Decode Repeated Groups

Complete the function below. The function receives the full standard input as a single string and returns the exact standard output lines.
Problem Decode strings that contain repeated parenthesized groups. Plain characters are copied directly. A group has the form (substring){k}, meaning the decoded substring inside the parentheses is repeated k times. Groups may be nested.
For each encoded string, output its decoded form.

Function
solveDecodeRepeatedGroups(input: String) → String[]
Complete solveDecodeRepeatedGroups. It has one parameter, String input. The first line is q, followed by q encoded strings. Return one decoded string per query.

Examples
Example 1
input = "3\nabs(cs){3}g\na(b(c){2}){2}\nx(yz){0}q"
return = ["abscscscsg","abccbcc","xq"]
The second case decodes the nested group b(c){2} as bcc, then repeats it twice.

Constraints
Repeat counts are non-negative integers written inside braces after a closing parenthesis.
Parentheses and braces are balanced in valid inputs.

The trick is a stack of partial results. Walk the string left to right. On a normal character, append it to the current buffer. On '(', push the current buffer onto the stack and start a fresh one. On ')', read the following '{', parse the digits until '}', then pop the previous buffer and set current = previous + current * k. That handles nesting without recursion. The pitfalls are real. A count of 0 must erase the group, as in x(yz){0}q giving xq. Counts can be multi-digit, so don't read a single character. Also, parse the first line as q and split the rest carefully, since the input arrives as one string. Recursion with an index pointer works too, but the stack is harder to break. If the live OA catches you mid-parse, StealthCoder is the hedge that gets you unstuck without the proctor seeing anything.

Drill it cold or hedge it with StealthCoder. Either way, don't walk into the OA hoping you remember the trick.