Game of Life — C++

// Time:  O(m * n)
// Space: O(1)

class Solution {
public:
    void gameOfLife(vector<vector<int>>& board) {
        const int m = board.size(), n = m ? board[0].size() : 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                int count = 0;
                // Count live cells in 3x3 block.
                for (int I = max(i - 1, 0); I < min(i + 2, m); ++I) {
                    for (int J = max(j - 1, 0); J < min(j + 2, n); ++J) {
                        count += board[I][J] & 1;
                    }
                }
                // if (count == 4 && board[i][j]) means:
                //     Any live cell with three live neighbors lives.
                // if (count == 3) means:
                //     Any live cell with two live neighbors.
                //     Any dead cell with exactly three live neighbors lives.
                if ((count == 4 && board[i][j]) || count == 3) {
                    board[i][j] |= 2;  // Mark as live. 
                }
            }
        }
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                board[i][j] >>= 1;  // Update to the next state.
            }
        }
    }
};