2D Array - DS

#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>


Given a 2D Array, :

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

We define an hourglass in to be a subset of values with indices falling in this pattern in 's graphical representation:

a b c
  d
e f g

There are hourglasses in , and an hourglass sum is the sum of an hourglass' values.
Task
Calculate the hourglass sum for every hourglass in , then print the maximum hourglass sum.
Note: If you have already solved the Java domain's Java 2D Array challenge, you may wish to skip this challenge.

Input Format

There are lines of input, where each line contains space-separated integers describing 2D Array ; every value in will be in the inclusive range of to .

Constraints

• 
  

Output Format

Print the largest (maximum) hourglass sum found in .

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

contains the following hourglasses:

1 1 1   1 1 0   1 0 0   0 0 0
  1       0       0       0
1 1 1   1 1 0   1 0 0   0 0 0

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

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

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

The hourglass with the maximum sum () is:

2 4 4
  2
1 2 4

int main(){
    int arr[6][6];
    for(int arr_i = 0; arr_i < 6; arr_i++){
       for(int arr_j = 0; arr_j < 6; arr_j++){
         
          scanf("%d",&arr[arr_i][arr_j]);
       } 
    }
    int sum =-5000;
    for (int i = 1;i<5;i++)
    {
        for (int j=1;j<5;j++)
        {
           int temp = 0;
            temp=arr[i][j]+arr[i-1][j-1]+arr[i-1][j]+arr[i-1][j+1]+arr[i+1][j-1]+arr[i+1][j]+arr[i+1][j+1];
            if (temp>sum)
            {sum = temp;}
        }
    }
    printf("%d",sum);
    return 0;
}