0186MediumTwo PointersString

Read the Caption Back to Front

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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 label printer emits a caption strip as the string caption. Whoever typed it was careless with the space bar, so the strip may begin with spaces, end with spaces, or hold several spaces between neighbouring words.

Treat a word as any run of characters that holds no space and cannot be extended. Hand back the caption with its words in the opposite order, joined by exactly one space each, with no space leading or trailing. Padding inside the original strip must not survive into your answer.

Letters keep their case and digits keep their value; nothing about a word itself changes, only where it sits.

Examples

Example 1

Input: caption = " gate 7 open "
Output: "open 7 gate"

The three words come back in the opposite order and the padding at both ends is dropped.

Example 2

Input: caption = "Delta"
Output: "Delta"

A single word has nothing to reorder, and its capital letter stays a capital.

Example 3

Input: caption = "load Bay 22 ready"
Output: "ready 22 Bay load"

The runs of two and three spaces each collapse to one space in the rebuilt caption.

Constraints

1 <= caption.length <= 10^4
caption holds English letters in either case, digits, and the space character ' '
caption holds at least one word

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 read_caption_back(caption: str) -> str:

Java

public String readCaptionBack(String caption)

The intended approach

Time O(n)Space O(n)

Read the strip from its last character towards its first. Skip over any space, then hold that position as the end of a word and walk a second cursor back until it falls off a space or off the front of the strip, which fixes the word's start. Slicing between the two cursors yields the words already in the order the answer wants, so joining them with one space each finishes it.

Pythonentry point: read_caption_back

def read_caption_back(caption):
    # Scan from the tail with two cursors, carving out each word as it is met so
    # the words are collected in the order they belong in the answer.
    # Time O(n). Space O(n) for the pieces of the answer being assembled.
    pieces = []
    end = len(caption) - 1
    while end >= 0:
        if caption[end] == " ":
            end -= 1
            continue
        start = end
        while start >= 0 and caption[start] != " ":
            start -= 1
        pieces.append(caption[start + 1:end + 1])
        end = start - 1
    return " ".join(pieces)

The obvious version

Sweep forward one character at a time, growing the word in hand and pushing it onto a list whenever a space closes it, with a final push for whatever is still in hand. Then glue the list back together walking it from the end.

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_read_back(caption):
    words = []
    current = ""
    for ch in caption:
        if ch == " ":
            if current:
                words.append(current)
                current = ""
        else:
            current += ch
    if current:
        words.append(current)
    out = ""
    for i in range(len(words) - 1, -1, -1):
        out += words[i]
        if i > 0:
            out += " "
    return out

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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