Maximum points by traversing from top left of Matrix to bottom right by two persons

Given a matrix grid[][] of order N*M with numbers 0-9 in the cells. The task is to find the maximum amount of money collected when two persons move from (0, 0) to (N-1, M-1) by moving only right and down. If both persons are at the same cell, then only one of them can pick the money at that location.

Examples:

Input: 
1 1 1 
1 0 1 
1 1 1
Output: 8
Explanation: Let 1 denote the places where person 1 collects the money and 2 denote where person 2 does so, then a possible solution is
1 1 1
2 0 1
2 2 1

Input: 
0 9 9 3 3 
2 9 3 3 3 
0 3 3 3 3 
4 1 1 1 1
Output: 52

Approach: The problem can be solved by using the recursion, by moving both the persons down and right in each of the cell and finding the maximum path sum in all the paths from (0, 0) to (N-1, M-1). So the idea is to find the cost of all possible paths and to find the maximum of them.

Below is the implementation of the above approach:

C++

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

const static int MAXR = 20, MAXC = 20;
int cache[MAXC][MAXR][MAXC][MAXR],
    dp[MAXC][MAXR][MAXC][MAXR];
int n, m;
vector<string> grid;

// Function to find maximum money collected
// when moving from (0, 0) to (N-1, M-1)
int maxMoney(int x1, int y1, int x2, int y2)
{
    // Out of bounds of grid
    if (x1 >= n || y1 >= m || x2 >= n || y2 >= m)
        return 0;
    if (cache[x1][y1][x2][y2] != 0)
        return dp[x1][y1][x2][y2];

    // Mark state as visited
    cache[x1][y1][x2][y2] = 1;

    // Collect money from the grid cell
    int money = grid[y1][x1] - '0';
    if (x1 != x2 || y1 != y2)
        money += grid[y2][x2] - '0';

    // Take maximum of all possibilities
    return dp[x1][y1][x2][y2]
           = money
             + max(
                   max(maxMoney(x1 + 1, y1, x2 + 1, y2),
                       maxMoney(x1, y1 + 1, x2 + 1, y2)),
                   max(maxMoney(x1 + 1, y1, x2, y2 + 1),
                       maxMoney(x1, y1 + 1, x2, y2 + 1)));
}

// Driver Code
int32_t main()
{
    // Given Input
    n = 3;
    m = 3;
    grid.push_back("111");
    grid.push_back("101");
    grid.push_back("111");

    // Function Call
    cout << maxMoney(0, 0, 0, 0);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{

static int MAXR = 20, MAXC = 20;
static int [][][][]cache = new int[MAXC][MAXR][MAXC][MAXR];
static int [][][][]dp = new int[MAXC][MAXR][MAXC][MAXR];
static int n, m;
static Vector<String> grid = new Vector<String>();

// Function to find maximum money collected
// when moving from (0, 0) to (N-1, M-1)
static int maxMoney(int x1, int y1, int x2, int y2)
{
    // Out of bounds of grid
    if (x1 >= n || y1 >= m || x2 >= n || y2 >= m)
        return 0;
    if (cache[x1][y1][x2][y2] != 0)
        return dp[x1][y1][x2][y2];

    // Mark state as visited
    cache[x1][y1][x2][y2] = 1;

    // Collect money from the grid cell
    int money = grid.get(y1).charAt(x1)- '0';
    if (x1 != x2 || y1 != y2)
        money += grid.get(y2).charAt(x2) - '0';

    // Take maximum of all possibilities
    return dp[x1][y1][x2][y2]
           = money
             + Math.max(
                   Math.max(maxMoney(x1 + 1, y1, x2 + 1, y2),
                       maxMoney(x1, y1 + 1, x2 + 1, y2)),
                   Math.max(maxMoney(x1 + 1, y1, x2, y2 + 1),
                       maxMoney(x1, y1 + 1, x2, y2 + 1)));
}

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

    // Given Input
    n = 3;
    m = 3;
    grid.add("111");
    grid.add("101");
    grid.add("111");

    // Function Call
    System.out.print(maxMoney(0, 0, 0, 0));

}
}

// This code is contributed by Princi Singh

Python3

# Python3 program for the above approach
MAXR, MAXC = 20, 20

cache  = [[[[0 for i in range(MAXR)] for j in range(MAXC)] for k in range(MAXR)] for l in range(MAXC)]
dp  = [[[[0 for i in range(MAXR)] for j in range(MAXC)] for k in range(MAXR)] for l in range(MAXC)]

grid = []

# Function to find maximum money collected
# when moving from (0, 0) to (N-1, M-1)
def maxMoney(x1, y1, x2, y2):
    # Out of bounds of grid
    if (x1 >= n or y1 >= m or x2 >= n or y2 >= m):
        return 0
    if (cache[x1][y1][x2][y2] != 0):
        return dp[x1][y1][x2][y2]

