Hourglass Array Submission error - arrays

Question

I am trying to solve the hourglass problem on hackerrank.you can find the details of problem here (https://www.hackerrank.com/challenges/2d-array).
On my machine code works fine and give correct results even for the testcase that gives error on hackerrank.
Here is the code:
maxSum = -70
#hourglass = []
arr = [[int(input()) for x in range(0,6)] for y in range(0,6)]
for row in range(0,6):
for col in range(0,6):
if (row + 2) < 6 and (col + 2) < 6 :
sum = arr[row][col] + arr[row][col+1] + arr[row][col+2] + arr[row+1][col+1] + arr[row+2][col] + arr[row+2][col+1] + arr[row+2][col+2]
if sum > maxSum:
#hourglass.append(arr[row][col])
#hourglass.append(arr[row][col+1])
#hourglass.append(arr[row][col+2])
#hourglass.append(arr[row+1][col+1])
#hourglass.append(arr[row+2][col])
#hourglass.append(arr[row+2][col+1])
#hourglass.append(arr[row+2][col+2])
maxSum = sum
print(maxSum)
#print(hourglass)
Following error rased while running code:
Traceback (most recent call last):
File "solution.py", line 4, in <module>
arr = [[int(input()) for x in range(0,6)] for y in range(0,6)]
File "solution.py", line 4, in <listcomp>
arr = [[int(input()) for x in range(0,6)] for y in range(0,6)]
File "solution.py", line 4, in <listcomp>
arr = [[int(input()) for x in range(0,6)] for y in range(0,6)]
ValueError: invalid literal for int() with base 10: '1 1 1 0 0 0'
The testcase for which error is raised is:
1 1 1 0 0 0
0 1 0 0 0 0
1 1 1 0 0 0
0 9 2 -4 -4 0
0 0 0 -2 0 0
0 0 -1 -2 -4 0

Answer 1

Solution in Python:
#!/bin/python3
import sys
arr = []
matt = []
v_sum = 0
for arr_i in range(6):
arr_t = [int(arr_temp) for arr_temp in input().strip().split(' ')]
arr.append(arr_t)
for i in range(len(arr)-2):
for j in range(len(arr)-2):
v_sum = 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]
matt.append(v_sum)
total = max(matt)
print (total)

Answer 2

In C# , I can provide you a very simple solution of famous hourglass problem. Below solution has been tested for 10 test cases.
class Class1
{
static int[][] CreateHourGlassForIndexAndSumIt(int p, int q, int[][] arr)
{
int[][] hourGlass = new int[3][];
int x = 0, y = 0;
for (int i = p; i <= p + 2; i++)
{
hourGlass[x] = new int[3];
int[] temp = new int[3];
int k = 0;
for (int j = q; j <= q + 2; j++)
{
temp[k] = arr[i][j];
k++;
}
hourGlass[x] = temp;
x++;
}
return hourGlass;
}
static int findSumOfEachHourGlass(int[][] arr)
{
int sum = 0;
for (int i = 0; i < arr.Length; i++)
{
for (int j = 0; j < arr.Length; j++)
{
if (!((i == 1 && j == 0) || (i == 1 && j == 2)))
sum += arr[i][j];
}
}
return sum;
}
static void Main(string[] args)
{
int[][] arr = new int[6][];
for (int arr_i = 0; arr_i < 6; arr_i++)
{
string[] arr_temp = Console.ReadLine().Split(' ');
arr[arr_i] = Array.ConvertAll(arr_temp, Int32.Parse);
}
int[] sum = new int[16];
int k = 0;
for (int i = 0; i < 4; i++)
{
for (int j = 0; j < 4; j++)
{
int[][] hourGlass = CreateHourGlassForIndexAndSumIt(i, j, arr);
sum[k] = findSumOfEachHourGlass(hourGlass);
k++;
}
}
//max in sum array
Console.WriteLine(sum.Max());
}
}
Thanks,
Ankit Bajpai

Answer 3

Consider the Array of dimension NxN
indexarr = [x for x in xrange(N-2)]
summ=0
for i in indexarr:
for j in indexarr:
for iter_j in xrange(3):
summ += arr[i][j+iter_j] + arr[i+2][j+iter_j]
summ += arr[i+1][j+1]
if i == 0 and j==0:
maxm=summ
if summ > maxm:
maxm = summ
summ = 0
print maxm

Answer 4

This is how I tacked it.
def gethourglass(matrix, row, col):
sum = 0
sum+= matrix[row-1][col-1]
sum+= matrix[row-1][col]
sum+= matrix[row-1][col+1]
sum+= matrix[row][col]
sum+= matrix[row+1][col-1]
sum+= matrix[row+1][col]
sum+= matrix[row+1][col+1]
return sum
def hourglassSum(arr):
maxTotal = -63
for i in range(1, 5):
for j in range(1, 5):
total = gethourglass(arr, i, j)
if total > maxTotal:
maxTotal = total
return maxTotal

Answer 5

Few test cases get failed as we ignore the constraints given for the given problem.
For example,
Constraints
1. -9<=arr[i][j]<=9, it means element of the given array will always between -9 to 9, it can not be 10 or anything else.
2. 0<=i,j<=5
So the max sum will be on range (-63 to 63).
Keep the maxSumValue according to the constraints given or you may use list, append all the sum values, then return the max list value.
Hope this helps in passing your all test cases.

Answer 6

The attractiveness of this algorithm bears a resemblance to CNN (Convolutional Neural Networks); with minor exceptions, such as: 3x3 Kernel size has fixed sparse points (i.e. the [size(3,1), size(1,1), size(3,1)] the second row was delimited by the corners/edges), striding/sliding is always 1 (but, in practice you might change to >=1 (e.g. deep CNN reduces number of filters of a NN, as a heuristic regularization approach to avoid overfitting), and padding was not taken into considerations (i.e. if a list meets its end, instead of continue to next rows(lists), it moves the next ranges to the being of the list, e.g. [0,1,2,3]:
[0,1,2] -> [1,2,3] -> [2,3,0] -> [3,0,1]).
def hourglassSum(arr):
Kernel_size = (3, 3)
stride = 1
memory = []
for i in range(0,Kernel_size[0]+1, stride):
for j in range(0, Kernel_size[1]+1, stride):
hour_glass_sum = sum(arr[i][j:3+j]) + arr[i+1][1+j] + sum(arr[i+2][j:3+j])
memory.append(hour_glass_sum)
return max(memory)