Maximum non-negative product of a path from top left to bottom right of given Matrix

Given an integer matrix mat[][] of dimensions N * M,  the task is to print the maximum product of matrix elements in the path from the top-left cell (0, 0) to the bottom-right cell (N – 1, M – 1) of the given matrix. Only possible moves from any cell (i, j) is (i + 1, j) or (i, j + 1). If the maximum product is negative, then print “-1”.

Examples:

Input: mat[][] = [[1, -2, 1], [1, -2, 1], [3, -4, 1]]
Output: 8
Explanation: The path from top left to bottom right with maximum product is (0, 0) ? (1, 0) -> (1, 1) -> (2, 1) -> (2, 2) = 1 * 1 * ( -2 ) * ( -4 ) * ( 1 ) = 8.
Therefore, the maximum positive product is 8.

Input: mat[][] = [[-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). Therefore, print -1.

Naive Approach: The simplest approach to solve to recursively traverse from the top-left cell and generate all possible paths from top left to the bottom-right cell by moving rightwards and downwards from each cell. Find the product for each path generated and print the maximum among them. 
Time Complexity: O(2^{max(N, M)})
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea is to use Dynamic Programming. Follow the steps below to solve the problem:

• Initialize matrices maxPath[][] and minPath[][], to store maximum and minimum product of paths respectively.
• The idea is to also keep track of the maximum negative product obtained so far, because if any negative product is encountered, then the negative product multiplied by maximum negative product can generate maximum positive product.
• To reach index (i, j), as the possible moves allowed are rightward and downward, so the maximum product can be maximum product or minimum product till (i – 1, j)th index or (i, j – 1)^thindex multiplied by the element at index (i, j).
• After completing the above steps, if the value of maxPath[M – 1][N – 1] is non-negative, then print it as the positive product path. Otherwise, print “-1”.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>

using namespace std;

// Function to find the maximum product
// from the top left and bottom right
// cell of the given matrix grid[][]
int maxProductPath(vector<vector<int>> grid){

    // Store dimension of grid
    int n = grid.size();
    int m = grid[0].size();

    // Stores maximum product path
    vector<vector<int>> maxPath(n, vector<int>(m, 0));

    // Stores minimum product path
    vector<vector<int>>  minPath(n, vector<int>(m, 0));

    // Traverse the grid and update
    // maxPath and minPath array
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < m; j++)
        {

            // Initialize to inf and -inf
            int mn = INT_MAX;
            int mx = INT_MIN;

            // Base Case
            if (i == 0 and j == 0)
            {
                mx = grid[i][j];
                mn = grid[i][j];
              }

            // Calculate for row:
            if (i > 0)
            {
                int tempmx = max((maxPath[i - 1][j] * grid[i][j]),
                             (minPath[i - 1][j] * grid[i][j]));

                // Update the maximum
                mx = max(mx, tempmx);
                int tempmn = min((maxPath[i - 1][j] * grid[i][j]),
                             (minPath[i - 1][j] * grid[i][j]));

                // Update the minimum
                mn = min(mn, tempmn);
              }

            // Calculate for column:
            if (j > 0)
            {
                int tempmx = max((maxPath[i][j - 1] * grid[i][j]),
                             (minPath[i][j - 1] * grid[i][j]));

                // Update the maximum
                mx = max(mx, tempmx);
                int tempmn = min((maxPath[i][j - 1] * grid[i][j]),
                             (minPath[i][j - 1] * grid[i][j]));

                // Update the minimum
                mn = min(mn, tempmn);
              }

            // Update maxPath and minPath
            maxPath[i][j] = mx;
            minPath[i][j] = mn;
          }
        }

    // If negative product
    if(maxPath[n - 1][m - 1] < 0)
        return -1;
    // Otherwise
    else
        return(maxPath[n - 1][m - 1]);
}

// Driver Code
int main(){

// Given matrix mat[][]
  vector<vector<int>> mat = {{1, -2, 1},
                            {1, -2, 1},
                            {3, -4, 1}};
  // Function Call
  cout<<(maxProductPath(mat));

}

// This code is contributed by mohit kumar 29

Java

// Java program for the above approach
import java.io.*;
class GFG
{

  // Function to find the maximum product
  // from the top left and bottom right
  // cell of the given matrix grid[][]
  static int maxProductPath(int[][] grid)
  {

    // Store dimension of grid
    int n = grid.length;
    int m = grid[0].length;

    // Stores maximum product path
    int[][] maxPath = new int[n][m];

    // Stores minimum product path
    int[][] minPath = new int[n][m];

