Simple Iterative approach, fully explained
Approach\nHere\'s how it works:\n\n1. It uses a for loop with a step of 2*k to iterate over the input string s in chunks of size 2*k. This ensures it processes 
# Approach\nHere\'s how it works:\n\n1. It uses a ```for``` loop with a step of ``2*k`` to iterate over the input string s in chunks of size ``2*k``. This ensures it processes consecutive groups of ``2k`` characters.\n\n2. Within each iteration, it performs the following operations:\n\n    - It slices the string from the beginning up to index ``i``, which represents the starting position of the current group.\n    - It then adds the reversed version of the slice ``s[i:i+k]`` to the result. ``s[i:i+k][::-1]`` reverses the current group of ``k`` characters.\n    - Finally, it adds the remaining part of the string after the reversed group by slicing from index ``i+k`` to the end.\n3. After processing all the groups, the method returns the modified string  ``s``.\nThe code reverses every consecutive group of ``k`` characters and keeps the rest of the string unchanged.\n\n# Complexity\n- #### Time Complexity:\n\n1. The for loop runs ``len(s) / (2*k)`` times, where ``len(s)`` is the length of the input string ``s``.\n2. Within each iteration, the slicing and concatenation operations are performed. Slicing takes $$O(k)$$ time, and concatenation takes $$O(len(s))$$ time in the ``worst case``.\n3. Reversing a substring of length ``k`` using ``[::-1]`` takes $$O(k)$$ time.\nConsidering the worst-case scenario, where every character in the string needs to be reversed, the overall time complexity is $$O(n * k)$$, where ``n`` is the length of the input string ``s``.\n\n- #### Space Complexity:\n\n1. The ``for`` loop and the three variables ``i``, ``s``, and ``k`` require constant space, so they don\'t contribute significantly to the space complexity.\n2. The slicing operation ``s[i:i+k]`` creates a new substring of length ``k``. In the worst case, this substring would be ``k``, so it takes $$O(k)$$ space for each iteration.\n3. The reversed substring ``s[i:i+k][::-1]`` also requires $$O(k)$$ space.\nConsidering the worst-case scenario, where every character in the string needs to be reversed, the overall space complexity is $$O(n + k)$$, where ``n`` is the length of the input string ``s``.\n\n# Code\n```\nclass Solution(object):\n    def reverseStr(self, s, k):\n        for i in range(0,len(s),2*k):\n            s = s[:i] + s[i:i+k][::-1] + s[i+k:]\n        return s\n```
['Python']