Find Next State of Cell Matrix in Python

Suppose we have a 2D binary matrix where a 1 means a live cell and a 0 means a dead cell. A cell's neigrs are its immediate horizontal, vertical and diagonal cells. We have to find the next state of the matrix using these rules

• Any living cell with two or three living neigrs lives.

• Any dead cell with three living neigrs becomes a live cell.

• All other cells die.

So, if the input is like

then the output will be

To solve this, we will follow these steps:

• n := row size of matrix, m := column size of matrix

• res := a matrix of size n x m, and fill with 0

• for i in range 0 to n, do

  • for j in range 0 to m, do

    • s := 0

    • if matrix[i, j] is same as 0, then

      • for k in range i - 1 to i + 1, do

        • or h in range j - 1 to j + 1, do

          • if 0 <= k < n and 0 <= h < m, then

            • s := s + matrix[k, h]

      • res[i, j] := [0, 1, true when s is same as 3]

    • otherwise,

      • for k in range i - 1 to i + 1, do

        • for h in range j - 1 to j + 1, do

          • if 0 <= k < n and 0 <= h < m, then

            • s := s + matrix[k, h]

      • if s is either 3 or 4, then

        • res[i, j] := 1

• return res

Let us see the following implementation to get better understanding:

Example

 Live Demo

class Solution:
   def solve(self, matrix):
      n, m = len(matrix), len(matrix[0])
      res = [[0 for j in range(m)] for i in range(n)]
      for i in range(n):
         for j in range(m):
            s = 0
            if matrix[i][j] == 0:
               for k in range(i - 1, i + 2):
                  for h in range(j - 1, j + 2):
                     if 0 <= k < n and 0 <= h < m:
                        s += matrix[k][h]
               res[i][j] = [0, 1][s == 3]
            else:
               for k in range(i - 1, i + 2):
                  for h in range(j - 1, j + 2):
                     if 0 <= k < n and 0 <= h < m:
                        s += matrix[k][h]
               if s in [3, 4]:
                  res[i][j] = 1
      return res

ob = Solution()
matrix = [
   [1, 1, 0, 0],
   [0, 1, 0, 0],
   [0, 1, 0, 1],
   [1, 1, 0, 1]
]

print(ob.solve(matrix))

Input

[[1, 1, 0, 0],
 [0, 1, 0, 0],
 [0, 1, 0, 1],
 [1, 1, 0, 1] ]

Output

[[1, 1, 0, 0],
 [0, 1, 0, 0],
 [0, 1, 0, 0],
 [1, 1, 0, 0]]