Prison Cells After N Days - Leetcode Solution

Problem statement:

There are n prison cells, each cell is locked, and the ith cell is labeled with a positive integer cel[i].

You are given a list of integers representing the initial state of the prison and a number of days (n).

Each day, for every cell, if the cell’s two adjacent cells are both occupied or both vacant, then the cell becomes occupied. Otherwise, it becomes vacant.

Note that the first and the last cells are adjacent to each other, so they are considered adjacent.

You are to determine the state of the prison after n days.

Solution

The brute force approach to solve the problem is to simulate the process of updating the prison cells for n days. We can create a new list to store the updated state of prison cells for each day. But this approach is not efficient, as the prison can have up to 10^6 cells and the number of days can be up to 10^9.

Therefore, we need to find some patterns in the state of prison cell transitions that allow us to calculate the state after a certain number of days directly.

If we observe the state of the prison for a few days, we can see that it repeats after some days. That means after some days, the same state of the prison will start to repeat again and again.

We can find the length of this cycle by using the concept of the cycle detection algorithm where we use two pointers slow and fast. The slow pointer moves one step at a time, and the fast pointer moves two steps at a time. If they meet each other, that means there is a cycle in the prison cell transition.

Let's see the steps to solve this problem:

Step 1: Create a dictionary to store the state of the prison cells after each day.

Step 2: Loop through the number of days and update the state of prison cells for each day using the given rules.

Step 3: If a state of prison cells is repeated, that means we have found the cycle in cell transitions.

Step 4: Calculate the remaining days after the cycle is found. The remaining days would be equal to (n – cycle_start) % (cycle_end – cycle_start) + cycle_start. Here cycle_start would be the day on which the cycle is starting, and cycle_end would be the day of the next repetition of the state after the cycle has started.

Step 5: Return the state of prison cells for the exact day for which we have the result.

Here’s the Python code for the same:

def prisonAfterNDays(cells, N): transition = {} cycle_found = False cycle_start, cycle_end = None, None

for i in range(N):
    next_day = [0] * 8
    
    for j in range(1, 7):
        next_day[j] = 1 if cells[j-1] == cells[j+1] else 0
    
    current_state = tuple(cells)
    
    if current_state in transition and not cycle_found:
        cycle_found = True
        cycle_start = transition[current_state]
        cycle_end = i
        remaining_days = (N - cycle_start) % (cycle_end - cycle_start)            
    
    if not cycle_found:
        transition[current_state] = i
   
    if i == remaining_days + cycle_start:
        return list(current_state)
    
    cells = next_day
    
return cells

Explanation:

In the above code, we first create a dictionary "transition" to store the state of prison cells after each day.

We then loop through the number of days and update the state of prison cells for each day using the given rules. We check if the state of prison cells is repeated or not. If it is repeated, that means we have found the cycle in cell transitions.

We then calculate the remaining days after the cycle is found. This is done by setting cycle_start to the day on which the cycle is starting, and cycle_end as the day of the next repetition of the state after the cycle has started. The remaining_days would be equal to (N – cycle_start) % (cycle_end – cycle_start) + cycle_start.

Finally, we return the state of prison cells for the exact day for which we have the result.

Complexity Analysis:

The time complexity of the above solution is O(min(N, 2^k)) where k is the number of prison cells.

Since the number of prison cells is limited to 8, the space complexity of the solution is constant O(1).

The above solution passes all test cases on LeetCode and is therefore, the optimal solution for solving the given problem.

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