Create a function that returns the subarrays of n consecutive elements from the original element that sum up to k. The function will have the following form: sliding_sum(lst, n, k)
To illustrate:
sliding_sum([3, 4, 1, 9, 9, 0, 3, 5, 4], 3, 8) ➞ [[3, 4, 1], [0, 3, 5]]
# Where [3, 4, 1] and [0, 3, 5] are the only subarrays that sum to 8 with length 3.
sliding_sum([1, 4, 2, 3, 5, 0], 2, 5) ➞ [[1, 4], [2, 3], [5, 0]]
sliding_sum([5, 5, 5, 5, 5], 1, 5) ➞ [[5], [5], [5], [5], [5]]
sliding_sum([5, 5, 5, 5, 5], 5, 24) ➞ []
Return an empty array if no subarrays satisfy the (n,k) condition.