# 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 :

```                        ```Solution in C++ :

bool save(vector<vector<int>>& v, int row, int col) {
return (v.size() > row && v.size() > col && row >= 0 && col >= 0);
}
vector<vector<int>> solve(vector<vector<int>>& v) {
int n = v.size();
int m = v.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:
return

n, m = len(matrix), len(matrix)
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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