## Problem Generate every valid generalized abbreviation of a lowercase word. Each character may either remain literal or be included in a positive skip count. Consecutive skipped characters must be represented by one number, never adjacent numeric tokens. Return all abbreviations in lexicographically ascending string order. ### Function Contract Implement `generateGeneralizedAbbreviations(word)` and return a string array. ### Constraints & Assumptions - `0 <= len(word) <= 15`. - `word` contains only lowercase English letters. - The empty word has one abbreviation: the empty string. - Every subset of skipped positions defines exactly one abbreviation, so the result has `2^n` elements. - Numeric tokens have no leading zero. ### Clarifying Questions to Ask - Should adjacent skipped positions be merged? Yes, into one count. - Is `"11"` a valid way to skip two adjacent characters? No; it is parsed as one count of eleven. - Is output order specified? Yes, lexicographically ascending. - May duplicate abbreviations occur? No for a lowercase-letter source under this encoding. ```hint Delay writing a skip count Carry the number of consecutive skipped characters in the recursion. Flush it only when retaining a letter or reaching the end. ``` ### Examples - `word = "word"` produces sixteen abbreviations, including `"word"`, `"1ord"`, `"w1rd"`, `"2rd"`, `"w2d"`, `"3d"`, and `"4"`, returned in lexical order. - `word = "a"` returns `["1", "a"]`. - `word = ""` returns `[""]`. ### Evaluation Focus - Produces every retain-or-skip choice exactly once. - Merges consecutive skipped characters into one count. - Handles the empty word and multi-digit counts. - Uses `O(n)` recursion state apart from the unavoidable `O(n * 2^n)` output, then applies the specified ordering. ### Extensions to Discuss 1. How could results be streamed without storing all of them? 2. What traversal order would produce lexical output without a final sort? 3. How would a maximum number of abbreviated characters prune the recursion?