https://www.hackerrank.com/challenges/2d-array/problem?isFullScreen=true

#1

#include
using namespace std;
int hourglass(int arr [][6],int r,int c)
{
if(r<=6&&c<=6)
{
int sum=0;
for(int row=r;row<r+2;row++)
{
for(int col=c;col<=c+2;col++)
{
if((col==c&&row==r+1)||((col==c+2)&&row==r+1))
{
continue;
}
else {
sum+=arr[row][col];
}
}
}
static int max=0;
if(sum>max)
{
max=sum;
}
if(r+3==5&&c+3==5)
{
return max;
}
int answer1=hourglass(arr,r,c+1);
int answer2=hourglass(arr,r+1,c);
int answer3=hourglass(arr,r+1,c+1);
if(answer1>answer2)
{
if(answer1>answer3)
{
return answer1;
}
else {
return answer3;
}
}
else {
if(answer2>answer3)
{
return answer2;
}
else {
return answer3;
}
}
}

}
int main()
{
int arr[6][6];
for(int i=0;i<6;i++)
{
for(int j=0;j<6;j++)
{
cin>>arr[i][j];
}
}
int answer=hourglass(arr,0,0);
cout<<answer;
}
plzz tell what is wrong in the code

#2

bro.
U went wrong at many few points, There are cases where the entire hour glass in composed of -ve no,
so return INT_MIN in those case instead of keeping static int max = 0 keep it INT_MIN’

then u could keep the max as a global variable, updating a diff sum variable in each recursive call, and maintaining max of both values

the recursive call for r+1 , c+1 is not required
Here is an updated version oof ur code

#3

maam could you please tell what is the significance of line 28

#4

Basically my thought process was that if we are moving to the 4th row then there need not be any computation done , so i returned INT_MIN ( considering the hour glass has all negative value) .

BUT even if u do not return INT_MIN at that point the function is returning SUM which would give the max_sum so far.
My logic would put the statement like max{sum , INT_MIN, fn(arr, r , c+1))
not return INT_MIN at that point would compute max{sum , sum , fn(arr, r , c+1)}

SO basically to be on the safer side i was returning INT_MIN, but its fine if u don`t. Since SUM is being returned which has the max value so far.

Hope i clarified ur doubt.