Program to find next state of next cell matrix state in Python?

PythonServer Side ProgrammingProgramming

Suppose we have a 2D binary matrix where a 1 means a live cell and a 0 means a dead cell. A cell's neighbors 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 neighbors lives.

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

  • All other cells die.

So, if the input is like

1100
0100
0101
1101

then the output will be

1100
0100
0100
1100

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]]
raja
Published on 10-Nov-2020 08:54:17
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