Count Substrings with Almost-Even Letter Counts

### Problem Count the nonempty contiguous substrings of `word` in which no more than one letter occurs an odd number of times. The input uses only lowercase letters `a` through `j`. Count each occurrence separately, even when two substrings contain identical text. ```cpp class Solution { public: long long wonderfulSubstrings(string word); }; ``` ### Examples **Example 1:** ```text word = "ccc" Output: 6 Explanation: All three length-one, both length-two, and the length-three substring qualify. ``` **Example 2:** ```text word = "abc" Output: 3 Explanation: Only the three single-character substrings have at most one odd count. ``` ### Constraints - `1 <= word.length <= 10^5`. - Characters belong to the first ten lowercase English letters, a through j. ### Clarifying Questions Zero odd-frequency letters is allowed. Two equal substring texts at different positions contribute separately. ```hint Inspect the structure How does XOR of two prefix parity masks describe the substring between them? ``` ### Reasoning checkpoints - Count exactly the substrings meeting the odd-frequency condition, with no empty substring or duplicate counting of one occurrence. - Handle repeated letters and a result that can exceed a signed 32-bit integer. ### Optional extensions How would expanding the permitted alphabet to 26 letters affect time and memory requirements?