Gold Mine Problem

Given a gold mine of n*m dimensions. Each field in this mine contains a positive integer which is the amount of gold in tons. Initially the miner is at first column but can be at any row. He can move only (right->,right up /,right down\) that is from a given cell, the miner can move to the cell diagonally up towards the right or right or diagonally down towards the right. Find out maximum amount of gold he can collect. 
Examples:

Input : mat[][] = {{1, 3, 3},
                   {2, 1, 4},
                  {0, 6, 4}};
Output : 12 
{(1,0)->(2,1)->(1,2)}

Input: mat[][] = { {1, 3, 1, 5},
                   {2, 2, 4, 1},
                   {5, 0, 2, 3},
                   {0, 6, 1, 2}};
Output : 16
(2,0) -> (1,1) -> (1,2) -> (0,3) OR
(2,0) -> (3,1) -> (2,2) -> (2,3)

Input : mat[][] = {{10, 33, 13, 15},
                  {22, 21, 04, 1},
                  {5, 0, 2, 3},
                  {0, 6, 14, 2}};
Output : 83

Source Flipkart Interview 

Method 1: Recursion

A simple method that is a direct recursive implementation 

C++

// C++ program to solve Gold Mine problem
#include<bits/stdc++.h>
using namespace std;

int collectGold(vector<vector<int>> gold, int x, int y, int n, int m) {

    // Base condition.
    if ((x < 0) || (x == n) || (y == m)) {
        return 0;
    }


    // Right upper diagonal
    int rightUpperDiagonal = collectGold(gold, x - 1, y + 1, n, m);

     // right
    int right = collectGold(gold, x, y + 1, n, m);

    // Lower right diagonal
    int rightLowerDiagonal = collectGold(gold, x + 1, y + 1, n, m);

    // Return the maximum and store the value
    return  gold[x][y] + max(max(rightUpperDiagonal, rightLowerDiagonal), right);
}

int getMaxGold(vector<vector<int>> gold, int n, int m)
{
    int maxGold = 0;

    for (int i = 0; i < n; i++) {

        // Recursive function call for  ith row.
        int goldCollected = collectGold(gold, i, 0, n, m);
        maxGold = max(maxGold, goldCollected);
    }

    return maxGold;
}

// Driver Code
int main()
{
    vector<vector<int>> gold { {1, 3, 1, 5},
        {2, 2, 4, 1},
        {5, 0, 2, 3},
        {0, 6, 1, 2}
    };
    int m = 4, n = 4;
    cout << getMaxGold(gold, n, m);
    return 0;
}

Java

// Java program to solve Gold Mine problem
class GFG {
  static int collectGold(int[][] gold, int x, int y,
                         int n, int m)
  {

    // Base condition.
    if ((x < 0) || (x == n) || (y == m)) {
      return 0;
    }

    // Right upper diagonal
    int rightUpperDiagonal
      = collectGold(gold, x - 1, y + 1, n, m);

    // right
    int right = collectGold(gold, x, y + 1, n, m);

    // Lower right diagonal
    int rightLowerDiagonal
      = collectGold(gold, x + 1, y + 1, n, m);

    // Return the maximum and store the value
    return gold[x][y]
      + Math.max(Math.max(rightUpperDiagonal,
                          rightLowerDiagonal),
                 right);
  }

  static int getMaxGold(int[][] gold, int n, int m)
  {
    int maxGold = 0;

    for (int i = 0; i < n; i++) {

      // Recursive function call for  ith row.
      int goldCollected
        = collectGold(gold, i, 0, n, m);
      maxGold = Math.max(maxGold, goldCollected);
    }

    return maxGold;
  }
  public static void main(String[] args)
  {
    int[][] gold = { { 1, 3, 1, 5 },
                    { 2, 2, 4, 1 },
                    { 5, 0, 2, 3 },
                    { 0, 6, 1, 2 } };
    int m = 4, n = 4;
    System.out.println(getMaxGold(gold, n, m));
  }
}

// This code is contributed by Karandeep Singh.

