0884EasyString

Flipping a Strip into Stripes

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Trains the technique from

LeetCode 1758Minimum Changes To Make Alternating Binary String

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 strip of cells reads s, a string of '0' and '1'. One flip changes a single cell to the other character.

The strip is striped when no two neighbouring cells hold the same character. Return the fewest flips that leave the strip striped.

Examples

Example 1

Input: s = "1100110011"
Output: 5

Aiming for the strip that starts with a one leaves five cells wrong, and aiming for the other leaves five as well, so five flips is the best either way.

Example 2

Input: s = "111000111"
Output: 3

Against the version starting with a one, the cells at positions 1, 3, 5, 7 are wrong, which is four flips. The other version needs five.

Example 3

Input: s = "010101010101"
Output: 0

The strip already alternates, so nothing needs flipping.

Constraints

1 <= s.length <= 10^4
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 min_operations(s: str) -> int:

Java

public int minOperations(String s)

The intended approach

Time O(n) for a strip of length n, one pass.Space O(1).

A striped strip is settled entirely by its first cell: from there every cell alternates, so only two strips of the right length are striped at all. That means the answer is the smaller of the two distances to those strips, and there is nothing to search. Walk the strip once comparing each cell against what the version starting with a zero would hold, counting the mismatches. The other version disagrees with the first everywhere, so its distance is the length minus that count, and the answer is the smaller of the two.

Pythonentry point: min_operations

def min_operations(s):
    n = len(s)
    off = 0
    for i in range(n):
        want = "0" if i % 2 == 0 else "1"
        if s[i] != want:
            off += 1
    other = n - off
    return off if off < other else other

The obvious version

Build every string of the same length over the two characters, keep the striped ones, count how many cells each differs from the strip in, and take the smallest count. Exponential in the length and it assumes nothing about which strips are striped.

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_min_operations(s):
    n = len(s)
    best = None
    for mask in range(1 << n):
        built = []
        for i in range(n):
            built.append("1" if mask >> i & 1 else "0")
        striped = True
        for i in range(1, n):
            if built[i] == built[i - 1]:
                striped = False
                break
        if not striped:
            continue
        cost = 0
        for i in range(n):
            if built[i] != s[i]:
                cost += 1
        if best is None or cost < best:
            best = cost
    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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