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Game of Life Step

Advance Conway's Game of Life by one generation.

Problema

Each cell is alive (1) or dead (0) and has up to 8 neighbors. In the next generation a live cell survives with 2 or 3 live neighbors, and a dead cell becomes alive with exactly 3. Cells outside the grid are dead. All cells update at the same time.

Implement next_generation(grid).

Input

grid: A rectangular grid of 0s and 1s.

Output

Return the next generation.

Vincoli

0 ≤ rows, columns ≤ 200

Esempi

Esempio 1
Input:
grid = [
    [0, 1, 0],
    [0, 1, 0],
    [0, 1, 0]
]

Output atteso:
expected = [
    [0, 0, 0],
    [1, 1, 1],
    [0, 0, 0]
]
Input:
std::vector<std::vector<long long>> grid = {
    {0, 1, 0},
    {0, 1, 0},
    {0, 1, 0}
};

Output atteso:
std::vector<std::vector<long long>> expected = {
    {0, 0, 0},
    {1, 1, 1},
    {0, 0, 0}
};
Input:
long[][] grid = new long[][] {
    new long[] {0, 1, 0},
    new long[] {0, 1, 0},
    new long[] {0, 1, 0}
};

Output atteso:
long[][] expected = new long[][] {
    new long[] {0, 0, 0},
    new long[] {1, 1, 1},
    new long[] {0, 0, 0}
};

Spiegazione

The vertical "blinker" turns horizontal: the center survives and its left and right neighbors are born, filling the middle row.

Esempio 2
Input:
grid = [
    [1, 1],
    [1, 1]
]

Output atteso:
expected = [
    [1, 1],
    [1, 1]
]
Input:
std::vector<std::vector<long long>> grid = {
    {1, 1},
    {1, 1}
};

Output atteso:
std::vector<std::vector<long long>> expected = {
    {1, 1},
    {1, 1}
};
Input:
long[][] grid = new long[][] {
    new long[] {1, 1},
    new long[] {1, 1}
};

Output atteso:
long[][] expected = new long[][] {
    new long[] {1, 1},
    new long[] {1, 1}
};

Spiegazione

Every cell of the 2×2 block has 3 neighbors, so the block stays the same.

Riferimenti

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