0,0,0,1,0,0]
5 [0,1,1,1,0,1,0,0]
6 [0,0,1,0,1,1,0,0]
7 [0,0,1,1,0,0,0,0]

Example 2: cells = [1,0,0,1,0,0,1,0], n = 1000000000

Cycle is detected after several iterations; final state after modulo reduction is [0,0,1,1,1,1,1,0].

Complexity Analysis

Measure Complexity Explanation
Time O(256) = O(1) Only up to 256 unique states exist, so cycle detection limits iterations.
Space O(256) = O(1) Hash map stores up to 256 states for cycle detection.

Since there are only 8 cells, the total number of states is small and fixed, so both time and space are effectively constant.

Test Cases

# Provided examples
assert Solution().prisonAfterNDays([0,1,0,1,1,0,0,1], 7) == [0,0,1,1,0,0,0,0]  # Example 1
assert Solution().prisonAfterNDays([1,0,0,1,0,0,1,0], 1000000000) == [0,0,1,1,1,1,1,0]  # Example 2

# Edge cases
assert Solution().prisonAfterNDays([0,0,0,0,0,0,0,0], 1) == [0,0,0,0,0,0,0,0]  # All zeros
assert Solution().prisonAfterNDays([1,1,1,1,1,1,1,1], 1) == [0,1,1,1,1,1,1,0]  # All ones
assert Solution().prisonAfterNDays([0,1,0,1,0,1,0,1], 10) == [0,0,1,1,1,1,0,0]  # Alternating pattern
Test Why