Given a string, count the number of palindromic contiguous substrings in the string.

The substrings with different start indexes or end indexes are counted as different substrings even they consist of same characters.

**Example:**

Input:"abbcbc"Output:9Explanation:["a", "b", "b", "c", "b", "c", "bb", "bcb", "cbc"]

The core challenge of this problem is to identify all substrings of a given string that are palindromic. A palindrome is a string that reads the same forward and backward. The significance of this problem lies in its applications in text processing, DNA sequence analysis, and other fields where pattern recognition is crucial.

Potential pitfalls include not accounting for overlapping substrings and not efficiently checking for palindromes, which can lead to suboptimal solutions.

To solve this problem, we can use a dynamic programming approach or expand around the center technique. Let's discuss both:

A naive solution would involve generating all possible substrings and checking each one for being a palindrome. This approach is not optimal due to its high time complexity of O(n^3).

A more efficient approach is to expand around each possible center of the palindrome. This method has a time complexity of O(n^2) and is more suitable for this problem.

Here is a step-by-step breakdown of the "Expand Around Center" algorithm:

- Initialize a count to zero.
- Iterate through each character in the string, treating each character and each pair of consecutive characters as potential centers of palindromes.
- For each center, expand outwards while the substring remains a palindrome, incrementing the count for each valid palindrome found.

```
public class PalindromeSubstrings {
// Function to count palindromic substrings
public int countSubstrings(String s) {
int n = s.length();
int count = 0;
// Expand around center approach
for (int center = 0; center < 2 * n - 1; center++) {
int left = center / 2;
int right = left + center % 2;
while (left >= 0 && right < n && s.charAt(left) == s.charAt(right)) {
count++;
left--;
right++;
}
}
return count;
}
// Main method to test the function
public static void main(String[] args) {
PalindromeSubstrings ps = new PalindromeSubstrings();
String input = "abbcbc";
System.out.println("Number of palindromic substrings: " + ps.countSubstrings(input));
}
}
```

The time complexity of the "Expand Around Center" approach is O(n^2) because we are expanding around each center, and there are 2n-1 such centers. The space complexity is O(1) as we are using a constant amount of extra space.

Potential edge cases include:

- Empty string: Should return 0.
- String with all identical characters: Should correctly count all possible palindromic substrings.

Examples:

Input: "" Output: 0 Input: "aaaa" Output: 10

To test the solution comprehensively, consider the following test cases:

- Simple cases with short strings.
- Strings with mixed characters.
- Strings with repeated characters.

Using a testing framework like JUnit can help automate and validate these test cases.

When approaching such problems, consider breaking down the problem into smaller parts and think about different ways to optimize the solution. Practice similar problems to improve your problem-solving skills and understand various algorithmic techniques.

Understanding and solving the problem of counting palindromic substrings is crucial for various applications. By using the "Expand Around Center" approach, we can achieve an efficient solution with a time complexity of O(n^2). Practice and exploration of similar problems will further enhance your skills.

For further reading and practice, consider the following resources: