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1594. Maximum Non Negative Product in a Matrix

Description

You are given a m x n 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 (m - 1, n - 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 is 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 shown (1 * 1 * -2 * -4 * 1 = 8).

Example 3:

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

 

Constraints:

  • m == grid.length
  • n == grid[i].length
  • 1 <= m, n <= 15
  • -4 <= grid[i][j] <= 4

Solutions

Solution 1

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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

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class Solution {
    private static final int MOD = (int) 1e9 + 7;

    public int maxProductPath(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        long[][][] dp = new long[m][n][2];
        dp[0][0][0] = grid[0][0];
        dp[0][0][1] = grid[0][0];
        for (int i = 1; i < m; ++i) {
            dp[i][0][0] = dp[i - 1][0][0] * grid[i][0];
            dp[i][0][1] = dp[i - 1][0][1] * grid[i][0];
        }
        for (int j = 1; j < n; ++j) {
            dp[0][j][0] = dp[0][j - 1][0] * grid[0][j];
            dp[0][j][1] = dp[0][j - 1][1] * grid[0][j];
        }
        for (int i = 1; i < m; ++i) {
            for (int j = 1; j < n; ++j) {
                int v = grid[i][j];
                if (v >= 0) {
                    dp[i][j][0] = Math.min(dp[i - 1][j][0], dp[i][j - 1][0]) * v;
                    dp[i][j][1] =