0493MediumStringEnumeration

Sign Segment Repair Pass

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Trains the technique from

LeetCode 3499Maximize Active Section with Trade I

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 shop sign is a strip of n segments described by the string strip, where '1' is a segment that is currently lit and '0' is one that is dark.

A technician gets one repair pass over the strip, and the pass has two steps taken in this order:

Pick a stretch of consecutive segments that are all lit as the strip stands, and switch every segment in that stretch off.
Pick a stretch of consecutive segments that are all dark in the strip step 1 left behind, and switch every segment in that stretch on.

Each stretch must be consecutive and must be uniform in the state its step calls for. Either stretch is allowed to be empty, and the two steps cannot be reordered or repeated.

Return the largest number of lit segments the sign can show once the pass is over.

Examples

Example 1

Input: strip = "1101000110"
Output: 9

Switching off the lit segment at index 3 welds the dark segment at index 2 to the dark stretch at indices 4 through 6, and switching that whole stretch of five on leaves nine lit.

Example 2

Input: strip = "000101000"
Output: 6

Switching off the lit segment at index 3 joins the dark stretch at indices 0 through 2 to the dark segment at index 4, and switching those four on leaves six lit.

Example 3

Input: strip = "10011101"
Output: 8

Switching off nothing and then switching on the dark stretch at indices 1 and 2 leaves seven lit.

Constraints

1 <= n == strip.length <= 10^5
strip[i] is either '0' or '1'.

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 best_lit_after_pass(strip: str) -> int:

Java

public int bestLitAfterPass(String strip)

The intended approach

Time O(n)Space O(1)

A stretch that is uniformly lit cannot span a dark segment, so step 1 always sits inside one run of lit segments. That means the dark block available to step 2 is either an existing dark run on its own, or two consecutive dark runs joined by switching off the entire lit run between them. Sweep the strip once counting lit segments and measuring dark runs, holding the previous dark run's length, and track the best of every single run and every neighbouring pair. Add that best gain to the original lit count.

Pythonentry point: best_lit_after_pass

def best_lit_after_pass(strip):
    # Step one can only reach inside a single run of lit segments, so the dark
    # block that step two sees is one dark run, or two dark runs welded together
    # by the lit run between them once that whole lit run is switched off. Walk
    # the strip once, keeping the length of the previous dark run to hand.
    # time O(n), space O(1)
    n = len(strip)
    lit = 0
    best_gain = 0
    prev_dark = 0
    i = 0
    while i < n:
        if strip[i] == '1':
            lit += 1
            i += 1
        else:
            j = i
            while j < n and strip[j] == '0':
                j += 1
            run = j - i
            if run > best_gain:
                best_gain = run
            if prev_dark and prev_dark + run > best_gain:
                best_gain = prev_dark + run
            prev_dark = run
            i = j
    return lit + best_gain

The obvious version

Enumerate every uniformly lit stretch as the step 1 choice, including the empty one. For each, build the intermediate strip, then enumerate every uniformly dark stretch of that intermediate strip as the step 2 choice and count the lit segments that result. Keeps the largest count seen.

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_best_lit_after_pass(strip):
    n = len(strip)
    start = list(strip)
    best = start.count('1')
    firsts = [None]
    for a in range(n):
        for b in range(a, n):
            ok = True
            for t in range(a, b + 1):
                if start[t] != '1':
                    ok = False
                    break
            if ok:
                firsts.append((a, b))
    for first in firsts:
        mid = list(start)
        if first is not None:
            for t in range(first[0], first[1] + 1):
                mid[t] = '0'
        here = mid.count('1')
        if here > best:
            best = here
        for a in range(n):
            for b in range(a, n):
                ok = True
                for t in range(a, b + 1):
                    if mid[t] != '0':
                        ok = False
                        break
                if ok and here + (b - a + 1) > best:
                    best = here + (b - a + 1)
    return best

How this problem was checked

Reference solution executed against 11 test cases — 11 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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