Conway's Game of Life - Amazon Top Interview Questions


Problem Statement :


You are given a two dimensional matrix where a 1 represents a live cell and a 0 represents a dead cell. A cell's (living or dead) neighbors are its immediate horizontal, vertical and diagonal cells. Compute 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.

Constraints

n, m ≤ 500 where n and m are the number of rows and columns in matrix

Example 1

Input

matrix = [
    [1, 1, 1, 0],
    [0, 1, 0, 1],
    [0, 1, 0, 0],
    [1, 1, 0, 1]
]

Output

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


Solution :



title-img 




                        Solution in C++ :

bool save(vector<vector<int>>& v, int row, int col) {
    return (v.size() > row && v[0].size() > col && row >= 0 && col >= 0);
}
vector<vector<int>> solve(vector<vector<int>>& v) {
    int n = v.size();
    int m = v[0].size();
    int p[] = {-1, -1, -1, 0, 0, 1, 1, 1};
    int q[] = {-1, 0, 1, -1, 1, -1, 0, 1};
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            if (v[i][j] > 0) {
                for (int k = 0; k < 8; k++) {
                    if (save(v, i + p[k], j + q[k]) && v[i + p[k]][j + q[k]] > 0) {
                        v[i][j]++;
                    }
                }
            } else {
                for (int k = 0; k < 8; k++) {
                    if (save(v, i + p[k], j + q[k]) && v[i + p[k]][j + q[k]] > 0) {
                        v[i][j]--;
                    }
                }
            }
        }
    }
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            if (v[i][j] > 0) {
                if (v[i][j] < 3)
                    v[i][j] = 0;
                else if (v[i][j] <= 4)
                    v[i][j] = 1;
                else if (v[i][j] > 4)
                    v[i][j] = 0;
            }
            if (v[i][j] <= 0) {
                if (v[i][j] == -3)
                    v[i][j] = 1;
                else
                    v[i][j] = 0;
            }
        }
    }
    return v;
}
                    




                        Solution in Python : 
                            
class Solution:
    def solve(self, matrix):
        if not matrix or not matrix[0]:
            return

        n, m = len(matrix), len(matrix[0])
        moves = [(1, 0), (-1, 0), (0, 1), (0, -1), (1, 1), (-1, -1), (1, -1), (-1, 1)]

        for i in range(n):
            for j in range(m):
                living_count = 0

                for x, y in moves:
                    di = i + x
                    dj = j + y

                    if di >= 0 and di < n and dj >= 0 and dj < m and abs(matrix[di][dj]) == 1:
                        living_count += 1

                if matrix[i][j] == 1 and (living_count < 2 or living_count > 3):
                    matrix[i][j] = -1

                if matrix[i][j] == 0 and living_count == 3:
                    matrix[i][j] = 2

        for i in range(n):
            for j in range(m):
                if matrix[i][j] > 0:
                    matrix[i][j] = 1
                else:
                    matrix[i][j] = 0

        return matrix


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