    # Mark state as visited
    cache[x1][y1][x2][y2] = 1

    # Collect money from the grid cell
    money = ord(grid[y1][x1]) - ord('0')
    if (x1 != x2 or y1 != y2):
        money += ord(grid[y2][x2]) - ord('0')

    dp[x1][y1][x2][y2] = money + max(max(maxMoney(x1 + 1, y1, x2 + 1, y2),  maxMoney(x1, y1 + 1, x2 + 1, y2)),
                   max(maxMoney(x1 + 1, y1, x2, y2 + 1),
                       maxMoney(x1, y1 + 1, x2, y2 + 1)))

    # Take maximum of all possibilities
    return dp[x1][y1][x2][y2]

# Given Input
n = 3
m = 3
grid.append("111")
grid.append("101")
grid.append("111")

# Function Call
print(maxMoney(0, 0, 0, 0))

# This code is contributed by suresh07.

C#

// C# program for the above approach
using System;
using System.Collections.Generic;

public class GFG{

static int MAXR = 20, MAXC = 20;
static int [,,,]cache = new int[MAXC,MAXR,MAXC,MAXR];
static int [,,,]dp = new int[MAXC,MAXR,MAXC,MAXR];
static int n, m;
static List<String> grid = new List<String>();

// Function to find maximum money collected
// when moving from (0, 0) to (N-1, M-1)
static int maxMoney(int x1, int y1, int x2, int y2)
{

    // Out of bounds of grid
    if (x1 >= n || y1 >= m || x2 >= n || y2 >= m)
        return 0;
    if (cache[x1,y1,x2,y2] != 0)
        return dp[x1,y1,x2,y2];

    // Mark state as visited
    cache[x1,y1,x2,y2] = 1;

    // Collect money from the grid cell
    int money = grid[y1][x1]- '0';
    if (x1 != x2 || y1 != y2)
        money += grid[y2][x2] - '0';

    // Take maximum of all possibilities
    return dp[x1,y1,x2,y2]
           = money
             + Math.Max(
                   Math.Max(maxMoney(x1 + 1, y1, x2 + 1, y2),
                       maxMoney(x1, y1 + 1, x2 + 1, y2)),
                   Math.Max(maxMoney(x1 + 1, y1, x2, y2 + 1),
                       maxMoney(x1, y1 + 1, x2, y2 + 1)));
}

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

    // Given Input
    n = 3;
    m = 3;
    grid.Add("111");
    grid.Add("101");
    grid.Add("111");

    // Function Call
    Console.Write(maxMoney(0, 0, 0, 0));

}
}

// This code is contributed by shikhasingrajput

Javascript

<script>
    // Javascript program for the above approach

    let MAXR = 20, MAXC = 20;
    let cache = new Array(MAXC);
    let dp = new Array(MAXC);
    for(let i = 0; i < MAXR; i++)
    {
        dp[i] = new Array(MAXR);
        cache[i] = new Array(MAXR);
        for(let j = 0; j < MAXC; j++)
        {
            dp[i][j] = new Array(MAXC);
            cache[i][j] = new Array(MAXC);
            for(let k = 0; k < MAXR; k++)
            {
                dp[i][j][k] = new Array(MAXR);
                cache[i][j][k] = new Array(MAXR);
                for(let l = 0; l < MAXR; l++)
                {
                    dp[i][j][k][l] = 0;
                    cache[i][j][k][l] = 0;
                }
            }
        }
    }
    let n, m;
    let grid = [];

    // Function to find maximum money collected
    // when moving from (0, 0) to (N-1, M-1)
    function maxMoney(x1, y1, x2, y2)
    {

        // Out of bounds of grid
        if (x1 >= n || y1 >= m || x2 >= n || y2 >= m)
            return 0;
        if (cache[x1][y1][x2][y2] != 0)
            return dp[x1][y1][x2][y2];

        // Mark state as visited
        cache[x1][y1][x2][y2] = 1;

        // Collect money from the grid cell
        let money = grid[y1][x1]- '0';
        if (x1 != x2 || y1 != y2)
            money += grid[y2][x2] - '0';

        // Take maximum of all possibilities
        dp[x1][y1][x2][y2]
               = money
                 + Math.max(
                       Math.max(maxMoney(x1 + 1, y1, x2 + 1, y2),
                           maxMoney(x1, y1 + 1, x2 + 1, y2)),
                       Math.max(maxMoney(x1 + 1, y1, x2, y2 + 1),
                           maxMoney(x1, y1 + 1, x2, y2 + 1)));
          return dp[x1][y1][x2][y2];
    }

    // Given Input
    n = 3;
    m = 3;
    grid.push("111");
    grid.push("101");
    grid.push("111");

    // Function Call
    document.write(maxMoney(0, 0, 0, 0));

// This code is contributed by divyeshrabadiya07.
</script>

Output

8

Time Complexity: O(2^N*2^M)
Auxiliary Space: O((N*M)²)