You are given a 6 * 6 2D array. An hourglass in an array is a portion shaped like this:

```
a b c
d
e f g
```

For example, if we create an hourglass using the number 1 within an array full of zeros, it may look like this:

```
1 1 1 0 0 0
0 1 0 0 0 0
1 1 1 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
```

Actually, there are many hourglasses in the array above. The three leftmost hourglasses are the following:

```
1 1 1 1 1 0 1 0 0
1 0 0
1 1 1 1 1 0 1 0 0
```

The sum of an hourglass is the sum of all the numbers within it. The sum for the hourglasses above are 7, 4, and 2, respectively.

In this problem you have to *print the largest sum among all the hourglasses* in the array.

**Input Format**

There will be exactly 6 lines, each containing 6 integers seperated by spaces. Each integer will be between -9 and 9 inclusive.

**Output Fromat**

Print the answer to this problem on a single line.

**Sample Input**

```
1 1 1 0 0 0
0 1 0 0 0 0
1 1 1 0 0 0
0 0 2 4 4 0
0 0 0 2 0 0
0 0 1 2 4 0
```

**Sample Output**

```
19
```

**Explanation**

The hourglass which has the largest sum is:

```
2 4 4
2
1 2 4
```

**Solution.java:**

import java.io.*; import java.util.*; import java.text.*; import java.math.*; import java.util.regex.*; public class Solution { public static int getarraysum(int[][] a,int ele_row_i,int ele_col_i) { int sum=0; //neglect fst int fst_row=ele_row_i+1; int fst_col=ele_col_i; //neglect_scnd int sec_row=ele_row_i+1; int sec_col=ele_col_i+2; for(int i=ele_row_i;i<ele_row_i+3;i++){ for(int j=ele_col_i;j<ele_col_i+3;j++){ if(i==fst_row && j==fst_col || i==sec_row && j==sec_col){ continue; }else{ sum=sum+(int)a[i][j]; } } }//end for return sum; } public static void main(String[] args) { Scanner in = new Scanner(System.in); int arr[][] = new int[6][6]; int flag_invalid_entry=0; int max_hour_sum=0; //input for(int i=0; i < 6; i++){ for(int j=0; j < 6; j++){ arr[i][j] = in.nextInt(); if(arr[i][j]<-9 || arr[i][j]>9){ flag_invalid_entry=1; } } } if(flag_invalid_entry==0){ for(int i=0;i<=3;i++){ for(int j=0;j<=3;j++){ int new_sum=getarraysum(arr,i,j); if(max_hour_sum==0 && flag_invalid_entry==0){ max_hour_sum=max_hour_sum+new_sum; flag_invalid_entry=1; }else if(max_hour_sum<new_sum){ max_hour_sum=new_sum; } //System.out.println(new_sum+":sum of hourglass,"+arr[i][j]+" i:"+ i +" j:" + j+"updated_sum:"+max_hour_sum); } } } System.out.println(""+max_hour_sum); } }

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