 Sarthak Mehra - 3 years ago 410
C++ Question

# Hourglass sum in 2D array

We are given a (6*6) 2D array of which we have to find largest sum of a hourglass in it.
For example, if we create an hourglass using the number 1 within an array full of zeros, it may look like this: The sum of an hourglass is the sum of all the numbers within it. The sum for the hourglasses above are 7, 4, and 2, respectively. I had written a code for it as follows.It is basically a competitive programming question and as I am new to the field,I have written the code with a very bad compplexity..perhaps so much that the program could not produce the desired output within the stipulated period of time.Below is my code:

``````    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];
}
} //numbers input

int temp; //temporary sum storing variable
int sum=INT_MIN; //largest sum storing variable
for(int i=0;i+2<6;i++) //check if at least3 exist at bottom
{
int c=0; //starting point of traversing column wise for row

while(c+2<6) //three columns exist ahead from index
{
int f=0; //test case variable
while(f!=1)
{ //if array does not meet requirements,no need of more execution

for(int j=c;j<=j+2;j++)
{ //1st and 3rd row middle element is 0 and 2nd row is non 0.
//condition for hourglass stucture
if((j-c)%2==0 && arr[i+1][j]==0||((j-c)%2==1 && arr[i+1][j]!=0)
//storing 3 dimensional subarray sum column wise
temp+=arr[i][j]+arr[i+1][j]+arr[i+2][j]; //sum storage
else
f=1; //end traversing further on failure

if(sum<temp)
sum=temp;

f=1;//exit condition
}//whiel loop of test variable

temp=0; //reset for next subarray execution
c++; /*begin traversal from one index greater column wise till
condition*/
}// while loop of c
}
}

cout<<sum;

return 0;
}
``````

This is my implementation of the code which failed to process in the time interval.Please suggest a better solution considering the time complexity and feel free to point out any mistakes from my side in understanding the problem.The question is from Hackerrank.
Here is the link if you need it anyways:
https://www.hackerrank.com/challenges/2d-array chema989

The solution for your problem is:

``````#include <cstdio>
#include <iostream>
#include <climits>

int main() {
int m;

for (int i = 0; i < 6; ++i) {
for (int j = 0; j < 6; ++j) {
std:: cin >> m[i][j];
}
}

// Compute the sum of hourglasses
long temp_sum = 0, MaxSum = LONG_MIN;
for (int i = 0; i < 6; ++i) {
for (int j = 0; j < 6; ++j) {
if (j + 2 < 6 && i + 2 < 6) {
temp_sum = m[i][j] + m[i][j + 1] + m[i][j + 2] + m[i + 1][j + 1] + m[i + 2][j] + m[i + 2][j + 1] + m[i + 2][j + 2];
if (temp_sum >= MaxSum) {
MaxSum = temp_sum;
}
}
}
}
fprintf(stderr, "Max Sum: %ld\n", MaxSum);

return 0;
}
``````

The algorithm is simple, it sums all the Hourglasses starting of the upper left corner and the last 2 columns and 2 rows are not processed because it can not form hourglasses.

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