Gold Mine Problem - GeeksforGeeks

27 Apr, 2022
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++
Java
Python3
Javascript
// C++ program to solve Gold Mine problem#include<bits/stdc++.h>using namespace std; int collectGold(vector<vector<int>> gold, int r, int c, int n, int m) {     // Base condition.    if ((r < 0) || (r == n) || (c == m)) {         return 0;    }        // Right upper diagonal    int rightUpperDiagonal = collectGold(gold, r - 1, c + 1, n, m);      // right    int right = collectGold(gold, r, c + 1, n, m);     // Lower right diagonal    int rightLowerDiagonal = collectGold(gold, r + 1, c + 1, n, m);      // Return the maximum and store the value    return  gold[r] + 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 Codeint 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 program to solve Gold Mine problemclass GFG {  static int collectGold(int[][] gold, int r, int c,                         int n, int m)  {     // Base condition.    if ((r < 0) || (r == n) || (c == m)) {      return 0;    }     // Right upper diagonal    int rightUpperDiagonal      = collectGold(gold, r - 1, c + 1, n, m);     // right    int right = collectGold(gold, r, c + 1, n, m);     // Lower right diagonal    int rightLowerDiagonal      = collectGold(gold, r + 1, c + 1, n, m);     // Return the maximum and store the value    return gold[r]      + 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.
# Python program to solve Gold Mine problemdef collectGold(gold, r, c, n, m):     # Base condition.    if ((r < 0) or (r == n) or (c == m)):         return 0     # Right upper diagonal    rightUpperDiagonal = collectGold(gold, r - 1, c + 1, n, m)      # right    right = collectGold(gold, r, c + 1, n, m)     # Lower right diagonal    rightLowerDiagonal = collectGold(gold, r + 1, c + 1, n, m)     # Return the maximum and store the value    return  gold[r] + 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 Codegold = [[1, 3, 1, 5],        [2, 2, 4, 1],        [5, 0, 2, 3],        [0, 6, 1, 2]] m,n = 4,4print(getMaxGold(gold, n, m)) # This code is contributed by shinjanpatra.
<script> // JavaScript program to solve Gold Mine problem  function collectGold(gold,r,c,n,m) {     // Base condition.    if ((r < 0) || (r == n) || (c == m)) {         return 0;    }        // Right upper diagonal    let rightUpperDiagonal = collectGold(gold, r - 1, c + 1, n, m);      // right    let right = collectGold(gold, r, c + 1, n, m);     // Lower right diagonal    let rightLowerDiagonal = collectGold(gold, r + 1, c + 1, n, m);      // Return the maximum and store the value    return  gold[r] + 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>
16
  Time complexity: O(3N*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++
Java
// C++ program to solve Gold Mine problem#include<bits/stdc++.h>using namespace std; int collectGold(vector<vector<int>> gold, int r, int c, int n, int m, vector<vector<int>> &dp) {     // Base condition.    if ((r < 0) || (r == n) || (c == m)) {         return 0;    }       if(dp[r] != -1){        return dp[r] ;    }     // Right upper diagonal    int rightUpperDiagonal = collectGold(gold, r - 1, c + 1, n, m, dp);      // right    int right = collectGold(gold, r, c + 1, n, m, dp);     // Lower right diagonal    int rightLowerDiagonal = collectGold(gold, r + 1, c + 1, n, m, dp);      // Return the maximum and store the value    return dp[r] = gold[r] + 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 Codeint 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 program to solve Gold Mine problemimport java.util.*;class Gold {  static int collectGold(int[][] gold, int r, int c,                         int n, int m, int[][] dp)  {     // Base condition.    if ((r < 0) || (r == n) || (c == m)) {      return 0;    }     if (dp[r] != -1) {      return dp[r];    }     // Right upper diagonal    int rightUpperDiagonal      = collectGold(gold, r - 1, c + 1, n, m, dp);     // right    int right = collectGold(gold, r, c + 1, n, m, dp);     // Lower right diagonal    int rightLowerDiagonal      = collectGold(gold, r + 1, c + 1, n, m, dp);     // Return the maximum and store the value    return gold[r]      + 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.
16
Time Complexity :O(m*n)
Space Complexity :O(m*n)
Method 3: Using Dp, TabulationCreate a 2-D matrix goldTable[][]) of the same as given matrix mat[][]. If we observe the question closely, we can notice following.  
Amount of gold is positive, so we would like to cover maximum cells of maximum values under given constraints.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
Amount of gold is positive, so we would like to cover maximum cells of maximum values under given constraints.
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++
Java
Python3
C#
PHP
Javascript
// 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-downint 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 Codeint 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 program to solve Gold Mine problemimport 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.
# 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-downdef 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 codegold = [[1, 3, 1, 5],    [2, 2, 4, 1],    [5, 0, 2, 3],    [0, 6, 1, 2]] m = 4n = 4 print(getMaxGold(gold, m, n)) # This code is contributed# by Soumen Ghosh.             
// C# program to solve Gold Mine problemusing 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 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-downfunction 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?>
<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>
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Time Complexity :O(m*n) Space Complexity :O(m*n)This article is contributed by Rakesh Kumar. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
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