Write a method `alternate_sq_sum()` (JS: `alternateSqSum` ) that takes an array of integers as input and finds the sum of squares of the elements at even positions (*i.e.,* 2nd, 4th, *etc.*) plus the sum of the rest of the elements at odd position.

NOTE:
The values at even *position* need to be squared. For a language with zero-based indices, this will occur at oddly-indexed locations. For instance, in Python, the values at indices 1, 3, 5, *etc.* should be squared because these are the second, fourth, and sixth positions in the list.
For Example:

```python
alternate_sq_sum([11, 12, 13, 14, 15]) #should return 379
```

Explanation:

Elements at indices 0, 2, 4 are 11, 13, 15 and they are at odd positions as 11 is at position #1, 13 is at position #3 and 15 at #5.

Elements at indices 1, 3 are 12 and 14 and they are at even position.
So we need to add 11, 13, 15 as they are and square of 12 and 14

--> 11 + 13 + 15 + 12^2 + 14^2
  = 11 + 13 + 15 + 144 + 196
  = 379

For empty arrays, result should be 0 (zero) (except for Haskell).

Difficulty: introductory

Starter code:

def alternate_sq_sum(arr):
	

Solutions:
def alternate_sq_sum(arr):
    return sum([ x**2 if i % 2 == 1 else x for i, x in enumerate(arr) ])

def alternate_sq_sum(arr):
    return sum(x * x for x in arr[1::2]) + sum(arr[::2])

def alternate_sq_sum(arr):
    return sum(n * (n if i % 2 else 1) for i, n in enumerate(arr))

def alternate_sq_sum(arr):
    return sum(num if not i%2 else num**2 for i,num in enumerate(arr))

def alternate_sq_sum(arr):
    return sum(v*v if i&1 else v for i,v in enumerate(arr))

def alternate_sq_sum(arr):
    return sum([x ** 2 if pos % 2 != 0 else x for pos, x in enumerate(arr)])

def alternate_sq_sum(arr):
    return sum(arr[::2]) + sum(map(lambda x : x*x, arr[1::2]))

def alternate_sq_sum(arr):
    return sum(n ** 2 if c % 2 == 1 else n for c, n in enumerate(arr))

def alternate_sq_sum(a):
    return sum(a[i]**2 if i%2 else a[i] for i in range(len(a))) or 0