⚡️ Speed up sorter() by 4211531.56 in PR #1

The function you provided, sorter, is already using Python's built-in sort function which has a time complexity of O(n log n), where n is a number of elements in the array. This is the fastest achievable sorting complexity for comparison-based sorts.

However, if you want to achieve a marginal speed increase, writing this in-place might help.

Here's an alternative version using list comprehension. Although this does not improve the time complexity, it gives a Pythonic touch:

```python
def sorter(arr):
    return sorted(arr)
```

Again, this command returns a new sorted list and does not modify the original list. If you want to sort the list in-place, you only have the original function:



Please note that sorting time complexity cannot be improved further than O(n log n) using comparison-based sorting algorithms. To really optimize this function, you would need a guarantee about the content of your data, for example, if your array only contained integers in a particular range, then you could use counting sort or radix sort, which can have a time complexity of O(n).

code_to_optimize/bubble_sort.py

@@ -0,0 +1,5 @@
+def sorter(arr):
+    arr.sort()
+    return arr
+
+