Path with maximum product in 2-d array

Last Updated : 06 Apr, 2022

Given a 2D matrix of size N*M. The task is to find the maximum product path from (0, 0) to (N-1, M-1). You can only move to right from (i, j) to (i, j+1) and down from (i, j) to (i+1, j).
Examples:

Input: arr[][] = { {1, 2, 3}, {4, 5, 6}, {7, 8, 9} }
Output: 2016
The path with maximum product is : 1->4->7->8->9 = 2016

Approach:
To choose which direction to go from the current position we have to check the direction that gives the maximum product. We will maintain two 2D arrays:

maxPath[i][j]: It will store the maximum product till (i, j)
minPath[i][j]: It will store the minimum product till (i, j)

Steps:

Initialise the maxPath[0][0] & minPath[0][0] to arr[0][0].
For all the remaining cells in maxPath[i][j] check whether the product of current cell(i, j) and previous cell (i-1, j) is maximum or not by:

considering rows:
maxValue1 = max(arr[i][j]*maxPath[i-1][j], arr[i][j]*minPath[i-1][j])
considering columns:
maxValue2 = max(maxPath[i][j - 1] * arr[i][j], minPath[i][j - 1] * arr[i][j])
as arr[i][j] can be negative and if minPath[i][j] is negative. Then,
arr[i][j]*minPath[i][j] is positive, which can be maximum product.

For all the remaining cells in minPath[i][j] check whether the product of current cell(i, j) and previous cell (i-1, j) is minimum or not by:

considering rows:
minValue1 = min(maxPath[i - 1][j] * arr[i][j], minPath[i - 1][j] * arr[i][j])
considering columns:
minValue2 = min(maxPath[i][j - 1] * arr[i][j], minPath[i][j - 1] * arr[i][j])

Update minPath[i][j] = min(minValue1, minValue2) & maxPath[i][j] = max(maxValue1, maxValue2)
Return maxPath[n-1][m-1] for maximum product.

Below is the implementation of the above approach:

C++

// C++ Program to find maximum
// product path from (0, 0) to
// (N-1, M-1)
#include <bits/stdc++.h>
using namespace std;
#define N 3
#define M 3

// Function to find maximum product
int maxProductPath(int arr[N][M])
{

    // It will store the maximum
    // product till a given cell.
    vector<vector<int> > maxPath(N, vector<int>(M, INT_MIN));

    // It will store the minimum
    // product till a given cell
    // (for -ve elements)
    vector<vector<int> > minPath(N, vector<int>(M, INT_MAX));

    for (int i = 0; i < N; i++) {
        for (int j = 0; j < M; j++) {

            int minVal = INT_MAX;
            int maxVal = INT_MIN;

            // If we are at topmost
            // or leftmost, just
            // copy the elements.
            if (i == 0 && j == 0) {
                maxVal = arr[i][j];
                minVal = arr[i][j];
            }

            // If we're not at the
            // above, we can consider
            // the above value.
            if (i > 0) {
                int tempMax = max(maxPath[i - 1][j] * arr[i][j],
                                  minPath[i - 1][j] * arr[i][j]);
                maxVal = max(maxVal, tempMax);

                int tempMin = min(maxPath[i - 1][j] * arr[i][j],
                                  minPath[i - 1][j] * arr[i][j]);
                minVal = min(minVal, tempMin);
            }

            // If we're not on the
            // leftmost, we can consider
            // the left value.
            if (j > 0) {
                int tempMax = max(maxPath[i][j - 1] * arr[i][j],
                                  minPath[i][j - 1] * arr[i][j]);
                maxVal = max(maxVal, tempMax);

                int tempMin = min(maxPath[i][j - 1] * arr[i][j],
                                  minPath[i][j - 1] * arr[i][j]);
                minVal = min(minVal, tempMin);
            }

            // Store max & min product
            // till i, j.
            maxPath[i][j] = maxVal;
            minPath[i][j] = minVal;
        }
    }

    // Return the max product path
    // from 0, 0 -> N-1, M-1.
    return maxPath[N - 1][M - 1];
}

// Driver Code
int main()
{
    int arr[N][M] = { { 1, -2, 3 },
                      { 4, -5, 6 },
                      { -7, -8, 9 } };

    // Print maximum product from
    // (0, 0) to (N-1, M-1)
    cout << maxProductPath(arr) << endl;
    return 0;
}

Java

// Java Program to find maximum
// product path from (0, 0) to
// (N-1, M-1)
class GFG
{

static final int N = 3;
static final int M = 3;

// Function to find maximum product
static int maxProductPath(int arr[][])
{

    // It will store the maximum
    // product till a given cell.
    int [][]maxPath = new int[N][M];
    