Python3

# Python program to solve Gold Mine problem
def collectGold(gold, x, y, n, m):

    # Base condition.
    if ((x < 0) or (x == n) or (y == m)):
        return 0

    # Right upper diagonal
    rightUpperDiagonal = collectGold(gold, x - 1, y + 1, n, m)

     # right
    right = collectGold(gold, x, y + 1, n, m)

    # Lower right diagonal
    rightLowerDiagonal = collectGold(gold, x + 1, y + 1, n, m)

    # Return the maximum and store the value
    return  gold[x][y] + max(max(rightUpperDiagonal, rightLowerDiagonal), right)


def getMaxGold(gold,n,m):

    maxGold = 0

    for i in range(n):

        # Recursive function call for  ith row.
        goldCollected = collectGold(gold, i, 0, n, m)
        maxGold = max(maxGold, goldCollected)

    return maxGold

# Driver Code
gold = [[1, 3, 1, 5],
        [2, 2, 4, 1],
        [5, 0, 2, 3],
        [0, 6, 1, 2]
]

m,n = 4,4
print(getMaxGold(gold, n, m))

# This code is contributed by shinjanpatra.

C#

// C# program to solve Gold Mine problem
using System;

public class GFG{

  static public int collectGold(int[,] gold, int x,
                                int y, int n, int m)
  {

    // Base condition.
    if ((x < 0) || (x == n) || (y == m)) {
      return 0;
    }

    // Right upper diagonal
    int rightUpperDiagonal = collectGold(gold, x - 1, y + 1, n, m);

    // right
    int right = collectGold(gold, x, y + 1, n, m);

    // Lower right diagonal
    int rightLowerDiagonal = collectGold(gold, x + 1, y + 1, n, m);

    // Return the maximum and store the value
    return gold[x,y] + Math.Max(Math.Max(rightUpperDiagonal,
                                         rightLowerDiagonal), right);
  }

  static public int getMaxGold(int[,] gold, int n, int m){
    int maxGold = 0;

    for (int i = 0; i < n; i++) {

      // Recursive function call for  ith row.
      int goldCollected = collectGold(gold, i, 0, n, m);
      maxGold = Math.Max(maxGold, goldCollected);
    }

    return maxGold;
  }

  // Driver Code
  static public void Main (){

    int[,] gold = new int[,] { { 1, 3, 1, 5 },
                              { 2, 2, 4, 1 },
                              { 5, 0, 2, 3 },
                              { 0, 6, 1, 2 } };

    int m = 4, n = 4;
    Console.Write(getMaxGold(gold, n, m));
  }
}

// This code is contributed by shruti456rawal

Javascript

<script>

// JavaScript program to solve Gold Mine problem


function collectGold(gold,x,y,n,m) {

    // Base condition.
    if ((x < 0) || (x == n) || (y == m)) {
        return 0;
    }


    // Right upper diagonal
    let rightUpperDiagonal = collectGold(gold, x - 1, y + 1, n, m);

     // right
    let right = collectGold(gold, x, y + 1, n, m);

    // Lower right diagonal
    let rightLowerDiagonal = collectGold(gold, x + 1, y + 1, n, m);

    // Return the maximum and store the value
    return  gold[x][y] + Math.max(Math.max(rightUpperDiagonal, rightLowerDiagonal), right);
}

function getMaxGold(gold,n,m)
{
    maxGold = 0;

    for (i = 0; i < n; i++) {

        // Recursive function call for  ith row.
        goldCollected = collectGold(gold, i, 0, n, m);
        maxGold = Math.max(maxGold, goldCollected);
    }

    return maxGold;
}

// Driver Code

let gold = [[1, 3, 1, 5],
        [2, 2, 4, 1],
        [5, 0, 2, 3],
        [0, 6, 1, 2]
];

let m = 4, n = 4;
document.write(getMaxGold(gold, n, m));

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

Output

16

  Time complexity: O(3^N*M)

Auxiliary Space: O(N*M)

Method 2: Memoization

Bottom-Up Approach: The second way is to take an extra space of size m*n and start computing values of states 

of right, right upper diagonal, and right bottom diagonal and store it in the 2d array.

