0502EasyStringCounting

Smudged Cradle Log

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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 window-washing cradle begins the day parked at its rest position on the side of a tower. The shift log moves records, in order, what the cradle did during each stage of the day. One character stands for one stage:

'U' means the cradle rose exactly one floor,
'D' means the cradle dropped exactly one floor,
'?' means the entry is smudged, and that stage was either a rise of one floor or a drop of one floor.

The offset at the end of the day is how many floors separate the cradle's final position from its rest position, counted without regard to direction. Each smudged stage may be read either way, and two smudged stages need not be read the same way.

Return the largest offset that is consistent with the log.

Examples

Example 1

Input: moves = "UUDD??"
Output: 2

Reading both smudged stages as drops leaves the cradle two floors below the rest position, so the offset is 2.

Example 2

Input: moves = "DDDU?"
Output: 3

Reading the smudged stage as a drop leaves the cradle three floors below the rest position, so the offset is 3.

Example 3

Input: moves = "??????"
Output: 6

Reading all six smudged stages as rises leaves the cradle six floors above the rest position, so the offset is 6.

Constraints

1 <= moves.length <= 50
moves consists only of the characters 'U', 'D' and '?'.

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 furthest_offset(moves: str) -> int:

Java

public int furthestOffset(String moves)

The intended approach

Time O(n): one pass over the log with three counters.Space O(1): three counters.

Count the clear rises, the clear drops and the smudged stages in one walk over the log. The clear stages leave a fixed net displacement, and since a smudged stage may be read in either direction, all of them can be read to move the cradle further from the rest position on the side the net displacement already sits. The answer is therefore the size of the net displacement plus the number of smudged stages.

Pythonentry point: furthest_offset

def furthest_offset(moves):
    # A single pass tallies the three kinds of stage. The clear stages fix a net
    # displacement; every smudged stage can be read to push the cradle one floor
    # further away from the rest position on whichever side that net already
    # favours, so the tallies alone settle the answer.
    # O(n) time, O(1) space.
    rises = drops = smudged = 0
    for ch in moves:
        if ch == 'U':
            rises += 1
        elif ch == 'D':
            drops += 1
        else:
            smudged += 1
    net = rises - drops
    if net < 0:
        net = -net
    return net + smudged

The obvious version

Tally the fixed displacement of the clear stages, then enumerate all 2^h readings of the h smudged stages with a bitmask, replay each reading to get a final position, and keep the largest distance seen. Exponential in the number of smudged stages.

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_furthest_offset(moves):
    fixed = 0
    holes = 0
    for ch in moves:
        if ch == 'U':
            fixed += 1
        elif ch == 'D':
            fixed -= 1
        else:
            holes += 1
    best = 0
    for mask in range(1 << holes):
        pos = fixed
        for b in range(holes):
            if (mask >> b) & 1:
                pos += 1
            else:
                pos -= 1
        if pos < 0:
            pos = -pos
        if pos > best:
            best = pos
    return best

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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