> 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-24/25-palindrome-string.md).

# 25. Palindrome String

The problem can be found at the following link: [Question Link](https://practice.geeksforgeeks.org/problems/palindrome-string0817/1)

## My Approach

I have implemented a two-pointer approach.

* I start with two pointers, one at the beginning of the string (`i`) and one at the end of the string (`j`).
* I iterate through the string, comparing characters at these two pointers. If they ever differ, I return `false` as it's not a palindrome.
* If I successfully reach the middle of the string without finding any differing characters, I return `true`.

## Time and Auxiliary Space Complexity

* **Time Complexity**: `O(n)`, where `n` is the length of the input string `S`. We traverse half of the string to determine if it's a palindrome.
* **Auxiliary Space Complexity**: `O(1)`, as we are using a constant amount of extra space to store `i` and `j`.

## Code (C++)

```cpp
class Solution {
public:
    int isPalindrome(string S) {
        int i = 0, j = S.size() - 1;
        while (i < j) {
            if (S[i] != S[j])
                return false;
            ++i;
            --j;
        }
        return true;
    }
};
```

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