0087HardStringDynamic ProgrammingStackBracket Sequences

Longest Clean Trace Window

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Trains the technique from

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 profiler writes one character into trace per event it sees: ( when a span opens and ) when a span closes. No other character is written, and a run with no events at all leaves trace empty.

A window is a stretch of neighbouring characters of trace. A window is clean when its opens and closes pair off completely inside it: every ) in the window is answered by an unused ( sitting earlier in the same window, and no ( in the window is left waiting when the window ends.

Return the length of the longest clean window. When no window of one character or more is clean, return 0.

Examples

Example 1

Input: trace = "))(())()"
Output: 6

Dropping the two stray closes at the front leaves the last six characters, which pair off completely.

Example 2

Input: trace = "()(()"
Output: 2

The first two characters pair off, and so do the last two, but no clean window reaches past the open at position 2.

Example 3

Input: trace = "(("
Output: 0

Both spans are still waiting when the trace ends, so nothing pairs off.

Constraints

0 <= trace.length <= 3 * 10^4
Every character of trace is either '(' or ')'

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 longest_clean_window(trace: str) -> int:

Java

public int longestCleanWindow(String trace)

The intended approach

Time O(n)Space O(n)

Keep a stack of the positions of opens still waiting, sitting on top of one barrier position that marks where the current clean stretch may start from. An open pushes its position; a close pops. If the pop empties the stack the close itself was stray, so it becomes the new barrier; otherwise everything after the position now on top has paired off, and the gap between it and the current index is a clean window. One pass, tracking the largest such gap.

Pythonentry point: longest_clean_window

def longest_clean_window(trace):
    pending = [-1]
    best = 0
    for i in range(len(trace)):
        if trace[i] == "(":
            pending.append(i)
        else:
            pending.pop()
            if not pending:
                pending.append(i)
            else:
                span = i - pending[-1]
                if span > best:
                    best = span
    return best

The obvious version

Try every starting position and walk forward keeping a running depth, abandoning the start as soon as the depth would go negative and recording the length whenever the depth returns to zero. O(n^2) with no stack at all.

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_longest_clean_window(trace):
    n = len(trace)
    best = 0
    for start in range(n):
        depth = 0
        for stop in range(start, n):
            if trace[stop] == "(":
                depth += 1
            else:
                depth -= 1
            if depth < 0:
                break
            if depth == 0 and stop - start + 1 > best:
                best = stop - start + 1
    return best

How this problem was checked

Reference solution executed against 15 test cases — 15 passed
3 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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