Count Three-Character Windows Without Repetition

### Problem Count length-three contiguous substrings of `s` whose characters are all different. Count every occurrence, including overlapping occurrences and repeated copies of the same three-letter string. ```cpp class Solution { public: int countGoodSubstrings(string s); }; ``` ### Examples **Example 1:** ```text s = "abcabc" Output: 4 Explanation: Windows abc, bca, cab, and abc all qualify. ``` **Example 2:** ```text s = "aabb" Output: 0 Explanation: Both length-three windows repeat a character. ``` ### Constraints - `1 <= s.length <= 100`. - Characters are lowercase English letters. ### Clarifying Questions Count occurrence positions rather than distinct substring texts. Fewer than three characters means no candidate window. ```hint Inspect the structure Are you counting distinct substring texts or every eligible starting position, including overlaps? ``` ### Reasoning checkpoints - Enforce pairwise distinctness across all three characters. - Count eligible start positions with overlap and repeated substring text handled correctly. ### Optional extensions How would a sliding frequency table support an arbitrary window length?