Merge pull request ephremdeme#150 from Druffl3/dev

Added hourglass Array Datastructure problem solution

Author: ephremdeme

Arrays/maxHourGlass.py

@@ -0,0 +1,59 @@
+# This python3 script accepts a nxn two dimensional matrix
+# and calculates the maximum hourglass sum possible from it.
+
+'''
+Example:
+In this 6x6 Matrix:
+
+1 1 1 0 0 0
+0 1 0 0 0 0
+1 1 1 0 0 0
+0 0 0 1 1 1
+0 0 0 0 1 0
+0 0 0 1 1 2
+
+The pattern:
+1 1 1
+0 1 0
+1 1 1
+
+makes 1 hour glass. And the sum of this hour glass is:
+1 + 1 + 1 + 1 + 1 + 1 + 1 = 7
+Similarly we need to find sum of all hourglasses in the Matrix
+And print the maximum sum.
+'''
+
+#Taking rank of the matrix from user
+n = int(input())
+l = []
+
+#Converting each string of row values into list and appending it to The
+#main two dimensional list 'l'
+for i in range(n):
+ il = list(map(int, input().split()))
+ l.append(il)
+
+#This will store our expected result
+finalSum = 0
+
+#Creating a 3x3 grid index to iterate over the two dimensional list And
+#calculate sum of the hourglasses.
+#-2 is added to make sure not to index beyond the actual list range.
+for r1 in range(n-2):
+ r2 = r1+1
+ r3 = r2+1
+ for c1 in range(n-2):
+ #to store sum of all hourglasses
+ res = 0
+ c2 = c1+1
+ c3 = c2+1
+ res = l[r1][c1] + l[r1][c2] + l[r1][c3] + l[r2][c2] + l[r3][c1] + l[r3][c2] + l[r3][c3]
+ #Maybe the first element is -ve, therefore need to store it no matter what.
+ if r1 == 0 and c1 == 0:
+ finalSum = res
+ #will always store the maximum result
+ if res > finalSum:
+ finalSum = res
+
+#print the maximum hourglass sum.
+print(finalSum)