    // It will store the minimum
    // product till a given cell
    // (for -ve elements)
    int [][]minPath = new int[N][M];
    
    for (int i = 0; i < N; i++) 
    {
        for (int j = 0; j < M; j++) 
        {

            int minVal = Integer.MAX_VALUE;
            int maxVal = Integer.MIN_VALUE;

            // If we are at topmost
            // or leftmost, just
            // copy the elements.
            if (i == 0 && j == 0)
            {
                maxVal = arr[i][j];
                minVal = arr[i][j];
            }

            // If we're not at the
            // above, we can consider
            // the above value.
            if (i > 0) 
            {
                int tempMax = Math.max(maxPath[i - 1][j] * arr[i][j],
                                minPath[i - 1][j] * arr[i][j]);
                maxVal = Math.max(maxVal, tempMax);

                int tempMin = Math.min(maxPath[i - 1][j] * arr[i][j],
                                minPath[i - 1][j] * arr[i][j]);
                minVal = Math.min(minVal, tempMin);
            }

            // If we're not on the
            // leftmost, we can consider
            // the left value.
            if (j > 0) {
                int tempMax = Math.max(maxPath[i][j - 1] * arr[i][j],
                                minPath[i][j - 1] * arr[i][j]);
                maxVal = Math.max(maxVal, tempMax);

                int tempMin = Math.min(maxPath[i][j - 1] * arr[i][j],
                                minPath[i][j - 1] * arr[i][j]);
                minVal = Math.min(minVal, tempMin);
            }

            // Store max & min product
            // till i, j.
            maxPath[i][j] = maxVal;
            minPath[i][j] = minVal;
        }
    }

    // Return the max product path
    // from 0, 0.N-1, M-1.
    return maxPath[N - 1][M - 1];
}

// Driver Code
public static void main(String[] args)
{
    int arr[][] = { { 1, -2, 3 },
                    { 4, -5, 6 },
                    { -7, -8, 9 } };

    // Print maximum product from
    // (0, 0) to (N-1, M-1)
    System.out.print(maxProductPath(arr) +"\n");
}
}

// This code is contributed by 29AjayKumar

Python3

# Python Program to find maximum
# product path from (0, 0) to
# (N-1, M-1)
import sys
N = 3;
M = 3;

# Function to find maximum product
def maxProductPath(arr):

    # It will store the maximum
    # product till a given cell.
    maxPath = [[0 for i in range(M)] for j in range(N)];

    # It will store the minimum
    # product till a given cell
    # (for -ve elements)
    minPath = [[0 for i in range(M)] for j in range(N)];

    for i in range(N):
        for j in range(M):

            minVal = sys.maxsize;
            maxVal = -sys.maxsize;

            # If we are at topmost
            # or leftmost, just
            # copy the elements.
            if (i == 0 and j == 0):
                maxVal = arr[i][j];
                minVal = arr[i][j];
            
            # If we're not at the
            # above, we can consider
            # the above value.
            if (i > 0):
                tempMax = max(maxPath[i - 1][j] * arr[i][j],\
                            minPath[i - 1][j] * arr[i][j]);
                maxVal = max(maxVal, tempMax);

                tempMin = min(maxPath[i - 1][j] * arr[i][j],\
                    minPath[i - 1][j] * arr[i][j]);
                minVal = min(minVal, tempMin);
            
            # If we're not on the
            # leftmost, we can consider
            # the left value.
            if (j > 0):
                tempMax = max(maxPath[i][j - 1] * arr[i][j],\
                    minPath[i][j - 1] * arr[i][j]);
                maxVal = max(maxVal, tempMax);

                tempMin = min(maxPath[i][j - 1] * arr[i][j],\
                    minPath[i][j - 1] * arr[i][j]);
                minVal = min(minVal, tempMin);
            
            # Store max & min product
            # till i, j.
            maxPath[i][j] = maxVal;
            minPath[i][j] = minVal;
        
    # Return the max product path
    # from 0, 0.N-1, M-1.
    return maxPath[N - 1][M - 1];

# Driver Code
if __name__ == '__main__':
    arr = [[ 1, -2, 3 ],[ 4, -5, 6 ],[ -7, -8, 9]];

    # Print maximum product from
    # (0, 0) to (N-1, M-1)
    print(maxProductPath(arr));

# This code is contributed by 29AjayKumar

// C# Program to find maximum
// product path from (0, 0) to
// (N-1, M-1)
using System;

class GFG
{
 
static readonly int N = 3;
static readonly int M = 3;
 
// Function to find maximum product
static int maxProductPath(int [,]arr)
{
 
    // It will store the maximum
    // product till a given cell.
    int [,]maxPath = new int[N, M];
     
