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Goldmine

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By Astrio Jather

1. You are given a number n, representing the number of rows.

2. You are given a number m, representing the number of columns.

3. You are given n*m numbers, representing elements of 2d array a, which represents a gold mine.

4. You are standing in front of left wall and are supposed to dig to the right wall. You can start from 

     any row in the left wall.

5. You are allowed to move 1 cell right-up (d1), 1 cell right (d2) or 1 cell right-down(d3).

6. Each cell has a value that is the amount of gold available in the cell.

7. You are required to identify the maximum amount of gold that can be dug out from the mine.

Input Format

A number n

A number m

e11

e12..

e21

e22..

.. n * m number of elements

Output Format

An integer representing Maximum gold available.


Constraints

1 <= n <= 10^2

1 <= m <= 10^2

0 <= e1, e2, .. n * m elements <= 1000

Sample Input

6

6

0 1 4 2 8 2

4 3 6 5 0 4

1 2 4 1 4 6

2 0 7 3 2 2

3 1 5 9 2 4

2 7 0 8 5 1

Sample Output

33


Solution:

import java.io.*;

import java.util.*;


public class Main {


    public static void main(String[] args) throws Exception {

        // write your code here

        Scanner sc = new Scanner(System.in);

        int n = sc.nextInt();

        int m = sc.nextInt();

        int[][] arr = new int[n][m];

        for(int i=0;i<n;i++)

            for(int j=0;j<m;j++)

                arr[i][j] = sc.nextInt();

        

        int dp[][] = new int[n][m];

        for(int i=0;i<n;i++)

            dp[i][0] = arr[i][0];

            

        for(int j=1;j<m;j++){

            for(int i=0;i<n;i++){

                int max = dp[i][j-1];

                if(i-1>=0)

                    max = Math.max(max,dp[i-1][j-1]);

                if(i+1<n)

                    max = Math.max(max,dp[i+1][j-1]);

                dp[i][j] = max + arr[i][j];

            }

        }

        

        int maxGold = 0;

        for(int i=0;i<n;i++){

            maxGold = Math.max(maxGold,dp[i][m-1]);

        }

        System.out.println(maxGold);

    }


}

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