0322EasyStringStackBracket Sequences

Strip Root Frame Markers

Tracked in this browser only

Trains the technique from

LeetCode 1021Remove Outermost Parentheses

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A sampling profiler writes one character every time control moves: ( when a function is entered and ) when it returns. The finished tape trace is therefore balanced, and it splits into root frames: a root frame is a balanced chunk of the tape that is not contained in any longer balanced chunk, so reading the tape left to right and cutting every time the open-count falls back to zero recovers the root frames in order.

The viewer already draws each root frame as its own row, so the pair of markers that encloses a root frame is redundant. Build the trimmed tape: take every root frame, drop its first and its last character, and concatenate what is left in the original order.

Return the trimmed tape as a string. It may be empty.

Examples

Example 1

Input: s = "(())(()())"
Output: "()()()"

The root frames are `(())` and `(()())`. Dropping the enclosing pair of each leaves `()` and `()()`, which join to `()()()`.

Example 2

Input: s = "((()))"
Output: "(())"

The whole tape is a single root frame, so only its first and last characters go away.

Example 3

Input: s = "()(())"
Output: "()"

The root frames are `()` and `(())`; the first contributes nothing and the second contributes `()`.

Example 4

Input: s = "(()(()))()"
Output: "()(())"

The root frames are `(()(()))` and `()`; the first contributes `()(())` and the second contributes nothing.

Constraints

1 <= s.length <= 10^5
s[i] is either '(' or ')'.
s is balanced: every '(' has a matching ')' after it and the open-count never drops below zero.

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def strip_frame_markers(trace: str) -> str:

Java

public String stripFrameMarkers(String trace)

The intended approach

Time O(n)Space O(n) for the returned string, O(1) beyond it

Track how deep the walk currently is. An opening marker is copied only when the depth is already positive before it, and a closing marker only when the depth is still positive after it, so the one pair that encloses each root frame is the only pair skipped. No explicit list of frames is ever built.

Pythonentry point: strip_frame_markers

def strip_frame_markers(trace):
    # Single left-to-right walk. A marker is copied only while the walk is
    # already inside some root frame, which drops exactly one pair per frame.
    # Time O(n). Space O(n) for the string that is returned.
    kept = []
    depth = 0
    for ch in trace:
        if ch == "(":
            if depth > 0:
                kept.append(ch)
            depth += 1
        else:
            depth -= 1
            if depth > 0:
                kept.append(ch)
    return "".join(kept)

The obvious version

Cut the tape into root frames by watching for the running balance to hit zero, slice the ends off each frame, then join them.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_strip_frame_markers(trace):
    pieces = []
    depth = 0
    start = 0
    for i, ch in enumerate(trace):
        depth += 1 if ch == "(" else -1
        if depth == 0:
            pieces.append(trace[start:i + 1])
            start = i + 1
    return "".join(p[1:-1] for p in pieces)

How this problem was checked

Reference solution executed against 7 test cases — 7 passed
4 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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