25. Palindrome String

The problem can be found at the following link: Question Link

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++)

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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