[BinarySearch] Maximum Product Path in 2D Matrix

You are given a two-dimensional list of integers matrix. You are currently at the top left corner and want to move to the bottom right corner. In each move, you can move down or right.

Return the maximum product of the cells visited by going to the bottom right cell. If the result is negative, return -1. Otherwise, mod the result by 10 ** 9 + 7.

Constraints

• 1 ≤ n, m ≤ 20 where n and m are the number of rows and columns in matrix
• -2 ≤ matrix[r][c] ≤ 2

Example 1

Input

matrix = [
    [2, 1, -2],
    [-1, -1, -2],
    [1, 1, 1]
]

Output

8

Explanation

We can take the following path: [2, 1, -2, -2, 1].

 

Dynamic programming:  This is essentially the 2D version of Maximum Subarray Product. We'll have 2 dp table: maxDp and minDp;  maxDp[i][j] is the max product path that ends at cell(i, j); minDp[i][j] is the min product path that ends at cell(i, j). The key here is that when matrix[i][j] is negative, we need to use the min of the previous 2 neighboring products to compute the max product; similiarly use the max of the previous 2 neighboring products to compute the min product at ends at cell(i, j).  The rest of the dp is pretty straightforward as shown in the following code.

 

 

class Solution {
    public int solve(int[][] matrix) {
        int n = matrix.length, m = matrix[0].length, mod = (int)1e9 + 7;
        if(n == 0) return 0;
        long[][] maxDp = new long[n][m], minDp = new long[n][m];
        maxDp[0][0] = matrix[0][0];
        minDp[0][0] = matrix[0][0];
        for(int i = 1; i < n; i++) {
            maxDp[i][0] = maxDp[i - 1][0] * matrix[i][0];
            minDp[i][0] = minDp[i - 1][0] * matrix[i][0];
        }
        for(int j = 1; j < m; j++) {
            maxDp[0][j] = maxDp[0][j - 1] * matrix[0][j];
            minDp[0][j] = minDp[0][j - 1] * matrix[0][j];
        }
        for(int i = 1; i < n; i++) {
            for(int j = 1; j < m; j++) {
                if(matrix[i][j] > 0) {
                    maxDp[i][j] = Math.max(maxDp[i - 1][j], maxDp[i][j - 1]) * matrix[i][j];     
                    minDp[i][j] = Math.min(minDp[i][j - 1], minDp[i - 1][j]) * matrix[i][j];
                }
                else if(matrix[i][j] < 0) {
                    maxDp[i][j] = Math.min(minDp[i - 1][j], minDp[i][j - 1]) * matrix[i][j];     
                    minDp[i][j] = Math.max(maxDp[i][j - 1], maxDp[i - 1][j]) * matrix[i][j];
                }
                else {
                    maxDp[i][j] = 0;
                    minDp[i][j] = 0;
                }
            }
        }
        return maxDp[n - 1][m - 1] >= 0 ? (int)(maxDp[n - 1][m - 1] % mod) : -1;
    }
}

 

 

 

Related Problems

Maximum Subarray Product