2D Array

2D Array - DS Hackerrank solutions 

Given a 2D 6×6 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 arr'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. Calculate the
hourglass sum for every hourglass in , then print the maximum hourglass sum.
For example, given the 2D array:

 -9 -9 -9 1 1 1
 0 -9 0 4 3 2
 -9 -9 -9 1 2 3
 0 0 8 6 6 0
 0 0 0 -2 0 0
 0 0 1 2 4 0

We calculate the following hourglass values:

    -63, -34, -9, 12,
    -10, 0, 28, 23,
    -27, -11, -2, 10,
     9, 17, 25, 18

 Our highest hourglass value is from the hourglass:

    0 4 3
    1
    8 6 6

Note: If you have already solved the Java domain's Java 2D Array challenge, you may wish to skip this
challenge.
Function Description
Complete the function hourglassSum in the editor below. It should return an integer, the maximum
hourglass sum in the array.
hourglassSum has the following parameter(s):
arr : an array of integers
Input Format

  Each of the lines of inputs contains space-separated integers .

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

arr 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

Solution of this problem is below:-

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

int main(){
    int i,j,a[6][6],max=0;
    for(i=0;i<6;i++){
        for(j=0;j<6;j++){
            scanf("%d",&a[i][j]);
        }
    }
    for(i=0;i<4;i++){
        for(j=0;j<4;j++){
            a[i][j]=a[i][j]+a[i][j+1]+a[i][j+2]+a[i+1][j+1]+a[i+2][j]+a[i+2][j+1]+a[i+2][j+2];  //This line is use for to find pattern like hour glass
            
        }   
    }
max=a[0][0];      // and this is use to find max sum 
for(i=0;i<4;i++){ 
for(j=0;j<4;j++){ 
if(max<=a[i][j]){ 
max=a[i][j];      
}
}
}

 printf("%d",max);
 return 0; 

} 

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