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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.
1def next_day(cells):
2 return [0] + [cells[i - 1] == cells[i + 1] for i in range(1, 7)] + [0]
3
4def prison_after_n_days(cells, N):
5 N = N % 14 if N % 14 != 0 else 14 # Cycle found of length 14
6 for _ in range(N):
7 cells = next_day(cells)
8 return cells
9
10# Example usage
11cells = [0, 1, 0, 1, 1, 0, 0, 1]
12N = 7
13result = prison_after_n_days(cells, N)
14print(result)The Python example demonstrates a concise approach using list comprehensions for daily state changes, with next_day computing the next configuration of cells. The state transition is optimized by observing cyclic behavior, ensuring that it only processes the minimum necessary iterations.
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.
#include <unordered_map>
#include <string>
#include <iostream>
std::string cellsToString(const std::vector<int> &cells) {
std::string result;
for (int cell : cells) {
result += std::to_string(cell);
}
return result;
}
std::vector<int> prisonAfterNDays(std::vector<int> cells, int n) {
std::unordered_map<std::string, int> seen;
int cycle = 0;
while (n > 0) {
std::string cellsStr = cellsToString(cells);
if (seen.find(cellsStr) != seen.end()) {
n %= cycle - seen[cellsStr];
}
seen[cellsStr] = cycle;
if (n > 0) {
--n;
std::vector<int> next(8, 0);
for (int i = 1; i < 7; ++i) {
next[i] = cells[i - 1] == cells[i + 1];
}
cells = next;
++cycle;
}
}
return cells;
}
int main() {
std::vector<int> cells = {0, 1, 0, 1, 1, 0, 0, 1};
int n = 7;
std::vector<int> result = prisonAfterNDays(cells, n);
for (int cell : result) {
std::cout << cell << " ";
}
return 0;
}Leveraging an unordered map in C++, this solution effectively identifies repeating configurations by mapping cell states to cycle iterations. The cycle is used to reduce the number of effective transformations applied.