C++

// C++ program to solve Gold Mine problem
#include<bits/stdc++.h>
using namespace std;

int collectGold(vector<vector<int>> gold, int x, int y, int n, int m, vector<vector<int>> &dp) {

    // Base condition.
    if ((x < 0) || (x == n) || (y == m)) {
        return 0;
    }

    if(dp[x][y] != -1){
        return dp[x][y] ;
    }

    // Right upper diagonal
    int rightUpperDiagonal = collectGold(gold, x - 1, y + 1, n, m, dp);

     // right
    int right = collectGold(gold, x, y + 1, n, m, dp);

    // Lower right diagonal
    int rightLowerDiagonal = collectGold(gold, x + 1, y + 1, n, m, dp);

    // Return the maximum and store the value
    return dp[x][y] = gold[x][y] + max(max(rightUpperDiagonal, rightLowerDiagonal), right);
}

int getMaxGold(vector<vector<int>> gold, int n, int m)
{
    int maxGold = 0;
    // Initialize the dp vector
    vector<vector<int>> dp(n, vector<int>(m, -1)) ;
    for (int i = 0; i < n; i++) {

        // Recursive function call for  ith row.
        int goldCollected = collectGold(gold, i, 0, n, m, dp);
        maxGold = max(maxGold, goldCollected);
    }

    return maxGold;
}

// Driver Code
int main()
{
    vector<vector<int>> gold { {1, 3, 1, 5},
        {2, 2, 4, 1},
        {5, 0, 2, 3},
        {0, 6, 1, 2}
    };
    int m = 4, n = 4;
    cout << getMaxGold(gold, n, m);
    return 0;
}

Java

// Java program to solve Gold Mine problem
import java.util.*;
class Gold {
  static int collectGold(int[][] gold, int x, int y,
                         int n, int m, int[][] dp)
  {

    // Base condition.
    if ((x < 0) || (x == n) || (y == m)) {
      return 0;
    }

    if (dp[x][y] != -1) {
      return dp[x][y];
    }

    // Right upper diagonal
    int rightUpperDiagonal
      = collectGold(gold, x - 1, y + 1, n, m, dp);

    // right
    int right = collectGold(gold, x, y + 1, n, m, dp);

    // Lower right diagonal
    int rightLowerDiagonal
      = collectGold(gold, x + 1, y + 1, n, m, dp);

    // Return the maximum and store the value
    return dp[x][y] = gold[x][y]
      + Math.max(Math.max(rightUpperDiagonal,
                          rightLowerDiagonal),
                 right);
  }

  static int getMaxGold(int[][] gold, int n, int m)
  {
    int maxGold = 0;
    int[][] dp = new int[n][m];
    for (int row = 0; row < n; row++) {
      Arrays.fill(dp[row], -1);
    }
    for (int i = 0; i < n; i++) {

      // Recursive function call for  ith row.
      int goldCollected
        = collectGold(gold, i, 0, n, m, dp);
      maxGold = Math.max(maxGold, goldCollected);
    }

    return maxGold;
  }
  public static void main(String[] args)
  {
    int[][] gold = { { 1, 3, 1, 5 },
                    { 2, 2, 4, 1 },
                    { 5, 0, 2, 3 },
                    { 0, 6, 1, 2 } };
    int m = 4, n = 4;
    System.out.println(getMaxGold(gold, n, m));
  }
}

// This code is contributed by Karandeep Singh.

