C++ Error in summing vector - c++

Question

Task:
In a 6x6 array, 16 3x3 array can be constructed. Find the largest sum of the 3x3 array (Source: https://www.hackerrank.com/challenges/30-2d-arrays/problem)
Example:
The largest 3x3 array is
2 4 4
0 2 0
1 2 4
So we print out 19
My Code:
#include <iostream>
#include <vector>
#include <algorithm>
#include <math.h>
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 maxVal = 0;
for (int y = 2; y < 6; y++)
{
for (int x = 2; x < 6; x++)
{
if ( y >=2 && x >=2)
{
int vert = y;
int hori = x;
int first = arr[vert-2][hori] + arr[vert-2][hori-1] + arr[vert-2][hori-2];
int second = arr[vert-1][hori-1] ;
int third = arr[vert][hori] + arr[vert][hori-1] + arr[vert][hori-2];
int sum = first + second + third;
if (sum > maxVal)
maxVal = sum;
}
}
}
cout << maxVal;
}
Edit:
I misunderstood the question, but I fixed the problem. Later on, I still got incorrect test-cases. The original solution seems to be wrong.
Problem: When I submit the code, some of test case return the wrong answer. I tried reading the code and nothing seems wrong.

Answer 1

The problem from the provided link is not about the 3x3 submatrices but about the hourglass shapes:
a b c
d
e f g
In order to get the correct values; your sum of the second row should be:
int second = arr[vert-1][hor-1]
And also, there is no need for the if statement as someone already mentioned in the comments.

Answer 2

Here is a cleaner code. Start from [1][1] and iterate till [length-1][length-1]. And on each iteration, calculate the sum of the surrounding hourglass.
Also, note that you need to sum the hourGlass and NOT the 3x3 subArray.
#include <iostream>
#include <vector>
#include <algorithm>
#include <math.h>
using namespace std;
int main()
{
// user input
//vector< vector<int> > myVector(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 >> myVector[arr_i][arr_j];
// sample input
vector<vector<int>> myVector = {
{ { 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 } },
};
int maxVal = -9999; // start with big negative
for (int y = 1; y < 5; y++)
{
int sum = 0;
for (int x = 1; x < 5; x++)
{
sum = myVector[y - 1][x - 1] + myVector[y - 1][x] + myVector[y - 1][x + 1]; // first row of the hourglass
sum += myVector[y][x]; // second row
sum += myVector[y + 1][x - 1] + myVector[y + 1][x] + myVector[y + 1][x + 1];// third row
if (sum > maxVal)
maxVal = sum;
}
}
cout << maxVal << endl;
system("pause");
return 0;
}

Answer 3

Code
#include <iostream>
#include <vector>
#include <algorithm>
#include <math.h>
using namespace std;
int main(){
int arr[7][7], sum[17], l = 0;
for (int i = 0; i < 6; i++){
for (int j = 0; j < 6; j++){
cin >> arr[i][j];
}
}
for (int i = 0; i < 4; i++){
for (int j = 0; j < 4; j++){
sum[l] = 0;
sum[l] = sum[l] + arr[i][j] + arr[i][j+1] + arr[i][j+2] + arr[i+1][j] + arr[i+1][j+1] + arr[i+1][j+2]+ arr[i+2][j] + arr[i+2][j+1] + arr[i+2][j+2];
l++;
}
}
sort(sum, sum + 17);
cout << sum[16];
}
Explanation
This is a simple problem with small values, so we do not need dynamic programming(YAY!).
Vectors are also quite useless here when we can do it correctly with arrays.
Basically I search every set of 3x3 squares, and find the maximum sum. Hope this helps!