LeetCode Streak Challenge - Day 1/30 - Reverse list

Problem 1: Reversing a list

Today's daily challenge question was about reversing lists. This topic has been discussed to death in many forums already. But I have found that trying to reproduce what I learnt/read, helps increase the retain the piece of information for longer in memory. Also I want to try and keep notes for each problem I solve so that I can revisit them later. There are different ways to go about reversing a list in Python:

1. list.reverse() -
   • Reversing is done in place - It does not create a new object and reverse it. Instead it modifies the original copy.
   • The function does not return the modified list. We cannot assign the returned value of the function to a variable. The return value is None.
   • Another term for operating in place is side-effect.
   • Time complexity - O(n) where n is the number of elements in the list.
   • Space complexity - O(n).
2. reversed(list) -
   • It returns an iterator that allows traversing the list in reverse order.
   • We can call a list on the returned iterator to get the reversed list.
   • Any changes to the original list will impact the iterator because the iterator points to the same list and not a copy.
   • Time complexity - O(n) where n is the number of elements in the list.
   • Space complexity - O(n).
3. Swap the first element with the last, the second element with the second last, and so on.
   • Time complexity - O(n).
   • Space complexity - O(n).
4. Slicing
   • list[start:stop:step]. The default value of step is one, which allows the slicing to extract items from start to stop-1 from left to right. Setting the step to -1 returns the objects in reverse order.
   • It returns a reversed copy of the list.
   • Time complexity - O(n).
   • Space complexity - O(n).
5. Loops
   • We can loop through the list in reverse order and copy each element to a new list. Needless to say, this is not in place.
   • Time complexity - O(n).
   • Space complexity - O(n).