Python3

# Python3 program to solve Gold Mine problem
def collectGold(gold, x, y, n, m, dp):

    # Base condition.
    if ((x < 0) or (x == n) or (y == m)):
        return 0

    if(dp[x][y] != -1):
        return dp[x][y]

    # Right upper diagonal
    rightUpperDiagonal = collectGold(gold, x - 1, y + 1, n, m, dp)

        # right
    right = collectGold(gold, x, y + 1, n, m, dp)

    # Lower right diagonal
    rightLowerDiagonal = collectGold(gold, x + 1, y + 1, n, m, dp)

    # Return the maximum and store the value
    dp[x][y] = gold[x][y] + max(max(rightUpperDiagonal, rightLowerDiagonal), right)
    return dp[x][y]


def getMaxGold(gold,n,m):

    maxGold = 0
    # Initialize the dp vector
    dp = [[-1 for i in range(m)]for j in range(n)]

    for i in range(n):

        # Recursive function call for  ith row.
        goldCollected = collectGold(gold, i, 0, n, m, dp)
        maxGold = max(maxGold, goldCollected)

    return maxGold

# Driver Code

gold = [ [1, 3, 1, 5],
        [2, 2, 4, 1],
        [5, 0, 2, 3],
        [0, 6, 1, 2] ]
m,n = 4,4
print(getMaxGold(gold, n, m))

# This code is contributed by Shinjanpatra

C#

// C# program to solve Gold Mine problem
using System;

public class Gold
{
  static int collectGold(int[,] gold, int x, int y,
                         int n, int m, int[,] dp)
  {
    // Base condition.
    if ((x < 0) || (x == n) || (y == m))
    {
      return 0;
    }

    if (dp[x,y] != -1)
    {
      return dp[x,y];
    }

    // Right upper diagonal
    int rightUpperDiagonal = collectGold(gold, x - 1, y + 1, n, m, dp);

    // right
    int right = collectGold(gold, x, y + 1, n, m, dp);

    // Lower right diagonal
    int rightLowerDiagonal = collectGold(gold, x + 1, y + 1, n, m, dp);

    // Return the maximum and store the value
    return gold[x,y] + Math.Max(Math.Max(rightUpperDiagonal,
                                         rightLowerDiagonal), right);
  }

  static int getMaxGold(int[,] gold, int n, int m)
  {
    int maxGold = 0;
    int[,] dp = new int[n, m];
    for (int row = 0; row < n; row++)
    {
      for (int col = 0; col < m; col++)
      {
        dp[row,col] = -1;
      }
    }
    for (int i = 0; i < n; i++)
    {
      // Recursive function call for  ith row.
      int goldCollected = collectGold(gold, i, 0, n, m, dp);
      maxGold = Math.Max(maxGold, goldCollected);
    }
    return maxGold;
  }
  static public void Main ()
  {
    int[,] gold = { { 1, 3, 1, 5 },
                   { 2, 2, 4, 1 },
                   { 5, 0, 2, 3 },
                   { 0, 6, 1, 2 } };
    int m = 4, n = 4;
    Console.Write(getMaxGold(gold, n, m));
  }
}

// This code is contributed by kothavvsaakash

Javascript

<script>

// JavaScript program to solve Gold Mine problem
function collectGold(gold, x, y, n, m, dp)
{

    // Base condition.
    if ((x < 0) || (x == n) || (y == m)) {
        return 0;
    }

    if(dp[x][y] != -1){
        return dp[x][y] ;
    }

    // Right upper diagonal
    let rightUpperDiagonal = collectGold(gold, x - 1, y + 1, n, m, dp);

     // right
    let right = collectGold(gold, x, y + 1, n, m, dp);

    // Lower right diagonal
    let rightLowerDiagonal = collectGold(gold, x + 1, y + 1, n, m, dp);

