0269EasyArray

Retained Revision Report

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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 build server stamps every artifact it publishes with a signed 32-bit revision number, and a background job prunes old ones, so what is left on disk has holes in it.

revisions holds the revision numbers still on disk, strictly increasing with no repeats. Gather them into stretches of consecutive numbers and describe each stretch with one string:

a stretch running from a through b where b is larger than a is written "a..b"
a stretch holding only the number a is written "a"

Return the descriptions in increasing revision order. Each retained revision must belong to exactly one stretch, and no stretch may stop while the very next number is also on disk, so the report is as short as it can be.

Examples

Example 1

Input: revisions = [-9, -8, -7, -3, 0, 1, 4]
Output: ["-9..-7", "-3", "0..1", "4"]

-9, -8 and -7 follow one another, so one description covers them. -3 sits alone because neither -4 nor -2 is on disk. 0 and 1 pair up, and 4 sits alone.

Example 2

Input: revisions = []
Output: []

Nothing is on disk, so the report is empty.

Example 3

Input: revisions = [2147483645, 2147483646, 2147483647]
Output: ["2147483645..2147483647"]

The three largest revision numbers follow one another and collapse into a single description.

Constraints

0 <= revisions.length <= 20
-2^31 <= revisions[i] <= 2^31 - 1
All the values of revisions are unique.
revisions is sorted in ascending order.

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 retained_revision_report(revisions: list[int]) -> list[str]:

Java

public List<String> retainedRevisionReport(int[] revisions)

The intended approach

Time O(n)Space O(n) for the returned report, O(1) beyond it

Walk the list once. Fix the first revision of a stretch, then advance a second index while each following revision is exactly one more than the one before it. When that stops, the stretch is complete: emit one number if it began and ended at the same revision, otherwise both endpoints joined by two dots, and restart from the next position.

Pythonentry point: retained_revision_report

def retained_revision_report(revisions):
    # One walk over the list: remember where the current run started and close
    # it as soon as the next revision is not exactly one higher.
    # Time: O(n).  Space: O(n) for the report that is returned.
    report = []
    n = len(revisions)
    i = 0
    while i < n:
        start = revisions[i]
        j = i
        while j + 1 < n and revisions[j + 1] == revisions[j] + 1:
            j += 1
        end = revisions[j]
        report.append(str(start) if start == end else str(start) + ".." + str(end))
        i = j + 1
    return report

The obvious version

Load every revision into a set. A revision starts a stretch when its predecessor is missing from the set; from there step forward while the next number is present, then emit that stretch.

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_retained_revision_report(revisions):
    held = set(revisions)
    report = []
    for v in revisions:
        if v - 1 in held:
            continue
        end = v
        while end + 1 in held:
            end += 1
        report.append(str(v) if end == v else str(v) + ".." + str(end))
    return report

How this problem was checked

Reference solution executed against 7 test cases — 7 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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