String Encoding / Decoding Variants

Two string parsing variants appear: encode/decode a list of strings using length prefixes and delimiters, and decode a compact alphabet-count expression where `#` forms two-digit letters and `(k)` repeats the previous letter. Requirements Variant A: Encode / Decode String List Input: ["foo", "lis", "jljl", "12345678901"] Encode to a single string using length prefixes: 3#foo3#lis4#jljl11#12345678901 Then implement decode to recover the original list exactly. Variant B: Alphabet Count Decoder Given a compact string such as: "12323#4(3)26#(2)" Decode letters where: 1 -> a, 2 -> b, ..., 9 -> i. 10# -> j, ..., 26# -> z. (k) means the previous decoded letter appears k times total or is repeated k times; clarify exact semantics. Return a length-26 count array. The sample maps to abcwdddzz, so the count array has one a, one b, one c, three d, and two z. Notes For length-prefix encode/decode, avoid delimiter escaping by trusting the numeric length. Decode by reading digits until the first #, then consume exactly length characters. Complexity is O(total characters). For alphabet-count decoding, scanning from right to left is clean because # tells you the previous two digits form one character. Parenthesized repeat counts can also be parsed from the right, applied to the next decoded symbol, then reset. Pitfalls: Strings may contain digits or #, so the delimiter must sit between length and payload (len#payload); appending the delimiter after the payload instead cannot be decoded when a payload is all digits. Empty strings need 0# plus no payload. Repeat semantics must be clarified: (3) may mean three total occurrences or three extra occurrences. Preparation Implement length-prefixed len#payload encoding and decoding from memory. Write the reverse scanner for the alphabet-count variant and test 10#, 26#, and repeated single-digit letters.