0790MediumStringSliding Window

Longest Complete Call Stretch

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Trains the technique from

LeetCode 1839Longest Substring Of All Vowels in Order

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 wetland recorder listens all night and tags every call it hears with one of five letters: a, e, i, o, u. The tags are written down in the order the calls arrived, giving the string calls.

The five tags have a fixed ranking, a then e then i then o then u. A stretch is a run of consecutive tags from the log. A stretch is complete when both of these hold:

all five tags appear in it at least once, and
reading it from left to right, no tag ever ranks lower than the tag before it (so each tag either repeats the previous one or ranks higher).

Return the number of calls in the longest complete stretch of calls. If the log holds no complete stretch, return 0.

Examples

Example 1

Input: calls = "aaeeiioouu"
Output: 10

The tags arrive as the blocks `aa`, `ee`, `ii`, `oo`, `uu`. All five tags appear and no tag ranks below the tag before it, so the whole log of ten calls is one complete stretch.

Example 2

Input: calls = "uaeiouu"
Output: 6

The last six tags read `a e i o u u`: all five tags appear and none ranks below its predecessor, so that stretch is complete and holds six calls. The opening `u` cannot belong to it, because the tag after it ranks lower.

Example 3

Input: calls = "aeio"
Output: 0

No call is ever tagged `u`, so no stretch holds all five tags.

Constraints

1 <= calls.length <= 5 * 10^5
Every character of calls is one of 'a', 'e', 'i', 'o', 'u'.

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_complete_stretch(calls: str) -> int:

Java

public int longestCompleteStretch(String calls)

The intended approach

Time O(n) for a log of n tags: each tag is looked at once and does constant work.Space O(1): five ranks plus a handful of counters.

Walk the log once, holding the length of the stretch that ends at the current tag and the number of distinct tags in it. A tag extends that stretch when it repeats the previous tag or ranks exactly one higher; anything else starts a fresh stretch at the current tag. Whenever the running stretch holds all five tags, its length competes for the best answer, so a stretch reaching the end of the log is measured without a closing pass.

Pythonentry point: longest_complete_stretch

def longest_complete_stretch(calls):
    # one pass: keep the stretch ending here that never steps backwards and only
    # ever steps to the next tag, plus how many distinct tags it holds
    step = {"a": 0, "e": 1, "i": 2, "o": 3, "u": 4}
    best = 0
    run = 0
    distinct = 0
    prev = ""
    for ch in calls:
        if run and (ch == prev or step[ch] == step[prev] + 1):
            run += 1
            if ch != prev:
                distinct += 1
        else:
            run = 1
            distinct = 1
        prev = ch
        if distinct == 5:
            if run > best:
                best = run
    return best

The obvious version

Apply the two rules literally to every stretch: from each starting tag, extend one tag at a time, stop as soon as a tag ranks lower than its predecessor, and keep the length whenever the set of tags seen reaches all five. O(n^2).

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_complete_stretch(calls):
    step = {"a": 0, "e": 1, "i": 2, "o": 3, "u": 4}
    n = len(calls)
    best = 0
    for i in range(n):
        seen = set()
        for j in range(i, n):
            if j > i and step[calls[j]] < step[calls[j - 1]]:
                break
            seen.add(calls[j])
            if len(seen) == 5 and j - i + 1 > best:
                best = j - i + 1
    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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