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: 9 Explanation: ["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 missing substrings that are palindromic or counting the same substring multiple times. It's important to ensure that each unique palindromic substring is counted correctly.
To solve this problem, we can start with a naive approach and then move to more optimized solutions.
The naive solution involves generating all possible substrings of the given string and checking each one to see if it is a palindrome. This approach is not optimal because it has a high time complexity due to the large number of substrings generated and checked.
We can optimize the solution using dynamic programming or expanding around the center technique.
In this approach, we use a 2D array to store whether a substring is a palindrome. We fill this table in a bottom-up manner.
This approach involves expanding around each character and each pair of characters in the string to find all palindromic substrings. This method is more efficient and has a time complexity of O(n^2).
1. Initialize a count to 0.
2. For each character in the string, expand around the center for both odd and even length palindromes.
3. Increment the count for each palindrome found.
// Function to count palindromic substrings
function countPalindromicSubstrings(s) {
let count = 0;
// Helper function to expand around the center
function expandAroundCenter(left, right) {
while (left >= 0 && right < s.length && s[left] === s[right]) {
count++;
left--;
right++;
}
}
// Iterate over each character and expand around center
for (let i = 0; i < s.length; i++) {
// Odd length palindromes
expandAroundCenter(i, i);
// Even length palindromes
expandAroundCenter(i, i + 1);
}
return count;
}
// Example usage
const input = "abbcbc";
console.log(countPalindromicSubstrings(input)); // Output: 9
The time complexity of the expand around center approach is O(n^2), where n is the length of the string. This is because we are expanding around each character and each pair of characters. The space complexity is O(1) as we are not using any extra space proportional to the input size.
Potential edge cases include:
Testing these edge cases ensures the robustness of the solution.
To test the solution comprehensively, consider the following test cases:
Using a testing framework like Jest can help automate and validate these test cases.
When approaching such problems, it's important to:
Understanding and solving the problem of counting palindromic substrings is crucial for various applications in computer science. By exploring different approaches and analyzing their complexities, we can develop efficient solutions. Practice and continuous learning are key to mastering such problems.
For further reading and practice, consider the following resources:
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