Decode String

Given a valid encoded string s, return its decoded form.
An encoded group has the form k[encodedString], meaning that encodedString is repeated exactly k times. Groups may be nested, and a repeat count may contain multiple digits.
The unencoded text contains lowercase English letters. Digits appear only as repeat counts immediately before bracketed groups.

Function
decodeString(s: String) → String

Examples
Example 1
s = "3[a]2[bc]"
return = "aaabcbc"
Repeat a three times and bc twice, then concatenate the two decoded parts.
Example 2
s = "3[a2[c]]"
return = "accaccacc"
The inner group becomes cc, so the outer group repeats acc three times.
Example 3
s = "2[abc]3[cd]ef"
return = "abcabccdcdcdef"
Decode the two adjacent repeated groups, then retain the trailing literal ef.

Constraints
1 <= s.length <= 30.
s contains lowercase English letters, digits, and square brackets.
s is a valid encoding with well-formed brackets.
Every repeat count is between 1 and 300.
The decoded output length does not exceed 10^5.

The trick is a stack that saves context at every open bracket. Walk the string once. Build the current number digit by digit (so 12[a] works). Build the current string from letters. On '[', push the current string and the current count, then reset both. On ']', pop the previous string and count, and set current to previous + current * count. At the end, current is your answer. The common pitfall is parsing a single digit and breaking on counts like 300 or 10. Another is recursing with wrong index handling when brackets nest. Build with a list and join, not repeated concatenation in a loop. The output cap of 10^5 means output-sized work is fine, so don't over-optimize. If you blank on the stack frame details during the live OA, StealthCoder can supply the solution while you keep typing naturally.

If this hits your live OA and you blank, StealthCoder solves it in seconds, invisible to the proctor.