> For the complete documentation index, see [llms.txt](https://gl01.gitbook.io/gfg-editorials/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://gl01.gitbook.io/gfg-editorials/2023/08-2023-aug-26/27-reverse-a-string.md).

# 27. Reverse a String

The problem can be found at the following link: [Question Link](https://practice.geeksforgeeks.org/problems/reverse-a-string/1)

## My Approach

Simply used a two-pointer approach to reverse the given string.

* I initialized two pointers, `i` and `j`, to the start and end of the string respectively.
* I swapped the characters at these pointers and moved `i` forward and `j` backward until they met in the middle of the string.

## Time and Auxiliary Space Complexity

* **Time Complexity**: The time complexity of this algorithm is `O(N/2)`, where `N` is the length of the input string. This is because we only need to swap characters up to the middle of the string.
* **Auxiliary Space Complexity**: The algorithm uses only a constant amount of extra space for the two pointers and the swapping operation, so the space complexity is `O(1)`.

## Code (C++)

```cpp
class Solution
{
public:
    string reverseWord(string str)
    {
        int i = 0, j = str.size() - 1;
        while (i < j)
        {
            swap(str[i], str[j]);
            ++i;
            --j;
        }

        return str;
    }
};
```

## Contribution and Support

For discussions, questions, or doubts related to this solution, please visit our [discussion section](https://github.com/getlost01/gfg-potd/discussions). We welcome your input and aim to foster a collaborative learning environment.

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