Here is the solution given in Python:

import math

```
def smallest_subarray_sum(s, arr):
min_length = math.inf
for window_start in range(0, len(arr)):
window_sum = 0
window_end = window_start
# shrink the window as small as possible until the 'window_sum' is smaller than 's'
while window_end < len(arr):
window_sum += arr[window_end] # add the next element
if window_sum >=s:
min_length = min(min_length, window_end - window_start + 1)
break
window_end += 1 # Sliding the window
if min_length == math.inf:
return 0
return min_length
```

First, the comment below the for loop says: “# shrink the window as small as possible until the ‘window_sum’ is smaller than ‘s’”. However, we are actually growing the window not shrinking it, and checking if the window_sum is >= s. To me, it seems the worst case runtime is closer to O(n^2) in this solution… for instance take arr = [1, 2, 3, 4, 5] and s = [100], then for each element we iterate over (n-i) elements where i is our current index (going from 0 to 4 in the example) and n is the length of the array (which is 5 in our example).