0564EasyArrayHash TableString

Catalogue Order Under a Custom Glyph Sequence

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Trains the technique from

LeetCode 953Verifying an Alien Dictionary

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 typesetting house keeps its catalogue sorted by a house glyph sequence rather than by the usual alphabet. The string order lists all 26 lowercase letters exactly once, from the earliest glyph to the latest.

Under that sequence, one entry comes before another when, at the first position where the two entries carry different letters, this entry's letter appears earlier in order. If one entry is a prefix of the other, the shorter one comes first. Two identical entries may sit next to each other in either arrangement.

Given the catalogue words in the order it is printed, return true if no entry comes before the entry ahead of it, and false otherwise.

Examples

Example 1

Input: words = ["hut", "hub", "bore"], order = "hqtbmzadcefgijklnoprsuvwxy"
Output: true

In this glyph sequence t comes before b, so "hut" precedes "hub", and h comes before b, so "hub" precedes "bore".

Example 2

Input: words = ["apple", "app"], order = "abcdefghijklmnopqrstuvwxyz"
Output: false

The glyph sequence is the ordinary alphabet here. "app" is a prefix of "apple", so the shorter entry has to be printed first, and it is not.

Example 3

Input: words = ["drum", "drums", "drop"], order = "abcdefghijklmnopqrstuvwxyz"
Output: false

"drum" before "drums" is fine by the prefix rule, but "drums" and "drop" first differ at position 2, where u comes after o.

Constraints

1 <= words.length <= 100
1 <= words[i].length <= 20
order.length == 26
order contains each lowercase English letter exactly once.
words[i] consists of lowercase English letters.

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 is_alien_sorted(words: list[str], order: str) -> bool:

Java

public boolean isAlienSorted(String[] words, String order)

The intended approach

Time O(total number of characters in words).Space O(1): a fixed table of 26 positions.

Turn `order` into a table giving each letter its position, so a comparison costs one lookup per character. Then check each neighbouring pair only: walk the two entries together until the letters differ, accept the pair the moment the earlier entry's letter ranks lower, reject it when the later entry's letter ranks lower, and when no difference is found treat the longer entry as the later one. Checking neighbours is enough because the relation carries through a chain.

Pythonentry point: is_alien_sorted

def is_alien_sorted(words, order):
    # O(total characters) time, O(1) space (a 26-slot rank table): compare each
    # neighbouring pair once, since the relation is transitive.
    rank = [0] * 26
    for pos, ch in enumerate(order):
        rank[ord(ch) - 97] = pos
    for i in range(len(words) - 1):
        a = words[i]
        b = words[i + 1]
        limit = min(len(a), len(b))
        decided = False
        for k in range(limit):
            ra = rank[ord(a[k]) - 97]
            rb = rank[ord(b[k]) - 97]
            if ra < rb:
                decided = True
                break
            if ra > rb:
                return False
        if not decided and len(a) > len(b):
            return False
    return True

The obvious version

Compare every pair of entries, not just neighbours, deciding each comparison by scanning the glyph sequence with a search for each letter's position.

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_is_alien_sorted(words, order):
    def less_or_equal(a, b):
        for k in range(min(len(a), len(b))):
            pa = order.index(a[k])
            pb = order.index(b[k])
            if pa < pb:
                return True
            if pa > pb:
                return False
        return len(a) <= len(b)

    for i in range(len(words)):
        for j in range(i + 1, len(words)):
            if not less_or_equal(words[i], words[j]):
                return False
    return True

How this problem was checked

Reference solution executed against 13 test cases — 13 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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