    // Traverse the grid and update
    // maxPath and minPath array
    for(int i = 0; i < n; i++)
    {
      for (int j = 0; j < m; j++)
      {
        // Initialize to inf and -inf
        int mn = Integer.MAX_VALUE;
        int mx = Integer.MIN_VALUE;

        // Base Case
        if(i == 0 && j == 0)
        {
          mx = grid[i][j];
          mn = grid[i][j];
        }

        // Calculate for row:
        if(i > 0)
        {
          int tempmx =Math.max((maxPath[i - 1][j] *
                                grid[i][j]),(minPath[i - 1][j] *
                                             grid[i][j]));

          // Update the maximum
          mx = Math.max(mx, tempmx);
          int tempmn = Math.min((maxPath[i - 1][j] *
                                 grid[i][j]),(minPath[i - 1][j] *
                                              grid[i][j]));

          // Update the minimum
          mn=Math.min(mn, tempmn);
        }

        // Calculate for column
        if(j > 0)
        {
          int tempmx = Math.max((maxPath[i][j - 1] *
                                 grid[i][j]),(minPath[i][j - 1] *
                                              grid[i][j]));

          // Update the maximum
          mx = Math.max(mx, tempmx);
          int tempmn = Math.min((maxPath[i][j - 1] *
                                 grid[i][j]),(minPath[i][j - 1] *
                                              grid[i][j]));

          // Update the minimum
          mn=Math.min(mn, tempmn);
        }

        // Update maxPath and minPath
        maxPath[i][j] = mx;
        minPath[i][j] = mn;
      }
    }

    // If negative product
    if(maxPath[n - 1][m - 1] < 0)
    {
      return -1;
    }

    // Otherwise
    else
    {
      return(maxPath[n - 1][m - 1]);
    }
  }

  // Driver Code
  public static void main (String[] args)
  {

    // Given matrix mat[][]
    int[][] mat = {{1, -2, 1}, {1, -2, 1}, {3, -4, 1}};

    // Function Call
    System.out.println(maxProductPath(mat));
  }
}

// This code is contributed by rag2127

Python3

# Python program for the above approach

# Function to find the maximum product
# from the top left and bottom right
# cell of the given matrix grid[][]
def maxProductPath(grid):

    # Store dimension of grid
    n, m = len(grid), len(grid[0])

    # Stores maximum product path
    maxPath = [[0 for i in range(m)] for j in range(n)]

    # Stores minimum product path
    minPath = [[0 for i in range(m)] for j in range(n)]

    # Traverse the grid and update
    # maxPath and minPath array
    for i in range(n):
        for j in range(m):

            # Initialize to inf and -inf
            mn = float("inf")
            mx = float("-inf")

            # Base Case
            if (i == 0 and j == 0):
                mx = grid[i][j]
                mn = grid[i][j]

            # Calculate for row:
            if i > 0:
                tempmx = max((maxPath[i - 1][j] * grid[i][j]),
                             (minPath[i - 1][j] * grid[i][j]))

                # Update the maximum
                mx = max(mx, tempmx)
                tempmn = min((maxPath[i - 1][j] * grid[i][j]),
                             (minPath[i - 1][j] * grid[i][j]))

                # Update the minimum
                mn = min(mn, tempmn)

            # Calculate for column:
            if (j > 0):
                tempmx = max((maxPath[i][j - 1] * grid[i][j]),
                             (minPath[i][j - 1] * grid[i][j]))

                # Update the maximum
                mx = max(mx, tempmx)
                tempmn = min((maxPath[i][j - 1] * grid[i][j]),
                             (minPath[i][j - 1] * grid[i][j]))

                # Update the minimum
                mn = min(mn, tempmn)

            # Update maxPath and minPath
            maxPath[i][j] = mx
            minPath[i][j] = mn

    # If negative product
    if(maxPath[n - 1][m - 1] < 0):
        return -1

    # Otherwise
    else:
        return(maxPath[n - 1][m - 1])

# Driver Code

# Given matrix mat[][]
mat = [[1, -2, 1],
        [1, -2, 1],
        [3, -4, 1]]

# Function Call
print(maxProductPath(mat))

C#

// C# program for the above approach
using System;
class GFG
{

  // Function to find the maximum product
  // from the top left and bottom right
  // cell of the given matrix grid[][]
  static int maxProductPath(int[,] grid)
  {

    // Store dimension of grid
    int n = grid.GetLength(0);
    int m = grid.GetLength(1);

    // Stores maximum product path
    int[,] maxPath = new int[n, m];

    // Stores minimum product path
    int[,] minPath = new int[n, m];

    // Traverse the grid and update
    // maxPath and minPath array
    for(int i = 0; i < n; i++)
    {
      for (int j = 0; j < m; j++)
      {

        // Initialize to inf and -inf
        int mn = Int32.MaxValue;
        int mx = Int32.MinValue;

        // Base Case
        if(i == 0 && j == 0)
        {
          mx = grid[i, j];
          mn = grid[i, j];
        }

        // Calculate for row:
        if(i > 0)
        {
          int tempmx = Math.Max((maxPath[i - 1, j] *
                                 grid[i, j]),
                                (minPath[i - 1, j] *
                                 grid[i, j]));

          // Update the maximum
          mx = Math.Max(mx, tempmx);
          int tempmn = Math.Min((maxPath[i - 1, j] *
                                 grid[i, j]),
                                (minPath[i - 1, j] *
                                 grid[i, j]));

          // Update the minimum
          mn = Math.Min(mn, tempmn);
        }

        // Calculate for column
        if(j > 0)
        {
          int tempmx = Math.Max((maxPath[i, j - 1] *
                                 grid[i, j]),
                                (minPath[i, j - 1] *
                                 grid[i, j]));

          // Update the maximum
          mx = Math.Max(mx, tempmx);
          int tempmn = Math.Min((maxPath[i, j - 1] *
                                 grid[i, j]),
                                (minPath[i, j - 1] *
                                 grid[i,j]));

          // Update the minimum
          mn = Math.Min(mn, tempmn);

        }

        // Update maxPath and minPath
        maxPath[i,j] = mx;
        minPath[i,j] = mn;
      }
    }
    // If negative product
    if(maxPath[n - 1, m - 1] < 0)
    {
      return -1;
    }

    // Otherwise
    else
    {
      return(maxPath[n - 1, m - 1]);
    }
  }

  // Driver Code
  static public void Main ()
  {

    // Given matrix mat[][]
    int[,] mat={{1, -2, 1}, {1, -2, 1}, {3, -4, 1}};

    // Function Call
    Console.WriteLine(maxProductPath(mat));
  }
}

// This code is contributed by avanitrachhadiya2155

Javascript

<script>

// JavaScript program for the above approach

// Function to find the maximum product
// from the top left and bottom right
// cell of the given matrix grid[][]
function maxProductPath(grid){

    // Store dimension of grid
    var n = grid.length;
    var m = grid[0].length;

    // Stores maximum product path
    var maxPath = Array.from(Array(n), ()=>Array(m).fill(0));

    // Stores minimum product path
    var minPath = Array.from(Array(n), ()=>Array(m).fill(0));

    // Traverse the grid and update
    // maxPath and minPath array
    for (var i = 0; i < n; i++)
    {
        for (var j = 0; j < m; j++)
        {

            // Initialize to inf and -inf
            var mn = 1000000000;
            var mx = -1000000000;

            // Base Case
            if (i == 0 && j == 0)
            {
                mx = grid[i][j];
                mn = grid[i][j];
              }

            // Calculate for row:
            if (i > 0)
            {
                var tempmx =
                Math.max((maxPath[i - 1][j] * grid[i][j]),
                         (minPath[i - 1][j] * grid[i][j]));

                // Update the maximum
                mx = Math.max(mx, tempmx);
                var tempmn =
                Math.min((maxPath[i - 1][j] * grid[i][j]),
                         (minPath[i - 1][j] * grid[i][j]));

                // Update the minimum
                mn = Math.min(mn, tempmn);
              }

            // Calculate for column:
            if (j > 0)
            {
                var tempmx =
                Math.max((maxPath[i][j - 1] * grid[i][j]),
                         (minPath[i][j - 1] * grid[i][j]));

                // Update the maximum
                mx = Math.max(mx, tempmx);
                var tempmn =
                Math.min((maxPath[i][j - 1] * grid[i][j]),
                         (minPath[i][j - 1] * grid[i][j]));

                // Update the minimum
                mn = Math.min(mn, tempmn);
              }

            // Update maxPath and minPath
            maxPath[i][j] = mx;
            minPath[i][j] = mn;
          }
        }

    // If negative product
    if(maxPath[n - 1][m - 1] < 0)
        return -1;
    // Otherwise
    else
        return(maxPath[n - 1][m - 1]);
}

// Driver Code

// Given matrix mat[][]
var mat = [[1, -2, 1],
           [1, -2, 1],
           [3, -4, 1]];

// Function Call
document.write(maxProductPath(mat));


</script>

Output:

8

Time Complexity: O(M*N)
Auxiliary Space: O(M*N)