    // It will store the minimum
    // product till a given cell
    // (for -ve elements)
    int [,]minPath = new int[N, M];
     
    for (int i = 0; i < N; i++) 
    {
        for (int j = 0; j < M; j++) 
        {
 
            int minVal = int.MaxValue;
            int maxVal = int.MinValue;
 
            // If we are at topmost
            // or leftmost, just
            // copy the elements.
            if (i == 0 && j == 0)
            {
                maxVal = arr[i, j];
                minVal = arr[i, j];
            }
 
            // If we're not at the
            // above, we can consider
            // the above value.
            if (i > 0) 
            {
                int tempMax = Math.Max(maxPath[i - 1,j] * arr[i,j],
                                minPath[i - 1,j] * arr[i,j]);
                maxVal = Math.Max(maxVal, tempMax);
 
                int tempMin = Math.Min(maxPath[i - 1,j] * arr[i,j],
                                minPath[i - 1,j] * arr[i,j]);
                minVal = Math.Min(minVal, tempMin);
            }
 
            // If we're not on the
            // leftmost, we can consider
            // the left value.
            if (j > 0) {
                int tempMax = Math.Max(maxPath[i,j - 1] * arr[i,j],
                                minPath[i,j - 1] * arr[i,j]);
                maxVal = Math.Max(maxVal, tempMax);
 
                int tempMin = Math.Min(maxPath[i,j - 1] * arr[i,j],
                                minPath[i,j - 1] * arr[i,j]);
                minVal = Math.Min(minVal, tempMin);
            }
 
            // Store max & min product
            // till i, j.
            maxPath[i, j] = maxVal;
            minPath[i, j] = minVal;
        }
    }
 
    // Return the max product path
    // from 0, 0.N - 1, M - 1.
    return maxPath[N - 1, M - 1];
}
 
// Driver Code
public static void Main(String[] args)
{
    int [,]arr = { { 1, -2, 3 },
                    { 4, -5, 6 },
                    { -7, -8, 9 } };
 
    // Print maximum product from
    // (0, 0) to (N - 1, M - 1)
    Console.Write(maxProductPath(arr) +"\n");
}
}

// This code is contributed by 29AjayKumar

JavaScript

<script>

// JavaScript Program to find maximum
// product path from (0, 0) to
// (N-1, M-1)

const N = 3;
const M = 3;

// Function to find maximum product
function maxProductPath(arr){

    // It will store the maximum
    // product till a given cell.
    let maxPath = new Array(N);
    for(let i=0;i<N;i++){
        maxPath[i] = new Array(M).fill(0);
     }

    // It will store the minimum
    // product till a given cell
    // (for -ve elements)
    let minPath = new Array(N);
    for(let i=0;i<N;i++){
       minPath[i] = new Array(M).fill(0);
    }

    for(let i=0;i<N;i++){
        for(let j=0;j<M;j++){

            let minVal = Number.MAX_VALUE;
            let maxVal = Number.MIN_VALUE;

            // If we are at topmost
            // or leftmost, just
            // copy the elements.
            if (i == 0 && j == 0){
                maxVal = arr[i][j];
                minVal = arr[i][j];
            }
            
            // If we're not at the
            // above, we can consider
            // the above value.
            if (i > 0){
                tempMax = Math.max(maxPath[i - 1][j] * arr[i][j],
                            minPath[i - 1][j] * arr[i][j]);
                maxVal = Math.max(maxVal, tempMax);

                tempMin = Math.min(maxPath[i - 1][j] * arr[i][j],
                    minPath[i - 1][j] * arr[i][j]);
                minVal = Math.min(minVal, tempMin);
            }
            // If we're not on the
            // leftmost, we can consider
            // the left value.
            if (j > 0){
                tempMax = Math.max(maxPath[i][j - 1] * arr[i][j],
                    minPath[i][j - 1] * arr[i][j]);
                maxVal = Math.max(maxVal, tempMax);

                tempMin = Math.min(maxPath[i][j - 1] * arr[i][j],
                    minPath[i][j - 1] * arr[i][j]);
                minVal = Math.min(minVal, tempMin);
            }
            
            // Store max & min product
            // till i, j.
            maxPath[i][j] = maxVal;
            minPath[i][j] = minVal;
        }
    }
        
    // Return the max product path
    // from 0, 0.N-1, M-1.
    return maxPath[N - 1][M - 1];
}

// Driver Code

let    arr = [[ 1, -2, 3 ],[ 4, -5, 6 ],[ -7, -8, 9]];

// Print maximum product from
// (0, 0) to (N-1, M-1)
document.write(maxProductPath(arr));

// This code is contributed by shinjanpatra

</script>

Output:

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

Path with maximum product in 2-d array

Aashish Chauhan

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Path with maximum product in 2-d array

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