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Formatted question description: https://leetcode.ca/all/1594.html

# 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 + 7If 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/
// 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/
// 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);
}
};


Java

• class Solution {
public int maxProductPath(int[][] grid) {
final int MODULO = 1000000007;
long maxProduct = -1;
int rows = grid.length, columns = grid[0].length;
long[][][] products = new long[rows][columns][2];
products[0][0][0] = products[0][0][1] = grid[0][0];
for (int i = 1; i < rows; i++) {
long product = products[i - 1][0][0] * grid[i][0];
if (product == 0 && maxProduct < 0)
maxProduct = 0;
products[i][0][0] = products[i][0][1] = product;
}
for (int j = 1; j < columns; j++) {
long product = products[0][j - 1][0] * grid[0][j];
if (product == 0 && maxProduct < 0)
maxProduct = 0;
products[0][j][0] = products[0][j][1] = product;
}
for (int i = 1; i < rows; i++) {
for (int j = 1; j < columns; j++) {
int num = grid[i][j];
if (num == 0 && maxProduct < 0)
maxProduct = 0;
long product1 = products[i - 1][j][0] * num;
long product2 = products[i - 1][j][1] * num;
long product3 = products[i][j - 1][0] * num;
long product4 = products[i][j - 1][1] * num;
long[] array = {product1, product2, product3, product4};
Arrays.sort(array);
long min = array[0], max = array[3];
if (max > 0)
products[i][j][0] = max;
else
products[i][j][0] = min;
if (min < 0)
products[i][j][1] = min;
else
products[i][j][1] = max;
}
}
maxProduct = Math.max(maxProduct, Math.max(products[rows - 1][columns - 1][0], products[rows - 1][columns - 1][1]));
return (int) (maxProduct % MODULO);
}
}

• // OJ: https://leetcode.com/problems/maximum-non-negative-product-in-a-matrix/
// 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);
}
};

• class Solution:
def maxProductPath(self, grid: List[List[int]]) -> int:
m, n = len(grid), len(grid[0])
mod = 10**9 + 7
dp = [[[grid[0][0]] * 2 for _ in range(n)] for _ in range(m)]
for i in range(1, m):
dp[i][0] = [dp[i - 1][0][0] * grid[i][0]] * 2
for j in range(1, n):
dp[0][j] = [dp[0][j - 1][0] * grid[0][j]] * 2
for i in range(1, m):
for j in range(1, n):
v = grid[i][j]
if v >= 0:
dp[i][j][0] = min(dp[i - 1][j][0], dp[i][j - 1][0]) * v
dp[i][j][1] = max(dp[i - 1][j][1], dp[i][j - 1][1]) * v
else:
dp[i][j][0] = max(dp[i - 1][j][1], dp[i][j - 1][1]) * v
dp[i][j][1] = min(dp[i - 1][j][0], dp[i][j - 1][0]) * v
ans = dp[-1][-1][1]
return -1 if ans < 0 else ans % mod

############

# 1594. Maximum Non Negative Product in a Matrix
# https://leetcode.com/problems/maximum-non-negative-product-in-a-matrix/

class Solution:
def maxProductPath(self, A: List[List[int]]) -> int:
m, n = len(A), len(A[0])
Max = [[0] * n for _ in range(m)]
Min = [[0] * n for _ in range(m)]
Max[0][0] = A[0][0]
Min[0][0] = A[0][0]
for j in range(1, n):
Max[0][j] = Max[0][j - 1] * A[0][j]
Min[0][j] = Min[0][j - 1] * A[0][j]

for i in range(1, m):
Max[i][0] = Max[i - 1][0] * A[i][0]
Min[i][0] = Min[i - 1][0] * A[i][0]

for i in range(1,m):
for j in range(1,n):
Max[i][j] = max(Max[i-1][j] * A[i][j], Min[i-1][j] * A[i][j], Max[i][j-1] * A[i][j], Min[i][j-1] * A[i][j])
Min[i][j] = min(Max[i-1][j] * A[i][j], Min[i-1][j] * A[i][j], Max[i][j-1] * A[i][j], Min[i][j-1] * A[i][j])

return Max[-1][-1] % int(1e9 + 7) if Max[-1][-1] >= 0 else -1

# revisited solutin
class Solution:
def maxProductPath(self, A: List[List[int]]) -> int:
M = 1000000007
row, col = len(A), len(A[0])

mp = [[0] * col for _ in range(row)]
mn = [[0] * col for _ in range(row)]
mp[0][0] = mn[0][0] = A[0][0]

for i in range(1,col):
mp[0][i] = A[0][i] * mp[0][i-1]
mn[0][i] = A[0][i] * mp[0][i-1]

for i in range(1,row):
mp[i][0] = A[i][0] * mp[i-1][0]
mn[i][0] = A[i][0] * mp[i-1][0]

for i in range(1,row):
for j in range(1,col):
if A[i][j] > 0:
mp[i][j] = max(mp[i-1][j], mp[i][j-1]) * A[i][j]
mn[i][j] = min(mn[i-1][j], mn[i][j-1]) * A[i][j]
else:
mp[i][j] = min(mn[i-1][j], mn[i][j-1]) * A[i][j]
mn[i][j] = max(mp[i-1][j], mp[i][j-1]) * A[i][j]

res = mp[row-1][col-1]

return res%M if res > -1 else -1