## Problem Given a string, count its palindromic substrings. Substrings with equal text but different start or end positions are counted separately. ### Function Contract Implement `countPalindromicSubstrings(text)`. ### Constraints & Assumptions - `0 <= len(text) <= 5,000`. - `text` contains Unicode code points; treating each runtime's character unit consistently is acceptable for the exercise. - Single-character substrings are palindromes. - Return a 64-bit integer. ### Clarifying Questions to Ask - Are subsequences included? No, only contiguous substrings. - Are duplicate textual substrings counted once? No, count occurrences by position. - Is the empty substring a palindrome for this count? No. - Is case folded? No, comparison is exact. ```hint Enumerate centers, not substrings Every palindrome has either one character or one gap as its unique center. Expand while the two characters match. ``` ### Examples ```text text = "abc" output = 3 text = "aaa" output = 6 ``` For `"aaa"`, the three length-one, two length-two, and one length-three occurrences all count. ### Evaluation Focus - Checks all `2n - 1` odd and even centers. - Counts occurrences rather than distinct values. - Handles empty input and repeated characters. - Runs in `O(n^2)` worst-case time with `O(1)` extra space. ### Extensions to Discuss 1. How does Manacher's algorithm reduce the running time to `O(n)`? 2. How would you count only distinct palindromic substrings? 3. What changes when comparison must use Unicode grapheme clusters?