1594

1594. Maximum Non Negative Product in a Matrix (Medium)

You are given a rows x cols matrix grid. Initially, you are located at the top-left corner (0, 0), and in each step, you can only move right or down in the matrix.

Among all possible paths starting from the top-left corner (0, 0) and ending in the bottom-right corner (rows - 1, cols - 1), find the path with the maximum non-negative product. The product of a path is the product of all integers in the grid cells visited along the path.

Return the maximum non-negative product modulo 109 + 7. If the maximum product is negative return -1.

Notice that the modulo is performed after getting the maximum product.

 

Example 1:

Input: grid = [[-1,-2,-3],
               [-2,-3,-3],
               [-3,-3,-2]]
Output: -1
Explanation: It's not possible to get non-negative product in the path from (0, 0) to (2, 2), so return -1.

Example 2:

Input: grid = [[1,-2,1],
               [1,-2,1],
               [3,-4,1]]
Output: 8
Explanation: Maximum non-negative product is in bold (1 * 1 * -2 * -4 * 1 = 8).

Example 3:

Input: grid = [[1, 3],
               [0,-4]]
Output: 0
Explanation: Maximum non-negative product is in bold (1 * 0 * -4 = 0).

Example 4:

Input: grid = [[ 1, 4,4,0],
               [-2, 0,0,1],
               [ 1,-1,1,1]]
Output: 2
Explanation: Maximum non-negative product is in bold (1 * -2 * 1 * -1 * 1 * 1 = 2).

 

Constraints:

• 1 <= rows, cols <= 15
• -4 <= grid[i][j] <= 4

Related Topics:
Dynamic Programming, Greedy

Solution 1. DP

// OJ: https://leetcode.com/problems/maximum-non-negative-product-in-a-matrix/
// Author: github.com/lzl124631x
// Time: O(MN)
// Space: O(MN)
class Solution {
public:
    int maxProductPath(vector<vector<int>>& G) {
        long ans = 0, mod = 1e9+7, M = G.size(), N = G[0].size();
        vector<vector<pair<long, long>>> dp(M + 1, vector<pair<long, long>>(N + 1, { INT_MAX, INT_MIN })); // min, max
        for (int i = 0; i < M; ++i) {
            for (int j = 0; j < N; ++j) {
                if (i == 0 && j == 0) dp[i][j] =  {G[i][j], G[i][j]};
                if (i != 0) {
                    dp[i][j].first = min(dp[i - 1][j].first * G[i][j], dp[i - 1][j].second * G[i][j]);
                    dp[i][j].second = max(dp[i - 1][j].first * G[i][j], dp[i - 1][j].second * G[i][j]);
                }
                if (j != 0) {
                    dp[i][j].first = min({ dp[i][j].first, dp[i][j - 1].first * G[i][j], dp[i][j - 1].second * G[i][j] });
                    dp[i][j].second = max({ dp[i][j].second, dp[i][j - 1].first * G[i][j], dp[i][j - 1].second * G[i][j] });
                }
            }
        }
        return dp[M - 1][N - 1].second < 0 ? -1 : (dp[M - 1][N - 1].second % mod);
    }
};

Solution 2. DP with Space Optimization

// OJ: https://leetcode.com/problems/maximum-non-negative-product-in-a-matrix/
// Author: github.com/lzl124631x
// Time: O(MN)
// Space: O(N)
class Solution {
public:
    int maxProductPath(vector<vector<int>>& G) {
        long ans = 0, mod = 1e9+7, M = G.size(), N = G[0].size();
        vector<pair<long, long>> dp(N + 1); // min, max
        for (int i = 0; i < M; ++i) {
            for (int j = 0; j < N; ++j) {
                long mn = INT_MAX, mx = INT_MIN;
                if (i == 0 && j == 0) mn = mx = G[i][j];
                if (i != 0) {
                    mn = min(dp[j].first * G[i][j], dp[j].second * G[i][j]);
                    mx = max(dp[j].first * G[i][j], dp[j].second * G[i][j]);
                }
                if (j != 0) {
                    mn = min({ mn, dp[j - 1].first * G[i][j], dp[j - 1].second * G[i][j] });
                    mx = max({ mx, dp[j - 1].first * G[i][j], dp[j - 1].second * G[i][j] });
                }
                dp[j] = { mn, mx };
            }
        }
        return dp[N - 1].second < 0 ? -1 : (dp[N - 1].second % mod);
    }
};