    // Return the maximum and store the value
    return dp[x][y] = gold[x][y] + Math.max(Math.max(rightUpperDiagonal, rightLowerDiagonal), right);
}

function getMaxGold(gold,n,m)
{
    let maxGold = 0;
    // Initialize the dp vector
    let dp = new Array(n);
    for(let i = 0; i < n; i++)
    {
        dp[i] = new Array(m).fill(-1);
    }
    for (let i = 0; i < n; i++)
    {

        // Recursive function call for  ith row.
        let goldCollected = collectGold(gold, i, 0, n, m, dp);
        maxGold = Math.max(maxGold, goldCollected);
    }

    return maxGold;
}

// Driver Code

let gold = [ [1, 3, 1, 5],
        [2, 2, 4, 1],
        [5, 0, 2, 3],
        [0, 6, 1, 2] ];
let m = 4, n = 4;
document.write(getMaxGold(gold, n, m));

// This code is contributed by Shinjanpatra

</script>

Output

16

Time Complexity :O(m*n)

Space Complexity :O(m*n)

Method 3: Using Dp, Tabulation
Create a 2-D matrix goldTable[][]) of the same as given matrix mat[][]. If we observe the question closely, we can notice following. 

1. Amount of gold is positive, so we would like to cover maximum cells of maximum values under given constraints.
2. In every move, we move one step toward right side. So we always end up in last column. If we are at the last column, then we are unable to move right

If we are at the first row or last column, then we are unable to move right-up so just assign 0 otherwise assign the value of goldTable[row-1][col+1] to right_up. If we are at the last row or last column, then we are unable to move right down so just assign 0 otherwise assign the value of goldTable[row+1][col+1] to right up. 
Now find the maximum of right, right_up, and right_down and then add it with that mat[row][col]. At last, find the maximum of all rows and first column and return it.

C++

// C++ program to solve Gold Mine problem
#include<bits/stdc++.h>
using namespace std;

const int MAX = 100;

// Returns maximum amount of gold that can be collected
// when journey started from first column and moves
// allowed are right, right-up and right-down
int getMaxGold(int gold[][MAX], int m, int n)
{
    // Create a table for storing intermediate results
    // and initialize all cells to 0. The first row of
    // goldMineTable gives the maximum gold that the miner
    // can collect when starts that row
    int goldTable[m][n];
    memset(goldTable, 0, sizeof(goldTable));

    for (int col=n-1; col>=0; col--)
    {
        for (int row=0; row<m; row++)
        {
            // Gold collected on going to the cell on the right(->)
            int right = (col==n-1)? 0: goldTable[row][col+1];

            // Gold collected on going to the cell to right up (/)
            int right_up = (row==0 || col==n-1)? 0:
                            goldTable[row-1][col+1];

            // Gold collected on going to the cell to right down (\)
            int right_down = (row==m-1 || col==n-1)? 0:
                             goldTable[row+1][col+1];

            // Max gold collected from taking either of the
            // above 3 paths
            goldTable[row][col] = gold[row][col] +
                              max(right, max(right_up, right_down));

        }
    }

    // The max amount of gold collected will be the max
    // value in first column of all rows
    int res = goldTable[0][0];
    for (int i=1; i<m; i++)
        res = max(res, goldTable[i][0]);
    return res;
}

// Driver Code
int main()
{
    int gold[MAX][MAX]= { {1, 3, 1, 5},
        {2, 2, 4, 1},
        {5, 0, 2, 3},
        {0, 6, 1, 2}
    };
    int m = 4, n = 4;
    cout << getMaxGold(gold, m, n);
    return 0;
}

Java

// Java program to solve Gold Mine problem
import java.util.Arrays;

class GFG {

    static final int MAX = 100;

    // Returns maximum amount of gold that
    // can be collected when journey started
    // from first column and moves allowed
    // are right, right-up and right-down
    static int getMaxGold(int gold[][],
                              int m, int n)
    {

