class Solution {
    /**
    A palindromic substring expands symmetrically from its center.
    Every palindrome has a center
    A string of length n has 2n − 1 possible centers

    For each index i, we consider:
        ✅ Odd-length palindrome
        Center at i
        Example: "aba"

        ✅ Even-length palindrome
        Center between i and i+1
        Example: "abba"
        
    Expand Around Center
        For each center:
            Move left pointer → left--
            Move right pointer → right++
        Continue while:
        Characters match
        Indices are valid
    Each valid expansion = 1 palindrome, so do count

    Count All Palindromes
        Loop through all indices:
            At each index:
                Count odd palindromes
                Count even palindromes
                Add both to total
    Return Result   
        Total count gives all palindromic substrings.
    */
    public int countSubstrings(String s) {
        int count = 0;
        for(int i=0; i<s.length(); i++)
        {
            int oddLength = countPalindromes(s, i, i);
            int evenLength = countPalindromes(s, i, i+1);
            count += (oddLength + evenLength);
        }
        return count;
    }

    public int countPalindromes(String s, int left, int right)
    {
        int count = 0;
        while(left >= 0 && right < s.length() && s.charAt(left) == s.charAt(right))
        {
            count++;
            left--;
            right++;
        }
        return count;
    }
}