|
| 1 | +""" |
| 2 | +Game of Life |
| 3 | +Given a matrix (array of arrays) representing the current state in Conway's Game of Life, return the next state of the matrix using these rules: |
| 4 | +
|
| 5 | +Each cell is either 1 (alive) or 0 (dead). |
| 6 | +A cell's neighbors are the up to eight surrounding cells (vertically, horizontally, and diagonally). |
| 7 | +Cells on the edges have fewer than eight neighbors. |
| 8 | +Rules for updating each cell: |
| 9 | +
|
| 10 | +Any live cell with fewer than two live neighbors dies (underpopulation). |
| 11 | +Any live cell with two or three live neighbors lives on. |
| 12 | +Any live cell with more than three live neighbors dies (overpopulation). |
| 13 | +Any dead cell with exactly three live neighbors becomes alive (reproduction). |
| 14 | +For example, given: |
| 15 | +
|
| 16 | +[ |
| 17 | + [0, 1, 0], |
| 18 | + [0, 1, 1], |
| 19 | + [1, 1, 0] |
| 20 | +] |
| 21 | +return: |
| 22 | +
|
| 23 | +[ |
| 24 | + [0, 1, 1], |
| 25 | + [0, 0, 1], |
| 26 | + [1, 1, 1] |
| 27 | +] |
| 28 | +Each cell updates according to the number of live neighbors. For instance, [0][0] stays dead (2 live neighbors), [0][1] stays alive (2 live neighbors), [0][2] dies (3 live neighbors), and so on. |
| 29 | +""" |
| 30 | + |
| 31 | +import unittest |
| 32 | + |
| 33 | + |
| 34 | +class GameOfLifeTest(unittest.TestCase): |
| 35 | + |
| 36 | + def test1(self): |
| 37 | + self.assertEqual(game_of_life([[0, 1, 0], [0, 1, 1], [1, 1, 0]]),[[0, 1, 1], [0, 0, 1], [1, 1, 1]]) |
| 38 | + |
| 39 | + def test2(self): |
| 40 | + self.assertEqual(game_of_life([[1, 1, 0, 0], [1, 0, 1, 0], [0, 1, 1, 1], [0, 0, 1, 0]]),[[1, 1, 0, 0], [1, 0, 0, 1], [0, 0, 0, 1], [0, 1, 1, 1]]) |
| 41 | + |
| 42 | + def test3(self): |
| 43 | + self.assertEqual(game_of_life([[1, 0, 0], [0, 1, 0], [0, 0, 1]]),[[0, 0, 0], [0, 1, 0], [0, 0, 0]]) |
| 44 | + |
| 45 | + def test4(self): |
| 46 | + self.assertEqual(game_of_life([[0, 1, 1, 0], [1, 1, 0, 1], [0, 1, 1, 0], [0, 0, 1, 0]]),[[1, 1, 1, 0], [1, 0, 0, 1], [1, 0, 0, 1], [0, 1, 1, 0]]) |
| 47 | + |
| 48 | + |
| 49 | + |
| 50 | +""" |
| 51 | +The algorithm is O(n x m) where n and m are dimensions of the board. |
| 52 | +Each cell is updated based on its neighbors, producing a new board state. |
| 53 | +Works for any rectangular grid, not just square ones. |
| 54 | +
|
| 55 | +""" |
| 56 | +def game_of_life(grid): |
| 57 | + |
| 58 | + rows, cols = len(grid), len(grid[0]) |
| 59 | + |
| 60 | + def count_neighbors(r, c): |
| 61 | + directions = [ |
| 62 | + (-1, -1), (-1, 0), (-1, 1), |
| 63 | + (0, -1), (0, 1), |
| 64 | + (1, -1), (1, 0), (1, 1) |
| 65 | + ] |
| 66 | + |
| 67 | + count = 0 |
| 68 | + for dr, dc in directions: |
| 69 | + nr, nc = r + dr, c + dc |
| 70 | + if 0 <= nr < rows and 0 <= nc < cols: |
| 71 | + count += grid[nr][nc] |
| 72 | + |
| 73 | + return count |
| 74 | + |
| 75 | + new_board = [[0] * cols for _ in range(rows)] |
| 76 | + |
| 77 | + for r in range(rows): |
| 78 | + for c in range(cols): |
| 79 | + neighbors = count_neighbors(r, c) |
| 80 | + if grid[r][c] == 1: |
| 81 | + if neighbors < 2 or neighbors > 3: |
| 82 | + new_board[r][c] = 0 |
| 83 | + else: |
| 84 | + new_board[r][c] = 1 |
| 85 | + else: |
| 86 | + if neighbors == 3: |
| 87 | + new_board[r][c] = 1 |
| 88 | + |
| 89 | + return new_board |
| 90 | + |
| 91 | + |
| 92 | + |
| 93 | +if __name__ == "__main__": |
| 94 | + print(game_of_life([[0, 1, 1, 0], [1, 1, 0, 1], [0, 1, 1, 0], [0, 0, 1, 0]])) |
| 95 | + unittest.main() |
| 96 | + |
| 97 | + |
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