        // Create a table for storing
        // intermediate results and initialize
        // all cells to 0. The first row of
        // goldMineTable gives the maximum
        // gold that the miner can collect
        // when starts that row
        int goldTable[][] = new int[m][n];

        for(int[] rows:goldTable)
            Arrays.fill(rows, 0);

        for (int col = n-1; col >= 0; col--)
        {
            for (int row = 0; row < m; row++)
            {

                // Gold collected on going to
                // the cell on the right(->)
                int right = (col == n-1) ? 0
                        : goldTable[row][col+1];

                // Gold collected on going to
                // the cell to right up (/)
                int right_up = (row == 0 ||
                               col == n-1) ? 0 :
                        goldTable[row-1][col+1];

                // Gold collected on going to
                // the cell to right down (\)
                int right_down = (row == m-1
                            || col == n-1) ? 0 :
                          goldTable[row+1][col+1];

                // Max gold collected from taking
                // either of the above 3 paths
                goldTable[row][col] = gold[row][col]
                 + Math.max(right, Math.max(right_up,
                                       right_down));

            }
        }

        // The max amount of gold collected will be
        // the max value in first column of all rows
        int res = goldTable[0][0];

        for (int i = 1; i < m; i++)
            res = Math.max(res, goldTable[i][0]);

        return res;
    }

    //driver code
    public static void main(String arg[])
    {
        int gold[][]= { {1, 3, 1, 5},
                        {2, 2, 4, 1},
                        {5, 0, 2, 3},
                        {0, 6, 1, 2} };

        int m = 4, n = 4;

        System.out.print(getMaxGold(gold, m, n));
    }
}

// This code is contributed by Anant Agarwal.

Python3

# Python program to solve
# Gold Mine problem

MAX = 100

# Returns maximum amount of
# gold that can be collected
# when journey started from
# first column and moves
# allowed are right, right-up
# and right-down
def getMaxGold(gold, m, n):

    # Create a table for storing
    # intermediate results
    # and initialize all cells to 0.
    # The first row of
    # goldMineTable gives the
    # maximum gold that the miner
    # can collect when starts that row
    goldTable = [[0 for i in range(n)]
                        for j in range(m)]

    for col in range(n-1, -1, -1):
        for row in range(m):

            # Gold collected on going to
            # the cell on the right(->)
            if (col == n-1):
                right = 0
            else:
                right = goldTable[row][col+1]

            # Gold collected on going to
            # the cell to right up (/)
            if (row == 0 or col == n-1):
                right_up = 0
            else:
                right_up = goldTable[row-1][col+1]

            # Gold collected on going to
            # the cell to right down (\)
            if (row == m-1 or col == n-1):
                right_down = 0
            else:
                right_down = goldTable[row+1][col+1]

            # Max gold collected from taking
            # either of the above 3 paths
            goldTable[row][col] = gold[row][col] + max(right, right_up, right_down)

    # The max amount of gold
    # collected will be the max
    # value in first column of all rows
    res = goldTable[0][0]
    for i in range(1, m):
        res = max(res, goldTable[i][0])

    return res

# Driver code
gold = [[1, 3, 1, 5],
    [2, 2, 4, 1],
    [5, 0, 2, 3],
    [0, 6, 1, 2]]

m = 4
n = 4

print(getMaxGold(gold, m, n))

# This code is contributed
# by Soumen Ghosh.

C#

// C# program to solve Gold Mine problem
using System;

class GFG
{
    static int MAX = 100;

    // Returns maximum amount of gold that
    // can be collected when journey started
    // from first column and moves allowed are
    // right, right-up and right-down
    static int getMaxGold(int[,] gold,
                            int m, int n)
    {

        // Create a table for storing intermediate
        // results and initialize all cells to 0.
        // The first row of goldMineTable gives
        // the maximum gold that the miner
        // can collect when starts that row
        int[,] goldTable = new int[m, n];

        for(int i = 0; i < m; i++)
            for(int j = 0; j < n; j++)
                goldTable[i, j] = 0;

        for (int col = n - 1; col >= 0; col--)
        {
            for (int row = 0; row < m; row++)
            {
                // Gold collected on going to
                // the cell on the right(->)
                int right = (col == n - 1) ? 0 :
                            goldTable[row, col + 1];

