✅ Explained - Simple and Clear Python3 Code✅
Intuition\nThe problem requires us to reverse the first k characters for every 2k characters in a given string. To solve this, we can iterate over the string an
# Intuition\nThe problem requires us to reverse the first k characters for every 2k characters in a given string. To solve this, we can iterate over the string and use a list to store the reversed segments. By considering the pattern of reversing every 2k characters, we can determine the positions where the reversal needs to occur.\n\n\n# Approach\n1.\tInitialize an empty list, "tab," to store the reversed segments of the string.\n2.\tIterate over the characters in the string using a for loop.\n3.\tFor each character at index i, check if (i//k)%2 is equal to 0. This condition ensures that we are within the first k characters of every 2k segment.\n4.\tIf the condition is true, insert the character at index i in the "tab" list, using the formula (i-i%k). This ensures that the first k characters are reversed and inserted at the correct position.\n5.\tIf the condition is false, append the character to the end of the "tab" list.\n6.\tOnce the loop is complete, initialize an empty string, "res," to store the final result.\n7.\tIterate over the elements in the "tab" list and concatenate them to the "res" string.\n8.\tReturn the "res" string as the reversed string with the specified pattern.\n\n\n\n# Complexity\n- Time complexity:\nThe algorithm iterates over each character in the string once, resulting in a time complexity of O(n), where n is the length of the string.\n\n\n- Space complexity:\nWe use additional space to store the reversed segments in the "tab" list. The space complexity is O(n), where n is the length of the string, as the size of the list grows linearly with the input.\n\n\n# Code\n```\nclass Solution:\n    def reverseStr(self, s: str, k: int) -> str:\n        tab=[]\n        for i in range(len(s)):\n            if (i//k)%2==0:\n               tab.insert(i-i%k,s[i])\n            else:\n                tab.append(s[i])\n        res=""\n        for t in tab:\n            res+=t\n        return res\n```
['Python3']