# Charlie Patterson

## Bits of code, reflections, and projects for the sake of learning

I tried out Pramp this morning and encountered an interesting question. The question can be solved with one line of code or by implementing a sorting algorithm. I implemented a selection sorting algorithm with the help of my partner and was delighted to learn about the one-liner approach afterwards. So in this post I’m going to review the problem and explain both approaches. Below is the problem prompt:

### The Question

Absolute Value Sort:

Given an array of integers `arr`, write a function `absSort(arr)` that sorts the array according to the absolute values of the numbers in `arr`. If two numbers have the same absolute value, sort them according to sign, where the negative numbers come before the positive numbers.

``````# Example
input: arr = [2, -7, -2, -2, 0]
output: [0, -2, -2, 2, -7]

# Constraints:
[time limit] 5000ms
[input] array.integer arr
0 <= arr.length <= 10
[output] array.integer
``````

### Implementing Selection Sort

Selection sort is not the most efficient sorting algorithm (O(n^2) vs. O(n log n) for merge sort), but it is relatively straight forward and easy to implement. In selection sort you break the array into two subarrays, sorted and unsorted. The sorted subarray is created by the first pass through the array. You begin the pass at the 0th index, storing that number and its index as the current_minimum, then compare it to every other number in the array, overwriting the current_minimum if you find a smaller number. When all numbers have been compared, you take the smallest number (stored in current_minimum) and swap that number and the number at the start of the pass (in this case the 0th index). Then you repeat the process. Over time the sorted subarray grows and overtakes the unsorted subarray, resulting in a final sorted array.

As you can see below, this algorithm is implemented with a nested for loop. The first for loop tracks the start of our current pass and the second for loop carries out the search for the next smallest number (using absolute value and a number’s sign).

``````def absSort(arr):

for i in range(len(arr)): # Starting index of current pass
cur_min = (arr[i],i) # reset min at start of each pass

for idx in range(i, len(arr)): # For each pass, compare each remaining elements
if abs(arr[idx]) < abs(cur_min):
cur_min = (arr[idx], idx)

elif abs(arr[idx]) == abs(cur_min):
if arr[idx] < 0: # check sign, negative goes first
cur_min = (arr[idx], idx)

else:
continue

# Perform swap after each pass
prev = arr[i]
arr[i] = cur_min
arr[cur_min] = prev

return arr
``````

### Using `sort()` or `sorted()` and Lambda Functions

Both `sort()` and `sorted()` take an optional keyword argument, “key”, that takes a function that determines sorting criteria. However, the two have some important differences. `sort()` is a list method that changes a list in-place. `sorted()` is a built-in function that takes an iterable as an argument and sorts its elements out-of-place. This means with `sort()` the original order of the elements in the list is non-recoverable, while `sorted()` does not alter the original iterable. Also keep in mind `sorted()` is less space efficient because it creates a copy of the input iterable.

The Absolute Value Sort problem does restrict us from changing the input array, so I’ll use sort(). The challenge is then to figure out how to write the key function that will allow us to sort by absolute value, but place negative numbers first if two numbers have the same absolute value. We can accomplish this by writing a lambda function that will produce a tuple to use as the key. The first value of the tuple is the absolute value of the element and the second value is the element. If the absolute value (the first element of the key) is the same, then the sorting algorithm will default to comparing the second element of the key, the original value of the element. This will cause the negative number to always be placed first.

Both sorting functions use the Timsort algorithm developed by early Python Core Developer Tim Peters, which has a time complexity of O(n log n). This is faster than our O(n^2) solution using Selection Sort.

``````def absSort(arr):
arr.sort(key=lambda x: (abs(x),x))
return arr
# not technically a one-liner with sort(), but space complexity is O(1)
``````

Remember that `arr.sort(...)` will return None, not the sorted list. That is why we return arr, not the statement above, which makes this function a “two-liner”.

``````def absSort(arr):
return sorted(arr, key=lambda x: (abs(x),x))
# space complexity is O(n)
``````