                // Gold collected on going to
                // the cell to right up (/)
                int right_up = (row == 0 || col == n - 1)
                            ? 0 : goldTable[row-1,col+1];

                // Gold collected on going
                // to the cell to right down (\)
                int right_down = (row == m - 1 || col == n - 1)
                                ? 0 : goldTable[row + 1, col + 1];

                // Max gold collected from taking
                // either of the above 3 paths
                goldTable[row, col] = gold[row, col] +
                                Math.Max(right, Math.Max(right_up,
                                                    right_down));
            }
        }

        // The max amount of gold collected will be the max
        // value in first column of all rows
        int res = goldTable[0, 0];
        for (int i = 1; i < m; i++)
            res = Math.Max(res, goldTable[i, 0]);
        return res;
    }

    // Driver Code
    static void Main()
    {
        int[,] gold = new int[,]{{1, 3, 1, 5},
                                {2, 2, 4, 1},
                                {5, 0, 2, 3},
                                {0, 6, 1, 2}
                                };
        int m = 4, n = 4;
        Console.Write(getMaxGold(gold, m, n));
    }
}

// This code is contributed by DrRoot_

PHP

<?php
// Php program to solve Gold Mine problem

// Returns maximum amount of gold that
// can be collected when journey started
// from first column and moves allowed are
// right, right-up and right-down
function getMaxGold($gold, $m, $n)
{
    $MAX = 100 ;

    // Create a table for storing intermediate
    // results and initialize all cells to 0.
    // The first row of goldMineTable gives the
    // maximum gold that the miner can collect
    // when starts that row
    $goldTable = array(array());
    for ($i = 0; $i < $m ; $i ++)
        for($j = 0; $j < $n ; $j ++)
            $goldTable[$i][$j] = 0 ;

    for ($col = $n - 1; $col >= 0 ; $col--)
    {
        for ($row = 0 ; $row < $m ; $row++)
        {

            // Gold collected on going to
            // the cell on the right(->)
            if ($col == $n - 1)
                $right = 0 ;
            else
                $right = $goldTable[$row][$col + 1];

            // Gold collected on going to
            // the cell to right up (/)
            if ($row == 0 or $col == $n - 1)
                $right_up = 0 ;
            else
                $right_up = $goldTable[$row - 1][$col + 1];

            // Gold collected on going to
            // the cell to right down (\)
            if ($row == $m - 1 or $col == $n - 1)
                $right_down = 0 ;
            else
                $right_down = $goldTable[$row + 1][$col + 1];

            // Max gold collected from taking
            // either of the above 3 paths
            $goldTable[$row][$col] = $gold[$row][$col] +
                                 max($right, $right_up,
                                             $right_down);
        }
    }

    // The max amount of gold collected will be the
    // max value in first column of all rows
    $res = $goldTable[0][0] ;
    for ($i = 0; $i < $m; $i++)
        $res = max($res, $goldTable[$i][0]);

    return $res;
}

// Driver code
$gold = array(array(1, 3, 1, 5),
              array(2, 2, 4, 1),
              array(5, 0, 2, 3),
              array(0, 6, 1, 2));

$m = 4 ;
$n = 4 ;

echo getMaxGold($gold, $m, $n) ;

// This code is contributed by Ryuga
?>

Javascript

<script>

    // JavaScript program to solve Gold Mine problem

    let MAX = 100;

    // Returns maximum amount of gold that
    // can be collected when journey started
    // from first column and moves allowed
    // are right, right-up and right-down
    function getMaxGold(gold, m, n)
    {

