1180EasyArrayHash TableStringSorting

Clearing Rearranged Labels Off the Shelf

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Trains the technique from

LeetCode 2273Find Resultant Array After Removing Anagrams

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 shelf holds the labels labels in order. While some label is a rearrangement of the label directly before it, the later of the two is taken off and the shelf closes up.

Return the labels that remain, in order. Whichever such pair is dealt with first, the shelf ends up the same.

Examples

Example 1

Input: labels = ["abc", "bca", "cab", "xyz"]
Output: ["abc", "xyz"]

The second and third labels both rearrange the first, so both come off, and the last shares none of its letters.

Example 2

Input: labels = ["a", "a", "b", "b", "a"]
Output: ["a", "b", "a"]

The repeats standing next to each other come off, but the final a follows a b so it stays.

Example 3

Input: labels = ["ab", "cd"]
Output: ["ab", "cd"]

Neither label rearranges the other, so the shelf is untouched.

Constraints

1 <= labels.length <= 100
1 <= labels[i].length <= 10
each label holds only 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 remove_anagrams(labels: list[str]) -> list[str]:

Java

public List<String> removeAnagrams(String[] labels)

The intended approach

Time O(n * m log m) for n labels of up to m letters.Space O(n) for the answer.

Being a rearrangement of one another is settled by the letters used and how often, so putting a label's letters in order gives a tag that two labels share exactly when either rearranges the other. That tag also carries across: if two labels share a tag with a third, they share it with each other. That is what makes one pass enough. Walk the shelf keeping the tag of the label most recently kept. A label whose tag matches goes, and a label whose tag differs stays and becomes the new tag to beat. Comparing against the last label kept rather than the last label seen is the right thing, though the carrying-across means both come to the same answer.

Pythonentry point: remove_anagrams

def remove_anagrams(labels):
    out = []
    last = None
    for word in labels:
        tag = "".join(sorted(word))
        if tag != last:
            out.append(word)
            last = tag
    return out

The obvious version

Do exactly as described: sweep the shelf looking for a label whose letters rearrange the one before it, take that label off, and start the sweep again, stopping when a whole sweep finds nothing. Costs a fresh sweep per removal.

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_remove_anagrams(labels):
    body = list(labels)
    changed = True
    while changed:
        changed = False
        for i in range(1, len(body)):
            if sorted(body[i]) == sorted(body[i - 1]):
                body.pop(i)
                changed = True
                break
    return body

How this problem was checked

Reference solution executed against 15 test cases — 15 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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