**2D Array - DS**

### Problem Statement :

Given a 6*6 2D Array, arr: 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 An hourglass in A is a subset of values with indices falling in this pattern in arr's graphical representation: a b c d e f g There are 16 hourglasses in arr. An hourglass sum is the sum of an hourglass' values. Calculate the hourglass sum for every hourglass in arr, then print the maximum hourglass sum. The array will always be 6*6. Example: arr= -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 The 16 hourglass sums are: -63, -34, -9, 12, -10, 0, 28, 23, -27, -11, -2, 10, 9, 17, 25, 18 The highest hourglass sum is 28 from the hourglass beginning at row 1, column 2: 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. hourglassSum has the following parameter(s): 1. int arr[6][6]: an array of integers Returns 2 .int: the maximum hourglass sum Input Format: Each of the 6 lines of inputs arr[i] contains 6 space-separated integers arr[i][j]. Constraints: -9<=arr[i][j]<=9 0<=i,j<=5 Output Format: Print the largest (maximum) hourglass sum found in arr.

### Solution :

` ````
Solution in C :
In C:
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() {
int i,j,k;
int arr[6][6],temp=-9999,a,b;
for(i=0;i<6;i++)
for(j=0;j<6;j++)
scanf("%d",&arr[i][j]);
for(i=0;i<=3;i++)
for(j=0;j<=3;j++)
{
a = arr[i][j]+arr[i][j+1]+arr[i][j+2]+arr[i+1][j+1]+arr[i+2][j]+arr[i+2][j+1]+arr[i+2][j+2];
if(temp < a)
temp = a ;
}
printf("%d",temp);
return 0;
}
In C++:
#include <cmath>
#include <cstdio>
#include <vector>
#include <climits>
#include <iostream>
#include <algorithm>
using namespace std;
int main() {
int a[6][6],s;
int m=INT_MIN;
for(int i=0;i<6;i++)
{
for(int j=0;j<6;j++)
{
cin>>a[i][j];
}
}
for(int i=0;i<4;i++)
{
for(int j=0;j<4;j++)
{
s=(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]);
if(s>m)
m=s;
}
}
cout<<m;
return 0;
}
In Java:
import java.io.*;
import java.util.*;
public class Solution {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int[][] array = new int[6][6];
for (int y = 0; y < 6; y++){
for (int x =0; x<6; x++){
array[x][y] = sc.nextInt();
}
}
int maxHourglass = getHourglass(array, 1,1);
for (int y=1; y<5; y++){
for (int x=1; x<5; x++){
int hourres = getHourglass(array, x, y);
if (hourres > maxHourglass){
maxHourglass = hourres;
}
}
}
System.out.println(maxHourglass);
}
public static int getHourglass(int[][] array, int x, int y) {
return array[x][y] + array[x-1][y-1] + array[x][y-1] + array[x+1][y-1] + array[x-1][y+1]
+ array[x][y+1] + array[x+1][y+1];
}
}
In Python 3:
def convert(x):
res1=[]
for i in range(len(x)):
res1.append(int(x[i]))
return res1
def cal_values(matrix):
a0 = matrix[0]
a1 = matrix[1]
a2 = matrix[2]
res3=[]
for kk in range(4):
temp1 = 0
temp1 = sum(a0[kk:kk+3])
temp1 += a1[kk + 1]
temp1 += sum(a2[kk:kk+3])
res3.append(temp1)
return max(res3)
ss=[]
result=[]
for i in range(6):
temp=input().split()
res2=convert(temp)
ss.append(res2)
for j in range(4):
ee=cal_values(ss[j:j+3])
result.append(ee)
print(max(result))
```

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