0837HardArrayStringBinary SearchSegment Tree

Most Lit Segments in a Window

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Trains the technique from

LeetCode 3501Maximize Active Section with Trade II

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 segments described by strip, where '1' is a segment that is lit and '0' is one that is dark.

Each entry queries[j] = [lo, hi] asks about the stretch of segments from lo to hi inclusive. The queries are independent: the strip itself is never changed, and every query starts from the strip as given.

Within the stretch a technician gets one repair pass, whose two steps are taken in this order.

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

The two runs need not be the same length, either may be empty, and neither may reach outside the stretch. The steps cannot be reordered or repeated.

Return an array best where best[j] is the largest number of lit segments the j-th stretch can show once its pass is over.

Examples

Example 1

Input: strip = "1100011", queries = [[0, 6], [2, 4], [0, 3]]
Output: [7, 3, 4]

For the whole strip, switching nothing off and switching the three dark segments on shows seven lit segments. For the stretch from 2 to 4 every segment is dark, and switching all three on shows three. For the stretch from 0 to 3 the two dark segments can be switched on, showing four.

Example 2

Input: strip = "0010100", queries = [[0, 6], [2, 4], [1, 5]]
Output: [5, 3, 4]

Across the whole strip, switching off the lit segment at position 3 leaves seven dark segments in a row, and switching them all on shows seven lit segments. The stretch from 2 to 4 reads lit, dark, lit and can show three. The stretch from 1 to 5 can show five.

Example 3

Input: strip = "111", queries = [[0, 2]]
Output: [3]

Every segment is already lit, so the best pass switches nothing off and nothing on.

Constraints

1 <= strip.length <= 10^5
strip[i] is either '0' or '1'
1 <= queries.length <= 10^5
queries[j].length == 2
0 <= queries[j][0] <= 99999
0 <= queries[j][1] <= 99999
Every query satisfies lo <= hi < strip.length

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_in_window(strip: str, queries: list[list[int]]) -> list[int]:

Java

public List<Integer> bestLitInWindow(String strip, int[][] queries)

The intended approach

Time O(n) to prepare, then O(runs spanned) per query.Space O(n) for the runs, the per-position run index and the running counts.

Split the whole strip into maximal runs of equal segments once, note which run each position belongs to, and keep a running count of lit segments so a stretch's starting total is one subtraction. For a stretch, step two lights a run of dark segments, and the dark runs available to it are either one of the strip's own dark runs clipped to the stretch, or two dark runs welded together because step one switched off the whole lit run between them. Welding is only available when that lit run lies wholly inside the stretch and both neighbouring dark runs are inside it too, since neither step may reach outside. Take the largest such run and add it to the starting total.

Pythonentry point: best_lit_in_window

def _prepare(s):
    n = len(s)
    runs = []
    i = 0
    while i < n:
        j = i
        while j < n and s[j] == s[i]:
            j += 1
        runs.append((s[i], i, j - 1))
        i = j
    which = [0] * n
    for idx, (_kind, a, b) in enumerate(runs):
        for p in range(a, b + 1):
            which[p] = idx
    live = [0] * (n + 1)
    for p in range(n):
        live[p + 1] = live[p] + (1 if s[p] == "1" else 0)
    return runs, which, live

def best_lit_in_window(strip, queries):
    # step one can only ever reach inside one lit run, so the dark stretch step
    # two sees is either a single dark run or two dark runs welded together by a
    # lit run that step one switched off entirely
    runs, which, live = _prepare(strip)
    out = []
    for lo, hi in queries:
        base = live[hi + 1] - live[lo]
        first = which[lo]
        last = which[hi]
        gain = 0
        for idx in range(first, last + 1):
            kind, a, b = runs[idx]
            if kind == "0":
                width = min(b, hi) - max(a, lo) + 1
                if width > gain:
                    gain = width
            elif a >= lo and b <= hi and idx - 1 >= first and idx + 1 <= last:
                _k1, la, lb = runs[idx - 1]
                _k2, ra, rb = runs[idx + 1]
                width = (min(lb, hi) - max(la, lo) + 1) + (min(rb, hi) - max(ra, lo) + 1)
                if width > gain:
                    gain = width
        out.append(base + gain)
    return out

The obvious version

For each stretch, try every all-lit run to switch off, including switching none off, then in the result try every all-dark run to switch on, and keep the largest lit count seen. Quartic in the stretch width, but it applies the two steps literally.

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_in_window(strip, queries):
    out = []
    for lo, hi in queries:
        window = list(strip[lo:hi + 1])
        width = len(window)
        best = sum(1 for ch in window if ch == "1")
        picks = [None]
        for a in range(width):
            for b in range(a, width):
                picks.append((a, b))
        for pick in picks:
            work = list(window)
            if pick is not None:
                a, b = pick
                if any(work[p] != "1" for p in range(a, b + 1)):
                    continue
                for p in range(a, b + 1):
                    work[p] = "0"
            lit = sum(1 for ch in work if ch == "1")
            if lit > best:
                best = lit
            for c in range(width):
                for d in range(c, width):
                    if any(work[p] != "0" for p in range(c, d + 1)):
                        continue
                    total = lit + (d - c + 1)
                    if total > best:
                        best = total
        out.append(best)
    return out

How this problem was checked

Reference solution executed against 14 test cases — 14 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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