Decode String

Given a valid encoded string s, return its fully decoded form.
The encoding rule is k[encoded_string], meaning that the content inside the brackets is repeated exactly k times. Encoded groups may be nested, and adjacent literal or encoded groups are concatenated.
For this exercise, assume repetition counts are positive decimal integers and literal characters are lowercase English letters.

Function
decodeString(s: String) → String

Examples
Example 1
s = "3[a2[c]]"
return = "accaccacc"
The inner group 2[c] becomes cc, so the outer group is 3[acc].
Example 2
s = "2[abc]3[cd]ef"
return = "abcabccdcdcdef"
Decode the two repeated groups independently, then append the literal suffix ef.

Constraints
1 <= s.length <= 10^5
s is a valid encoding with balanced brackets.
Every repetition count is in [1, 300].
Literal characters are lowercase English letters.
The decoded output length is at most 10^5.

The trick is a stack. Scan left to right. Build the current number digit by digit (num = num * 10 + digit), so 300 parses correctly. Build the current string from letters. On '[', push the current string and the number onto the stack, then reset both. On ']', pop the previous string and the count, and set current = previous + current * count. At the end, current is your answer. The common pitfall is reading only one digit for the count, or forgetting to reset num after pushing. Another is repeated string concatenation in a way that blows up time, but the output cap of 10^5 keeps this safe. A recursive parser works too, with an index pointer. If you freeze on the pop order during the live OA, StealthCoder is the hedge that shows the stack logic while you keep typing.

If you see this problem in your OA tomorrow, the play is to recognize the pattern in 30 seconds. StealthCoder buys you that recognition.