C++ 📈 || Beats 100% 🚀 || Reverse String ll || Brute Force 🔥|| CLearCode and Efficient Approach
IntuitionMy first thought for solving this problem was to use an iterative approach that processes the string in chunks. The problem explicitly states a pattern
# Intuition
My first thought for solving this problem was to use an iterative approach that processes the string in chunks. The problem explicitly states a pattern: reverse the first k characters in every 2*k chunk of the string. This means the task can be broken into manageable segments while ensuring we handle cases like string length being less than k or between k and 2*k.

# Approach
1.Iterate Through the String:

->Use a loop to traverse the string in steps of 2*k.
->For each step, focus only on reversing the first k characters in that chunk.

2.Reversal Logic:

->Define the start index as the beginning of the current chunk.
->Calculate the end index as min(start + k - 1, s.size() - 1) to avoid going out of bounds.
->Reverse the characters between these indices using a two-pointer approach.

3.Handle Edge Cases:

->If the string has fewer than k characters, reverse the entire string.
->If the string has between k and 2*k characters, reverse the first k characters and leave the rest unchanged.

4.Return the Modified String:

->After processing all chunks, return the modified string.

# Complexity
- Time complexity:
O(n)

The string is traversed once, and reversing a chunk of k characters takes 𝑂(𝑘). Since 𝑘 is a constant and each character is processed once, the overall complexity is linear relative to the string length.

- Space complexity:
O(1)

The operation is performed in-place, so no additional space is required apart from a few variables for indices and pointers.

# Code
```cpp []
class Solution {
public:
    string reverseStr(string s, int k) {
        for (int i = 0; i < s.size(); i += 2 * k) {

            int start = i; 
            int end = min(i + k - 1, (int)s.size() - 1); 
            while (start < end) {
                swap(s[start], s[end]);
                start++;
                end--;
            }
        }
        return s;
    }
};

```
['String', 'C++']