added Prison Cells After N Days medium

Author: minhyeong-joe

Medium/PrisonCellsAfterNDays/README.md

@@ -0,0 +1,44 @@
+
+# Prison Cells After N Days
+[Leetcode Link](https://leetcode.com/problems/prison-cells-after-n-days/)
+
+## Problem:
+
+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.)
+
+We describe the current state of the prison in the following way: `cells[i] == 1` if the `i`-th cell is occupied, else `cells[i] == 0`.
+
+Given the initial state of the prison, return the state of the prison after `N` days (and `N` such changes described above.)
+
+## Example:
+
+```
+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]
+```
+```
+Input: cells = [1,0,0,1,0,0,1,0], N = 1000000000
+Output: [0,0,1,1,1,1,1,0]
+```
+
+## Note:
+
+1. `cells.length == 8`
+2. `cells[i]` is in `{0, 1}`
+3. `1 <= N <= 10^9`

Medium/PrisonCellsAfterNDays/solution.py

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+from typing import List
+
+
+def prisonAfterNDays(cells: List[int], N: int) -> List[int]:
+ # the pattern repeats every 15 days
+ N = (N-1) % 14 + 1
+ prev = cells
+ current = list()
+ for day in range(N):
+ for i in range(len(prev)):
+ # first and last cell is always vacant because it does not have two adjacent neighbors
+ if i == 0 or i == len(prev)-1:
+ current.append(0)
+ # insert 0 or 1 based on neighboring cells
+ else:
+ isOccupied = 1 if not(prev[i-1] ^ prev[i+1]) else 0
+ current.append(isOccupied)
+ # after each day, set current to be prev, and new list as current
+ prev = current
+ # print("Day", day+1, ":", current)
+ current = list()
+ return prev
+
+
+# test driver
+cells = [0, 1, 0, 1, 1, 0, 0, 1]
+N = 7
+print("Input: cells =", cells, ", N =", N)
+print("Output:", prisonAfterNDays(cells, N))
+print()
+
+cells = [1, 0, 0, 1, 0, 0, 1, 0]
+N = 1000000000
+print("Input: cells =", cells, ", N =", N)
+print("Output:", prisonAfterNDays(cells, N))
+print()

README.md

@@ -45,6 +45,7 @@ Languages used: Java and Python
  - [Search in Rotated Sorted Array](Medium/SearchInRotatedSortedArray)
  - [Car Pooling](Medium/CarPooling)
  - [Palindromic Substrings](Medium/PalindromicSubstrings)
+ - [Prison Cells After N Days](Medium/PrisonCellsAfterNDays)
 - Hard
 
 ---