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This approach involves simulating each day's changes directly. Given the constraints of the problem, this can be manageable for relatively small values of 'n'. We iterate over each day, applying the rules to determine the state of each cell. Importantly, the edge cells are always vacant as they lack two neighbors. By updating the state iteratively, we can reach the final configuration after 'n' days.
Time complexity: O(1), as the complexity per cycle of states is fixed. During each cycle, we just iterate over the 8 states.
Space complexity: O(1), only a fixed array is used for state computation.
1#include <vector>
2#include <iostream>
3
4std::vector<int> nextDay(const std::vector<int> &cells) {
5 std::vector<int> nextCells(8, 0);
6 for (int i = 1; i < 7; ++i) {
7 nextCells[i] = cells[i - 1] == cells[i + 1];
8 }
9 return nextCells;
10}
11
12std::vector<int> prisonAfterNDays(std::vector<int> cells, int n) {
13 n = n % 14 == 0 ? 14 : n % 14; // Handle periodicity
14 for (int i = 0; i < n; ++i) {
15 cells = nextDay(cells);
16 }
17 return cells;
18}
19
20int main() {
21 std::vector<int> cells = {0, 1, 0, 1, 1, 0, 0, 1};
22 int n = 7;
23 std::vector<int> result = prisonAfterNDays(cells, n);
24 for (int cell : result) {
25 std::cout << cell << " ";
26 }
27 return 0;
28}In the C++ solution, the use of vectors allows for easy manipulation of cell states. The function nextDay generates the next day's state based on current day rules. The outer function prisonAfterNDays manages the state transitions, applying the cyclic pattern to avoid redundant calculations when n is extremely large.
This alternative approach builds on recognizing patterns within the transformations. Given only 8 cells and binary states, the maximum number of possible unique states is 27 = 128. Through simulation, we can identify that the pattern cycles or repeats after at most 14 days, reducing the need to simulate all given 'n' days. By finding cycles, one can compute the state that would correspond to the final day directly, drastically minimizing operations.
Time complexity: O(128), as we explore states to identify cycles, bounded by maximum unique states of 128.
Space complexity: O(128), accommodates tracking up to 128 states.
using System.Collections.Generic;
public class PrisonCells {
public static string CellsToString(int[] cells) {
return string.Join("", cells);
}
public static int[] PrisonAfterNDays(int[] cells, int N) {
var seen = new Dictionary<string, int>();
int cycle = 0;
while (N > 0) {
string cellState = CellsToString(cells);
if (seen.ContainsKey(cellState)) {
N %= cycle - seen[cellState];
}
seen[cellState] = cycle;
if (N > 0) {
N--;
int[] next = new int[8];
for (int i = 1; i < 7; ++i) {
next[i] = cells[i - 1] == cells[i + 1] ? 1 : 0;
}
Array.Copy(next, cells, 8);
cycle++;
}
}
return cells;
}
public static void Main(string[] args) {
int[] cells = new int[] {0, 1, 0, 1, 1, 0, 0, 1};
int N = 7;
int[] result = PrisonAfterNDays(cells, N);
Console.WriteLine(string.Join(", ", result));
}
}The C# solution forms an efficient mechanism to recognize cycles through hashing previously seen states. Tanks to hash mapping, we rapidly conclude redundant iterations, reflecting upon cycle patterns.