Provide functional-programming style solution for 2D Array - DS challenge Problem Statement:
  
  Given a 6×6 2D Array, arr:

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

  
  We define an hourglass in A to be a subset of values with indices
  falling in this pattern in arr's graphical representation:

a b c
  d
e f g

  
  There are 16 hourglasses in arr, and an hourglass sum is the
  sum of an hourglass' values. Calculate the hourglass sum for every
  hourglass in arr, then print the maximum hourglass sum.
  
  For example, given the 2D array:

-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

  
  We calculate the following 16 hourglass values:

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

  
  Our highest hourglass value is 28 from the hourglass:

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. It should
  return an integer, the maximum hourglass sum in the array.
  
  hourglassSum has the following parameter(s):
  
  
  arr: an array of integers
  
  
  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
  
  arr contains the following hourglasses: ![enter image
  description here]2 The hourglass with the maximum sum (19) is:

2 4 4
  2
1 2 4



Following is my imperative style Solution:

def hourglassSum(arr: Array[Array[Int]]): Int = {
  def sum(rowNumber: Int, columnNumber: Int): Int = {
    var oneHourglassSum = 0
    for {
      i &lt;- 0 to 2
      j &lt;- 0 to 2
    } yield {
      val cumulativeSum = if (i == 1 &amp;&amp; (j == 0 || j == 2)) 0
      else arr(i + rowNumber)(j + columnNumber)
      oneHourglassSum += cumulativeSum
    }
    oneHourglassSum
  }

  def max(x: Int, y: Int): Int = if (x &gt; y) x else y

  var hourglassMaxSum = 0
  for {
    rowOffset &lt;- 0 to 3
    columnOffset &lt;- 0 to 3
  } yield {
    hourglassMaxSum = max(sum(rowOffset, columnOffset), hourglassMaxSum)
  }
  hourglassMaxSum
}
There's no need to write a max() method. The Standard Library provides for that.

A for comprehension, with a yield clause, produces a result. So instead of using a mutable var to capture and collect the results of the yield you should capture them directly.

val res = for {...} yield {...}


Or, if you want to process the results more directly, i.e. without the intermediate variable, you can use the somewhat awkward parentheses construct.

def hourglassSum(arr: Array[Array[Int]]): Int = {
  (for {
    x &lt;- 0 to 3
    y &lt;- 0 to 3
  } yield {
    arr(y)(x)   + arr(y)(x+1)   + arr(y)(x+2) +
                  arr(y+1)(x+1) +
    arr(y+2)(x) + arr(y+2)(x+1) + arr(y+2)(x+2)
  }).max
}