Maximum Non Negative Product in a Matrix — C++

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

// dp with rolling window
class Solution {
public:
    int maxProductPath(vector<vector<int>>& grid) {
        static const int MOD = 1e9 + 7;
        vector<vector<int64_t>> max_dp(2, vector<int64_t>(size(grid[0])));
        vector<vector<int64_t>> min_dp(2, vector<int64_t>(size(grid[0])));
        for (int i = 0; i < size(grid); ++i) {
            for (int j = 0; j < size(grid[0]); ++j) {
                if (i == 0 && j == 0) {
                    max_dp[0][0] = min_dp[0][0] = grid[0][0];
                    continue;
                }
                auto curr_max = max(i > 0 ? max_dp[(i - 1) % 2][j] : max_dp[i % 2][j - 1],
                                    j > 0 ? max_dp[i % 2][j - 1] : max_dp[(i - 1) % 2][j]);
                auto curr_min = min(i > 0 ? min_dp[(i - 1) % 2][j] : min_dp[i % 2][j - 1],
                                    j > 0 ? min_dp[i % 2][j - 1] : min_dp[(i - 1) % 2][j]);
                if (grid[i][j] < 0) {
                    swap(curr_max, curr_min);
                }
                max_dp[i % 2][j] = curr_max * grid[i][j];
                min_dp[i % 2][j] = curr_min * grid[i][j];
            }
        }
        return max_dp[(size(grid) - 1) % 2].back() >= 0 ? max_dp[(size(grid) - 1) % 2].back() % MOD : -1;
    }
};