        // Create a table for storing
        // intermediate results and initialize
        // all cells to 0. The first row of
        // goldMineTable gives the maximum
        // gold that the miner can collect
        // when starts that row
        let goldTable = new Array(m);

        for(let i = 0; i < m; i++)
        {
            goldTable[i] = new Array(n);
            for(let j = 0; j < n; j++)
            {
                goldTable[i][j] = 0;
            }
        }

        for (let col = n-1; col >= 0; col--)
        {
            for (let row = 0; row < m; row++)
            {

                // Gold collected on going to
                // the cell on the right(->)
                let right = (col == n-1) ? 0
                        : goldTable[row][col+1];

                // Gold collected on going to
                // the cell to right up (/)
                let right_up = (row == 0 ||
                               col == n-1) ? 0 :
                        goldTable[row-1][col+1];

                // Gold collected on going to
                // the cell to right down (\)
                let right_down = (row == m-1
                            || col == n-1) ? 0 :
                          goldTable[row+1][col+1];

                // Max gold collected from taking
                // either of the above 3 paths
                goldTable[row][col] = gold[row][col]
                 + Math.max(right, Math.max(right_up,
                                       right_down));

            }
        }

        // The max amount of gold collected will be
        // the max value in first column of all rows
        let res = goldTable[0][0];

        for (let i = 1; i < m; i++)
            res = Math.max(res, goldTable[i][0]);

        return res;
    }

    let gold = [ [1, 3, 1, 5],
                  [2, 2, 4, 1],
                  [5, 0, 2, 3],
                  [0, 6, 1, 2] ];

    let m = 4, n = 4;

    document.write(getMaxGold(gold, m, n));

</script>

Output

16

Time Complexity :O(m*n) 
Space Complexity :O(m*n)

Space Complex Solution: In the above-given method we require O(m x n) space. This will not be suitable if the length of strings is greater than 2000 as it can only create 2D array of 2000 x 2000. To fill a row in DP array we require only one row the upper row. For example, if we are filling the i = 10 rows in DP array we require only values of 9th row. So we simply create a DP array of 2 x str1 length. This approach reduces the space complexity. Here is the Python3 implementation of the above-mentioned problem.

Python3

# Python program to solve
# Gold Mine problem

MAX = 100

# Returns maximum amount of
# gold that can be collected
# when journey started from
# first column and moves
# allowed are right, right-up
# and right-down
def getMaxGold(gold, m, n):

    # Create a table for storing
    # intermediate results
    # and initialize all cells to 0.
    # The first row of
    # goldMineTable gives the
    # maximum gold that the miner
    # can collect when starts that row
    goldTable = [[0 for i in range(2)]
                        for j in range(m)]

    for col in range(n-1, -1, -1):
        for row in range(m):

            # Gold collected on going to
            # the cell on the right(->)
            if (col == n-1):
                right = 0
            else:
                right = goldTable[row][(col+1)%2]

            # Gold collected on going to
            # the cell to right up (/)
            if (row == 0 or col == n-1):
                right_up = 0
            else:
                right_up = goldTable[row-1][(col+1)%2]

            # Gold collected on going to
            # the cell to right down (\)
            if (row == m-1 or col == n-1):
                right_down = 0
            else:
                right_down = goldTable[row+1][(col+1)%2]

            # Max gold collected from taking
            # either of the above 3 paths
            goldTable[row][col%2] = gold[row][col] + max(right, right_up, right_down)

    # The max amount of gold
    # collected will be the max
    # value in first column of all rows
    res = goldTable[0][0]
    for i in range(1, m):
        res = max(res, goldTable[i][0])

    return res

# Driver code
gold = [[1, 3, 1, 5],
    [2, 2, 4, 1],
    [5, 0, 2, 3],
    [0, 6, 1, 2]]

m = 4
n = 4

print(getMaxGold(gold, m, n))

# This code is contributed
# by Bhagirath Sarvaiya.

Time Complexity :O(m*n) 
Auxiliary Space :O(m*2)

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