# Hacker Rank Solution For Day 11: 2D Arrays

Problem:- Write a Hackerrank Solution For Day 11: 2D Arrays or Hacker Rank Solution Program In C++ For " Day 11: 2D Arrays " or Hackerrank 30 days of code Java Solution:Day 11: 2D Arrays solution or Hackerrank solution for 30 Days of Code Challenges or Hackerrank 30 days of code Java Solution, Day 11: 2D Arrays solution, or C/C++ Logic & Problem Solving: Day 11: 2D Arrays.

Logic:- So basically the problem is clear we have to find hourglass sum or maximum hourglass sum in a given 2D array. Hourglass is a sum of input in 2D array those shapes are in 'A' graphical. We can understand this by taking an example of a 2D array. Below are an example and solution of this problem. before solving this problem lets first know What is Array?: An array is a group of similar data types, so basically we have to find the maximum sum of A shapes in a Matrix.

Example:- The 6*6 array is given below.

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

an array contains the following hourglasses, and we have to find the maximum sum of 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

As we can see the above hourglasses maximum sum is formed by below hourglass.

2 4 4
2
1 2 4

add all the value 2 + 4 + 4 +2 + 1 + 2 +4 = 19 so this is a solution of above hour glass.

All solution provided here are in C++ (CPP) if any reader wants these solutions in C, and Java comments below or sends a mail with your query like " day n solution in C / C++ / Java. Check the end of the post solutions with the full explanation.

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Tip:- Try to solve this problem with out using vector just use a simple array, and try to the modified the solution to this problem.

Solution:-

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <string>
#include <bitset>
#include <cstdio>
#include <limits>
#include <vector>
#include <climits>
#include <cstring>
#include <cstdlib>
#include <fstream>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
using namespace std;

int main()
{
vector< vector<int> > arr(6,vector<int>(6));
for(int arr_i = 0;arr_i < 6;arr_i++)
{
for(int arr_j = 0;arr_j < 6;arr_j++)
{
cin >> arr[arr_i][arr_j];
}
}
int maxsum=-64;
int hoursum;
for(int i=0;i<4;i++)
{
for(int j=0;j<4;j++)
{
hoursum=arr[i+1][j+1];
for(int k=0;k<3;k++)
{
hoursum = hoursum + arr[i][j+k] + arr[i+2][j+k];
}
if(hoursum > maxsum)
maxsum = hoursum;
}
}
cout<<maxsum;
return 0;
}

Output:-

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