How do you solve Prison Cells After N Days on LeetCode?

Prison Cells After N Days (LeetCode #957) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: The optimal solution leverages the fact that the prison cell states repeat every 14 days. By using this observation, we can reduce the number of iterations significantly, especially for large n.

What is the brute force solution for Prison Cells After N Days?

The brute force approach for Prison Cells After N Days is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach simulates each day one by one, applying the rules to determine the state of each cell. This is straightforward but inefficient for large values of n. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Prison Cells After N Days?

Prison Cells After N Days uses the following concepts: Array, Hash Table, Math, Bit Manipulation. The recommended patterns to study are: .

#957

Prison Cells After N Days

Medium

ArrayHash TableMathBit Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution leverages the fact that the prison cell states repeat every 14 days. By using this observation, we can reduce the number of iterations significantly, especially for large n.

⚙️

Algorithm

3 steps

1Step 1: Use a loop to simulate the state changes until either n is reduced to 0 or we find a repeating pattern.
2Step 2: Store the states in a dictionary to detect cycles.
3Step 3: If a cycle is found, calculate the effective day using n % cycle_length.

solution.py14 lines

1def prisonAfterNDays(cells, n):
2    seen = {}
3    while n > 0:
4        state = tuple(cells)
5        if state in seen:
6            cycle_length = seen[state] - n
7            n %= cycle_length
8        seen[state] = n
9        new_cells = [0] * 8
10        for i in range(1, 7):
11            new_cells[i] = 1 if cells[i-1] == cells[i+1] else 0
12        cells = new_cells
13        n -= 1
14    return cells

ℹ

Complexity note: The time complexity is O(n) in the worst case, but due to the cycle detection, it can be significantly less for large n. The space complexity is O(n) due to the storage of seen states.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.