2D Array - DS solution

 Given a  2D Array, :

1 1 1 0 0 0
0 1 0 0 0 0
1 1 1 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0

An hourglass in  is a subset of values with indices falling in this pattern in 's graphical representation:

a b c
  d
e f g

There are  hourglasses in . An hourglass sum is the sum of an hourglass' values. Calculate the hourglass sum for every hourglass in , then print the maximum hourglass sum. The array will always be .

Example

-9 -9 -9  1 1 1 
 0 -9  0  4 3 2
-9 -9 -9  1 2 3
 0  0  8  6 6 0
 0  0  0 -2 0 0
 0  0  1  2 4 0

The  hourglass sums are:

-63, -34, -9, 12, 
-10,   0, 28, 23, 
-27, -11, -2, 10, 
  9,  17, 25, 18

The highest hourglass sum is  from the hourglass beginning at row , column :

0 4 3
  1
8 6 6

Note: If you have already solved the Java domain's Java 2D Array challenge, you may wish to skip this challenge.

Function Description

Complete the function hourglassSum in the editor below.

hourglassSum has the following parameter(s):

  • int arr[6][6]: an array of integers

Returns

  • int: the maximum hourglass sum

Input Format

Each of the  lines of inputs  contains  space-separated integers .

Constraints

Output Format

Print the largest (maximum) hourglass sum found in .

Sample Input

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

Sample Output

19

Explanation

 contains the following hourglasses:

image

The hourglass with the maximum sum () is:

2 4 4
  2
1 2 4
solution :-
#!/bin/python3

import math
import os
import random
import re
import sys



def hourglassSum(arr):
    # Write your code here
    maxsum=-63
    for i in range(4):
        for j in range(4):
            top=sum(arr[i][j:j+3])
            mid=(arr[i+1][j+1])
            bot=sum(arr[i+2][j:j+3])
            hg=top+mid+bot
            maxsum=max(hg,maxsum)
    return maxsum

if __name__ == '__main__':
    fptr = open(os.environ['OUTPUT_PATH'], 'w')

    arr = []

    for _ in range(6):
        arr.append(list(map(int, input().rstrip().split())))

    result = hourglassSum(arr)

    fptr.write(str(result) + '\n')

    fptr.close()

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