Count Palindromic Substrings

shriyanshi24jindal

Mar 28th, 2026

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Java 7.39 KB | None | 0 0

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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;
}
}

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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;
    }
}