957. Prison Cells After N Days LeetCode Solution

In this guide, you will get 957. Prison Cells After N Days LeetCode Solution with the best time and space complexity. The solution to Prison Cells After N Days problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Table of Contents

  1. Problem Statement
  2. Complexity Analysis
  3. Prison Cells After N Days solution in C++
  4. Prison Cells After N Days solution in Java
  5. Prison Cells After N Days solution in Python
  6. Additional Resources
957. Prison Cells After N Days LeetCode Solution image

Problem Statement of Prison Cells After N Days

There are 8 prison cells in a row and each cell is either occupied or vacant.
Each day, whether the cell is occupied or vacant changes according to the following rules:

If a cell has two adjacent neighbors that are both occupied or both vacant, then the cell becomes occupied.
Otherwise, it becomes vacant.

Note that because the prison is a row, the first and the last cells in the row can’t have two adjacent neighbors.
You are given an integer array cells where cells[i] == 1 if the ith cell is occupied and cells[i] == 0 if the ith cell is vacant, and you are given an integer n.
Return the state of the prison after n days (i.e., n such changes described above).

Example 1:

Input: cells = [0,1,0,1,1,0,0,1], n = 7
Output: [0,0,1,1,0,0,0,0]
Explanation: The following table summarizes the state of the prison on each day:
Day 0: [0, 1, 0, 1, 1, 0, 0, 1]
Day 1: [0, 1, 1, 0, 0, 0, 0, 0]
Day 2: [0, 0, 0, 0, 1, 1, 1, 0]
Day 3: [0, 1, 1, 0, 0, 1, 0, 0]
Day 4: [0, 0, 0, 0, 0, 1, 0, 0]
Day 5: [0, 1, 1, 1, 0, 1, 0, 0]
Day 6: [0, 0, 1, 0, 1, 1, 0, 0]
Day 7: [0, 0, 1, 1, 0, 0, 0, 0]

See also  95. Unique Binary Search Trees II LeetCode Solution

Example 2:

Input: cells = [1,0,0,1,0,0,1,0], n = 1000000000
Output: [0,0,1,1,1,1,1,0]

Constraints:

cells.length == 8
cells[i] is either 0 or 1.
1 <= n <= 109

Complexity Analysis

  • Time Complexity:
  • Space Complexity: O(1)

957. Prison Cells After N Days LeetCode Solution in C++

class Solution {
 public:
  vector<int> prisonAfterNDays(vector<int>& cells, int n) {
    vector<int> firstDayCells;
    vector<int> nextDayCells(cells.size());

    for (int day = 0; n-- > 0; cells = nextDayCells, ++day) {
      for (int i = 1; i + 1 < cells.size(); ++i)
        nextDayCells[i] = cells[i - 1] == cells[i + 1];
      if (day == 0)
        firstDayCells = nextDayCells;
      else if (nextDayCells == firstDayCells)
        n %= day;
    }

    return cells;
  }
};
/* code provided by PROGIEZ */

957. Prison Cells After N Days LeetCode Solution in Java

class Solution {
  public int[] prisonAfterNDays(int[] cells, int n) {
    int[] firstDayCells = new int[cells.length];
    int[] nextDayCells = new int[cells.length];

    for (int day = 0; n-- > 0; cells = nextDayCells.clone(), ++day) {
      for (int i = 1; i + 1 < cells.length; ++i)
        nextDayCells[i] = cells[i - 1] == cells[i + 1] ? 1 : 0;
      if (day == 0)
        firstDayCells = nextDayCells.clone();
      else if (Arrays.equals(nextDayCells, firstDayCells))
        n %= day;
    }

    return cells;
  }
}
// code provided by PROGIEZ

957. Prison Cells After N Days LeetCode Solution in Python

class Solution:
  def prisonAfterNDays(self, cells: list[int], n: int) -> list[int]:
    nextDayCells = [0] * len(cells)
    day = 0

    while n > 0:
      n -= 1
      for i in range(1, len(cells) - 1):
        nextDayCells[i] = 1 if cells[i - 1] == cells[i + 1] else 0
      if day == 0:
        firstDayCells = nextDayCells.copy()
      elif nextDayCells == firstDayCells:
        n %= day
      cells = nextDayCells.copy()
      day += 1

    return cells
 # code by PROGIEZ

Additional Resources

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