0934MediumHash TableStringPrefix Sum

Longest Even Stretch After One Swap

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Trains the technique from

LeetCode 3900Longest Balanced Substring After One Swap

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 tape reads s, a string of '0' and '1'.

You may swap the characters at two positions of the tape, at most once. Swapping two equal characters is allowed and simply changes nothing.

A stretch of the tape is even when it holds as many '0' as '1'. Return the length of the longest even stretch the tape can be made to hold.

Examples

Example 1

Input: s = "01110"
Output: 4

Left alone, the longest even stretch is just two characters. The four characters from position 1 to 4 hold three ones and one zero, two too many ones, and a zero sits outside them at the front, so swapping that zero in for one of the ones leaves that stretch even.

Example 2

Input: s = "1111"
Output: 0

Every character is a one, so no swap changes anything and no stretch is ever even.

Example 3

Input: s = "0011"
Output: 4

The whole tape is already even, so no swap is needed.

Constraints

1 <= s.length <= 10^5
s[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 longest_balanced(s: str) -> int:

Java

public int longestBalanced(String s)

The intended approach

Time O(n log n) for a tape of length n: one pass with a search over stored positions per step.Space O(n) for the positions of each running total.

Score each `'1'` as one and each `'0'` as minus one and take running totals; a stretch is even exactly when its two end totals agree, which is the usual reading of this kind of question. The swap adds one wrinkle. Swapping two characters inside a stretch leaves its counts alone, and so does swapping two outside it, so the only swap that helps trades a character inside for one outside, which shifts the stretch's tally by exactly two. So a stretch is reachable when its tally is zero outright, or when its tally is two and a `'0'` sits outside it to trade for one of its `'1'`s, or when its tally is minus two and a `'1'` sits outside. For a fixed right end, the longest stretch with a wanted tally comes from the earliest matching running total, so keep the positions of each running total in order and search them. When the tally is two, the trade is impossible only if the stretch swallows every `'0'` on the tape, which is easy to test: either a `'0'` sits at or past the right end, or the start has to come after the first `'0'`, and a search over the stored positions finds the earliest start that qualifies.

Pythonentry point: longest_balanced

def longest_balanced(s):
    import bisect
    n = len(s)
    running = [0] * (n + 1)
    for i in range(n):
        running[i + 1] = running[i] + (1 if s[i] == "1" else -1)
    where = {}
    for i in range(n + 1):
        where.setdefault(running[i], []).append(i)
    first_zero = -1
    last_zero = -1
    first_one = -1
    last_one = -1
    for i in range(n):
        if s[i] == "0":
            if first_zero < 0:
                first_zero = i
            last_zero = i
        else:
            if first_one < 0:
                first_one = i
            last_one = i
    best = 0
    for end in range(1, n + 1):
        # a stretch (start, end] is even outright
        spots = where.get(running[end])
        if spots:
            start = spots[0]
            if start < end and end - start > best:
                best = end - start
        # tally of two: trade a one inside for a zero outside
        for want, first_out, last_out in ((2, first_zero, last_zero),
                                          (-2, first_one, last_one)):
            spots = where.get(running[end] - want)
            if not spots or first_out < 0:
                continue
            if last_out >= end:
                start = spots[0]
            else:
                at = bisect.bisect_right(spots, first_out)
                if at == len(spots):
                    continue
                start = spots[at]
            if start < end and end - start > best:
                best = end - start
    return best

The obvious version

Try leaving the tape alone and then every single swap of two positions, and for each resulting tape measure the longest even stretch by checking every stretch outright. Quartic in the tape length and it reasons about nothing.

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_balanced(s):
    n = len(s)

    def longest(text):
        best = 0
        for start in range(n):
            ones = 0
            zeros = 0
            for end in range(start, n):
                if text[end] == "1":
                    ones += 1
                else:
                    zeros += 1
                if ones == zeros and end - start + 1 > best:
                    best = end - start + 1
        return best

    best = longest(s)
    letters = list(s)
    for i in range(n):
        for j in range(i + 1, n):
            letters[i], letters[j] = letters[j], letters[i]
            here = longest(letters)
            if here > best:
                best = here
            letters[i], letters[j] = letters[j], letters[i]
    return best

How this problem was checked

Reference solution executed against 20 test cases — 20 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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